problem solving(due by sep. 11)

Problem Solving(Due by Sep. 11)

Caution, This Induction May Induce Vomiting!

1. a) Observe that 2 3 4

1 2 2 33

,

3 4 51 2 2 3 3 4

3

, and

4 5 6

1 2 2 3 3 4 4 53

.

Use inductive reasoning to make a conjecture about the value of

1 2 2 3 3 4 1n n .

Use your conjecture to determine the value of 1 2 2 3 3 4 100,000 100,001 .

b) Observe that 2

3 12

1 3 , 2

5 12

1 3 5 , and 2

7 12

1 3 5 7 .

Use inductive reasoning to make a conjecture about the value of 1 3 5 2 1n .

Use your conjecture to determine the value of 1 3 5 1,999,999 .

Oh Brother! No, Oh Sister!

2. A boy has twice as many sisters as brothers, and each sister has two more sisters than

brothers. How many brothers and how many sisters are in the family?

{Hint: Let b be the number of boys and g the number of girls. Now write down some

equations.}

Exactly How Do You Want Your Million?

3. Find a positive number that you can add to 1,000,000 that will give you a larger value than if

you multiplied this number by 1,000,000? Find all such numbers.

{Hint: Let the positive number be x, and solve 1,000,000 1,000,000x x .}

Interesting Is In The Eye Of The Beholder

4. There is an interesting five-digit number. With a 1 after it, it is three times as large as with a

1 before it. What is the number?

{Hint: If x is the five-digit number, then , 1 10 1,1 100,000x abcde abcde x abcde x .}

Twenty-one, But Not Blackjack.

5. Find the 21-digit number so that when you write the digit 1 in front and behind, the new

number is 99 times the original number.

{Hint: If x is the 21-digit number then x abcdefghijklmnopqrstu ,

and 1 10 1abcdefghijklmnopqrstu x ,

and 1 1 10,000,000,000,000,000,000,000 10 1abcdefghijklmnopqrstuv x }

Stand On Your Heads And Get It Together.

6. a) The sum of two numbers is 50, and their product is 25. Find the sum of their reciprocals.

{Hint: 50x y , 25xy , so divide the first equation by the second equation.}

b) The sum of the squares of two numbers is 50, and their product is 25. Find the ratio of the

two numbers.

The Last Two Standing

7. What are the final two digits of 20157 ?

{Hint: Look for a pattern:

Power of 7 Final two digits 27 49 49

37 343 43 47 2401 01

57 16807 07 67 117649 49

}

Don’t Give Up; Don’t Ever Give Up!

8. Given that 11 11f and

13

1

f xf x

f x

for all x , find 2015f . First find

14 , 17 , 20 ,f f f .

{Hint: Look for a pattern: n f(n)

11 11

14 5

6

17 1

11

20 6

5

23 11

}

Mind Your Four’s And Two’s

9. What is the value of x if 20,000 20,0008 8 2x ?

{Hint: Factor 20,000 20,0008 8 and use the fact that 38 2 .}

A Lot Of Weeks, But How Many Days Left Over?

10. What is the remainder when 15,1102 is divided by 7?

{Hint: Look for a pattern in the remainders:

Power of 2 Remainder when divided by 7 12 2 2 22 4 4 32 8 1

42 16 2 52 32 4

}

Can You Just Tell Me How Old Your Children Are!

11. A student asked his math teacher, “How many children do you have, and how old are they?”

“I have 3 girls,” replied the teacher. “The product of their ages is 72, and the sum of their

ages is the same as the room number of this classroom.” Knowing that number, the student

did some calculations and said, “There are two solutions.” “Yes, that is so,” said the

teacher, “but I still hope that the oldest will some day win a math prize at this school.” The

student then gave the ages of the three girls. What are the ages?

{Hint: Triple factors of 72 Sum of the factors

1,1,72 74

1,2,36 39

1,3,24 28

1,4,18 23

1,6,12 19

1,8,9 18

2,2,18 22

2,3,12 17

2,4,9 15

2,6,6 14

3,3,8 14

3,4,6 13

}

Cover All Your Bases, If It’s Within Your Power.

12. Solve for x if 2 9 20

2 5 5 1x x

x x

.

{Hint: Any number raised to the zero power, except zero itself, equals 1. 1 raised to any

power is equal to 1. -1 raised to an even power is equal to 1.}

Who Needs Logarithms?

13. If 2 15x and 15 32y , then find the value of xy .

{Hint: Substitute the first equation into the second equation, and use an exponent property.}

I Refuse To Join Any Club That Would Have Me As A Member.

14. A club found that it could achieve a membership ratio of 2 Aggies for each Longhorn either

by inducting 24 Aggies or by expelling x Longhorns. Find x.

{Hint: Let L be the number of Longhorns and A be the number of Aggies, to get

2 24 2,

1 1

A A

L L x

.}

I Cannot Tell A Fib(onacci), My Name Is Lucas.

15. If 1 1f , 2 3f , and 1 2f n f n f n for 3,4,5,n

a) What is the value of 12f ?

{Hint: 3 2 1 1 3 4f f f , 4 3 2 4 3 7f f f ,…Keep going.}

Amazingly, f can be represented as n nf n x y for 1,2,3,4,5,n

b) Find the values of x and y .

{Hint: 1 , 1 1f x y f , 2 22 , 2 3f x y f , this should be enough to find

values of x and y.}

c) Alicia always climbs steps 1, 2, or 4 at a time. For example, she climbs 4 steps by

1-1-1-1, 1-1-2, 1-2-1,2-1-1,2-2, or 4. In how many ways can she climb 10 steps?

{Hint: If f n represents the number of ways to climb to the nth step, then

1 2 4f n f n f n f n .}

Seven Heaven or Seven…

16. Find the largest power of 7 that divides 343!. 343! 1 2 3 4 342 343

{Hint: The multiples of 7 occurring in the expansion of 343! are 7,14,21,28, ,7 49 .

The multiples of 27 49 occurring in the expansion of 343! are 49,98, ,49 7

The multiple of 37 343 occurring in the expansion of 343! is just 343.

There are no multiples of higher powers of 7 occurring in the expansion of 343!}

A Special Case Of The Chinese Remainder Theorem

17. The positive integer n, when divided by 3, 4, 5, 6, and 7, leaves remainders of 2, 3, 4, 5, and

6, respectively. Find the smallest possible value of n.

{Hint: 3 2, 4 3, 5 4, 6 5, 7 6n a n b n c n d n e .

This means that

1 3 1 , 1 4 1 , 1 5 1 , 1 6 1 , 1 7 1n a n b n c n d n e . So

1n is a common multiple of 3, 4, 5, 6, and 7. What’s the least common multiple?}

Highs And Lows In The Classroom.

18. a) The grades on six tests all range from 0 to 100 inclusive. If the average for the six tests is

93, what is the lowest possible grade on any one of the tests?

{Hint: 1 2 3 4 5 61 2 3 4 5 693 558

6

T T T T T TT T T T T T

1 2 3 4 5 6558T T T T T T , so 1T will be as small as possible when

2 3 4 5 6T T T T T is as large as possible.}

b) If the average for the six tests is 16, what is the highest possible grade on any one of the

tests?

Just Your Average Joe.

19. If Joe gets 97 on his next math quiz, his average will be 90. If he gets 73, his average will

be 87. How many quizzes has Joe already taken?

{Hint: Let n be the number of quizzes he has already taken, and T the total number of points

he has already earned on the quizzes. Then 97 73

90, 871 1

T T

n n

.}

A Whole Lotta Zeroes

20. How many zeroes are at the end of the number 127!? 127! 1 2 3 4 126 127

{Hint: Zeroes come from factors of 10. Factors of 10 come from 5’s and 2’s. See the hint

for problem #16.}

The Last One Standing

21. Find the ones digit of 2421 378313 17 .


Powers of 13 One’s-digit Powers of 17 One’s digit 113 3 117 7 213 9 217 9 313 7 317 3 413 1 417 1 513 3 517 7 613 9 617 9

}

Happy 2015!

22. Find the 2015th digit in the decimal representation of 1

7.

{Hint: 1

70.142857 , so use a pattern.}

Zero, The Something That Stands For Nothing.

23. How many zeroes are at the end of the number 3,000 6,000 4,0002 5 4 ?

{Hint: Zeroes come from factors of 10. Factors of 10 come from 5’s and 2’s.}

Looky Here Son, This Is A Problem, Not A Chicken.

24. Foghorn C sounds every 34 seconds, and foghorn D sounds every 38 seconds. If they sound

together at noon, what time will it be when they next sound together?

Foghorn C 12:00 12:00:34 12:01:08 12:01:42 Foghorn D 12:00 12:00:38 12:01:16

sound

together

{Hint: Every time they sound together after noon will have to be both a multiple of 34

seconds after noon and a multiple of 38 seconds after noon.}

A European Sampler.

25. A box contains 8 French books, 12 Spanish books, 9 German books, 15 Russian books, 18

Italian books, and 10 Chinese books. What is the fewest number of books you can select

from the box without looking to be guaranteed of selecting at least 10 books of the same

language?

{Hint: What is the largest number of books you can select and still not have 10 books of the

same language? The answer to the problem is 1 more than the answer to the

previous question.}

I Hope That You Are A Digital Computer?

26. What is the ones digit of 1! 2! 3! 999! ?

{Hint: See what happens to the one’s digits:

Factorial One’s-digit One’s digit of the sum

1! 1 1

2! 2 3

3! 6 9

4! 4 3

}

The Collapse Of Rationalism.

27. Find the exact value of 1 1 1 1

1 2 2 3 3 4 999,999 1,000,000

.

{Hint: Rationalize the denominators. For example:1 1 2 1 2

2 111 2 1 2

.}

The Beast With Many Fingers And Toes.

28. How many digits does the number 6,666 20,0008 5 have?

{Hint: Zeroes come from factors of 10. Factors of 10 come from 5’s and 2’s.}

Don’t Get Stumped; Use The Fundamental Theorem Of Arithmetic.

29. Forrest Stump heard that there are only three numbers between 2 and

2,000,000,000,000,000,000 which are perfect squares, perfect cubes, and perfect fifth

powers. He decided to look for them, and so far he has checked out every number up to

about 100,000 and is beginning to get discouraged. What are the numbers he is trying to

find?

{Hint: Every positive whole number greater than 1 can be written as a product of prime

factors. If N is a positive whole number greater than 2, then 31 22 3 5 kn nn n

kN p . In order for N to be a perfect square, all the positive

exponents would have to be multiples of 2; in order for N to be a perfect cube, all

the positive exponents would have to be multiples of 3; and in order for N to be a

perfect fifth power, all the positive exponents would have to be multiples of 5. So all

the positive exponents would have to be common multiples of 2, 3, and 5.}

Sorry, I Can’t Give You Change For A Dollar.

30. a) What is the largest amount of money in current U.S. coins(pennies, nickels, dimes,

quarters, but no half-dollars or dollars) you can have and still not have change for a

dollar?

{Hint: It’s more than 99 cents. For instance: 3 quarters and 3 dimes is $1.05, but you

can’t make change for a dollar.}

b) A collection of coins is made up of an equal number of pennies, nickels, dimes, and

quarters. What is the largest possible value of the collection which is less than $2?

c) Trina has two dozen coins, all dimes and nickels, worth between $1.72 and $2.11. What

is the least number of dimes she could have?

Destination Cancellation.

31. Express as a fraction, in lowest terms, the value of the following product of 1,999,999

factors 1 1 1 1

1 1 1 12 3 4 2,000,000

.


11

2

1

2

1 1 1 21 1

2 3 2 3

1

3

1 1 1 1 2 31 1 1

2 3 4 2 3 4

1

4

}

Officer, I Got The License Plate Number, But I Was Lying On My Back.

32. The number on a license plate consists of five digits. When the license plate is turned

upside-down, you can still read a number, but the upside-down number is 78,633 greater

than the original license number. What is the original license number?

{Hint: The digits that make sense when viewed upside-down are 0, 1, 6, 8, and 9.

1st digit 2nd digit 3rd digit 4th digit 5th digit

Upside-down plate

Original plate

Difference of the plates 7 8 6 3 3

}

Two Smokin’ Good Problems In One.

33. a) Mrs. Puffem, a heavy smoker for many years finally decided to stop smoking altogether.

“I’ll finish the 27 cigarettes I have left,” she said to herself, “and never smoke another

one.” It was Mrs. Puffem’s practice to smoke exactly two-thirds of each complete

cigarette(the cigarettes are filterless). It did not take her long to discover that with the

aid of some tape, she could stick three butts together to make a new complete cigarette.

With 27 cigarettes on hand, how many complete cigarettes can she smoke before she

gives up smoking forever, and what portion of a cigarette will remain?

{Hint: With 27 complete cigarettes, she can smoke 27 complete ones and assemble 9

new complete ones…, keep going.}

b) Hobo Hal makes his cigars by connecting 5 cigar butts. Hal smokes 4

5 of a cigar and

then stops, leaving a cigar butt. If Hal finds 625 cigar butts, how many cigars will he be

able to make and smoke?

Just Gimme An A.

34. a) A class of fewer than 45 students took a test. The results were mixed. One-third of the

class received a B, one-fourth received a C, one-sixth received a D, one-eight of the

class received an F, and the rest of the class received an A. How many students in the

class got an A?

{Hint: The number of students in the class must be a multiple of 3, 4, 6, and 8, and must

be smaller than 45.}

b) The library in Johnson City has between 1000 and 2000 books. Of these, 25% are

fiction, 1

13 are biographies, and

1

17 are atlases. How many books are either biographies

or atlases?

Working Backwards In Notsuoh.

35. A castle in the far away land of Notsuoh was surrounded by four moats. One day the castle

was attacked and captured by a fierce tribe from the north. Guards were stationed at each

bride over the moats. Johann, from the castle, was allowed to take a number of bags of gold

as he went into exile. However, the guard at the first bridge took half of the bags of gold

plus one more bag. The guards at the second third and fourth bridges made identical

demands, all of which Johann met. When Johann finally crossed all the bridges, he had just

one bag of gold left. With how many bags of gold did Johann start?

{Hint: Sometimes working backwards is a good idea. If Johann has 1 bag of gold left, then

how many did he have when he approached the fourth guard? 12

1 1bags leftapproached the 4th guard

taken by the 4th guard

x x }

Grazin’ In The Grass Is A Gas, Baby, Can You Dig It?

36. A horse is tethered by a rope to a corner on the outside of a square corral that is 10 feet on

each side. The horse can graze at a distance of 18 feet from the corner of the corral where

the rope is tied. What is the total grazing area for the horse?

{Hint:

}

Life Is Like A Box Of Chocolate Covered Cherries.

37. a) Assume that chocolate covered cherries come in boxes of 5, 7, and 10. What is the

largest number of chocolate covered cherries that cannot be ordered exactly?

{Hint: If you can get five consecutive amounts of cherries, then you can get all amounts

larger. Here’s why: Suppose you can get the amounts 23,24,25,26,27 , then by the

addition of the box of size 5, you can also get 28,29,30,31,32 , and another addition

of the box of size 5 produces 33,34,35,36,37 and so on. This would also be true of

seven consecutive amounts and ten consecutive amounts, but five consecutive

amounts would occur first. So look for amounts smaller than the first five

consecutive amounts.}

b) Do the same problem, except the cherries come in boxes of 6, 9, and 20.

18

18

8

8

Easy Come, But Not So Easy Go.

38. A man whose end was approaching summoned his sons and said, “Divide my money as I

shall prescribe.” To his eldest son he said, “You are to have $1,000 and a seventh of what

is left.” To his second son he said, “Take $2,000 and a seventh of what remains.” To the

third son he said, “You are to take $3,000 and a seventh of what is left.” Thus he gave each

son $1,000 more than the previous son and a seventh of what remained, and to the last son

all that was left. After following their father’s instructions with care, the sons found that

they had shared their inheritance equally. How many sons were there, and how large was

the estate?

{Hint: Let M be the total value of the estate.

First Son Second Son

1,0001,000

7

M

1,0003,000

72,0007

MM

Since each son’s share is the same, solve an equation to determine the value of M,

and use it to find the number of sons.}

Round And Round With Sarah And Hillary

39. Sarah and Hillary are racing cars around a track. Sarah can make a complete circuit in 72

seconds, and Hillary completes a circuit in 68 seconds.

a) If they start together at the starting line, how many seconds will it take for Hillary to pass

Sarah at the starting line for the first time.

{Hint: Every time Sarah reaches the starting line must be a multiple of 72 seconds, and

every time Hillary reaches the starting line must be a multiple of 68 seconds. So

they will both be at the starting line at common multiples of 72 and 68.}

b) If they start together, how many laps will Sarah have completed when Hillary has

completed one more lap than Sarah?

{Hint: Let n be the number of laps completed by Sarah, then 72 68 1n n .}

Some divisibility rules for positive integers:

1) A positive integer is divisible by 2 if its one’s digit is even.

Here’s why:

Suppose we have the three-digit integer abc, then its value can be expressed as

100 10a b c , but 100 10 2 50 5a b c a b c , so if the one’s digit, c is even(divisible

by 2) then the integer abc will also be divisible by 2.

2) A positive integer is divisible by 3 if the sum of its digits is divisible by 3. This process may

be repeated.

Examples: 243 is divisible by 3, but 271 is not.

Here’s why:


100 10a b c , but the sum of the digits

100 10 99 9 3 33 3a b c a b a b c a b a b c , so if the

sum of the digits, a b c , is divisible by 3 then the integer abc will also be divisible by 3.

3) A positive integer is divisible by 4 if the ten’s and one’s digits form a two-digit integer

divisible by 4.


Here’s why:

Suppose we have the four-digit integer abcd, then its value can be expressed as

1000 100 10a b c d , but

the two-digit integer cd

1000 100 10 1000 100 10 4 250 25 10a b c d a b c d a b c d , so if

the ten’s and one’s two-digit integer, cd, is divisible by 4 then the integer abcd will also be

divisible by 4.

4) A positive integer is divisible by 5 if its one’s digit is either a 5 or a 0.

Here’s why:


100 10a b c , but 100 10 5 20 2a b c a b c , so if the one’s digit, c is divisible by 5,

then the integer abc will also be divisible by 5, but the only digits divisible by 5 are 0 and 5.

5) A positive integer is divisible by 6 if it’s both divisible by 2 and divisible by 3.

6) A positive integer is divisible by 7 if when you remove the one’s digit from the integer and

then subtract twice the one’s digit from the new integer, you get an integer divisible by 7.

This process may be repeated.

Examples: 714 is divisible by 7 since 71 – 8 = 63, but 423 is not since 42 – 6 = 36.

Here’s why:


100 10a b c , but

new integer minus twice one's digit

new integer minus twice one's digit

100 10 90 9 3 10 2

9 10 2 21 10 2

10 10 2 7 3

a b c a b c a b c

a b c c a b c

a b c c

, so if the new integer minus twice the one’s digit, 10 2a b c , is divisible by 7 then so is the

original integer abc.

Or

A positive integer is divisible by 7 if when you remove the one’s digit from the integer and

then subtract nine times the one’s digit from the new integer, you get an integer divisible by

7. This process may be repeated.


Here’s why:


100 10a b c , but

new integer minus 9 times one's digit

new integer minus 9 times one's digit

100 10 90 9 10 10 9

9 10 9 91 10 9

10 10 9 7 13

a b c a b c a b c

a b c c a b c

a b c c

, so if the new integer minus nine times the one’s digit, 10 9a b c , is divisible by 7 then so

is the original integer abc.

Or

A positive integer with more than three digits is divisible by 7 if when you split the digits

into groups of three starting from the right and alternately add and subtract these three digit

numbers you get a result which is divisible by 7.

Examples: 1412236 is divisible by 7 since 1 – 412 + 236 = -175, but 130747591 is not since

130 – 747 + 591 = -26.

Here’s why:

Suppose we have the five-digit integer abcde, then its value can be expressed as

10000 1000 100 10a b c d e , but

integer ab integer cde


10000 1000 100 10 10010 1001 10 100 10

2 100 10 2 10

7 143 10 10 100 10

a b c d e a b a b c d e

c d e a b

a b a b c d e

, so if the

integer ab minus the integer cde is divisible by 7 then the integer abcde will also be divisible

by 7.

7) A positive integer is divisible by 8 if the hundred’s, ten’s, and one’s digits form a three-digit

integer divisible by 8.

Examples: 1240 is divisible by 8, since 240 is, 3238 is not, since 238 is not even divisible by

4.

Here’s why:


10000 1000 100 10a b c d e , but

the three-digit integer cde

10000 1000 100 10 10000 1000 100 10

8 1250 125 100 10

a b c d e a b c d e

a b c d e

, so if the hundred’s,

ten’s, and one’s three-digit integer, cde, is divisible by 8 then the integer abcde will also be

divisible by 8.

8) A positive integer is divisible by 9 if the sum of its digits is divisible by 9. This process may

be repeated.


Here’s why:


100 10a b c , but the sum of the digits

100 10 99 9 9 11a b c a b a b c a b a b c , so if the

sum of the digits, a b c , is divisible by 9 then the integer abc will also be divisible by 9.

9) A positive integer is divisible by 11 if when you remove the one’s digit from the integer and

then subtract the one’s digit from the new integer, you get an integer divisible by 11. This

process may be repeated.


Here’s why:


100 10a b c , but

new integer minus one's digit

new integer minus one's digit

100 10 90 9 2 10

9 10 11 10

10 10 11

a b c a b c a b c

a b c c a b c

a b c c

, so if the new integer minus the one’s digit, 10a b c , is divisible by 11 then so is the

original integer abc.

Or

A positive integer is divisible by 11 if when you subtract the sum of the ten’s digit and every

other digit to the left from the sum of the one’s digit and every other digit to the left you get a

number divisible by 11. This process may be repeated.

Examples: 9031 is divisible by 11 since 1 0 3 9 11 , but 423 is not since

4 3 2 5 .

Here’s why:


10,000 1,000 100 10a b c d e , but

alternating from one's alternating from ten's

10,000 1,000 100 10 9,999 1 1,001 1 99 1 11 1

9,999 1,001 99 11

11 909 91 9

a b c d e a b c d e

a b c d a c e b d

a b c d a c e b d

, so if the sum of the alternating digits from the one’s digit minus the sum of the alternating

digits from the ten’s digit, a c e b d , is divisible by 11 then so is the original

integer abcde.

Or

A positive integer with more than three digits is divisible by 11 if when you split the digits

into groups of three starting from the right and alternately add and subtract these three digit

numbers you get a result which is divisible by 11.

Examples: 1412290 is divisible by 7 since 1 – 412 + 290 = -121, but 130747591 is not since

130 – 747 + 591 = -26.

Here’s why:


10000 1000 100 10a b c d e , but



10000 1000 100 10 10010 1001 10 100 10

2 100 10 2 10

11 91 10 10 100 10

a b c d e a b a b c d e

c d e a b

a b a b c d e

, so if the

integer ab minus the integer cde is divisible by 11 then the integer abcde will also be

divisible by 11.

Divisibility And Conquer.

40. Of the following three numbers determine which are divisible by 2, which are divisible by

3, which are divisible by 4, which are divisible by 6, which are divisible by 8, and which are

divisible by 9: 9993 10 736 , 9993 10 534 , 9993 10 952 .

{Hint: Divisibility rules.}

Keep On Dividing And Conquering.

41. Of the following three numbers determine which are divisible by 3, which are divisible by

6, which are divisible by 7, which are divisible by 9, and which are divisible by 11:

1358024680358024679, 864197523864197523, 964197523864197522.

{Hint: Divisibility rules.}

Seven Come Thirteen, Not Eleven.

42. Show that the second divisibility rule for 7 can also be used as a divisibility rule for 13.

Modify the explanation to show why it works for 13.

Cogswell Cogs Or Spacely Sprockets?

43. In a machine, a small gear with 45 teeth is engaged with a large gear with 96 teeth. How

many more revolutions will the smaller gear have made than the larger gear the first time

the two gears are in their starting position?

{Hint: A revolution of the smaller gear is a multiple of 45 teeth, and a revolution of the

larger gear is a multiple of 96 teeth. So the gears are again in the starting positions

at common multiples of 45 and 96.}

My Tile Cutter Is Broken, So What Now?

44. The figure shows that twenty-four 8" 12" rectangular tiles can be used to tile a 48" 48"

square without cutting tiles.

a) Is there a smaller sized square that can be tiled without cutting using 8" 12" tiles? If so,

find it.

{Hint: The dimension of the square tile must be a multiple of both dimensions of the

rectangular tiles.}

b) What is the smallest square that can be tiled without cutting using 9" 12" tiles?

{Hint: See the previous hint.}

Solving Without Completely Solving.

45. If 2a b and 2 2 3a b , then find 8 8a b .

{Hint: Start squaring and substituting.}

Odds, Evens, What’s The Difference?

46. What do you get if the sum of the first 8,000,000,000 positive odd integers is subtracted

from the sum of the first 8,000,000,000 positive even integers?

{Hint: 2 4 6 8 16,000,000,000 1 3 5 7 15,999,999,999 }

The Great Luggage Caper.

47. Great Aunt Christine is going for her annual holiday to Barbados. She sends her butler John

down to the airport with her collection of suitcases, each of which weighs either 18 or 84

pounds, and is informed that the total weight checked-in is 652 pounds. Show that this is

impossible without listing and checking all the possible combinations of 18 and 84 pound

suitcases.

{Hint: The total weight of the suitcases is of the form 18 84x y , where x and y are

nonnegative integers. So the expression must be divisible by the greatest common

factor of 18 and 84.}

Gerry Benzel’s Favorite Problem.

48. A bottle and a cork together cost $1.10. If the bottle costs $1.00 more than the cork, what

does the cork cost?

{Hint: Let x be the cost of the cork and y the cost of the bottle.}

Getting Solutions Without Actually Solving.

49. Notice that

2x a x b x a b x ab

3 2x a x b x c x a b c x ab ac bc x abc

4 3 2x a x b x c x d x a b c d x ab ac ad bc bd cd x

abc abd acd bcd x abcd .

a) Use inductive reasoning to determine the value of the coefficient of 1nx and the constant

term in the expansion of the following product: 1 2 nx a x a x a .

b) Use the previous result to determine the sum of the seventeenth powers of the 17

solutions of the equation 17 3 1 0x x .

{Hint: The Fundamental Theorem of Algebra guarantees that the equation 17 3 1 0x x has seventeen solutions(counting duplicates). The seventeen

solutions of 17

1 2 17 3 1 0x a x a x a x x are 1 2 17, , ,a a a . So

adding the seventeen equations together yields:

17

1 1

17

2 2

17

17 17

17 17 17

1 2 17 1 2 17

3 1 0

3 1 0

3 1 0

3 17 0

a a

a a

a a

a a a a a a

. Now use the previous result.}

Dots And Dashes, But It’s Not Morse Code.

50. The diagram shows a sequence of shapes 1 2 3 4, , , ,T T T T . Each shape consists of a number

of squares. A dot is placed at each point where there is a corner of one or more squares.

1T 2T

3T 4T

Shape number of rows, n number of squares, S number of dots, D 2D n

1T 1 1 4 3

2T 2 4 10 6

3T 3 9 18 9

4T 4 16 28 12

a) Use inductive reasoning to find a formula for S in terms of n.

b) How many squares are in shape 25T ?

c) Use inductive reasoning to find a formula for D in terms of n.

{Hint: Notice that 2D n in the last column is always a multiple of 3.}

d) How many dots are in shape 25T ?

It Squares; It Cubes; It Does It All!

51. If 3 3 10a b and 5a b , then what is the value of 2 2a ab b ?

{Hint: 2 2 3 3a b a ab b a b .}

Two Squares Don’t Get Along – A Difference Of Squares!

52. If 20,000 20,000 5x y , 10,000 10,000 4x y , 5,000 5,000 3x y , 2,500 2,500 2x y , and 2,500 2,500 1x y , then find the value of 40,000 40,000x y .

Hint: 2 2a b a b a b .

How Touching!

53. Four circles, each of which has a diameter of 2 feet, touch as shown. Find the area of the

shaded portion.

{Hint: The area of the shaded portion would be the area of the square minus the area of the

four circular sectors.

}

The method of finite differences can be used to produce formulas for lists of numbers. For

example, if the list of numbers is 5,7,9,11,13,…, then the list of first finite differences is the list

of differences of consecutive numbers: 7 5 9 7 11 9 13 11

2 , 2 , 2 , 2 ,

or more simply, 2,2,2,2,….

Whenever the list of first finite differences is a repetition of the same number, it means that the

original list of numbers can be produced by a formula of the form an b , where n represents

the position in the list. Here’s why:

original a b 2a b 3a b 4a b first finite differences a a a

In fact, the repeated number in the first finite differences will always be the coefficient of n.

In our example, it must be that 2a and 5a b . So we get that 2a and 3b , and a

formula for the list of numbers 5,7,9,11,13,… is 2 3.n Sometimes the list of first differences

is not a repetition of the same number. An example of this is the list 3,9,19,33,51,73,…. The

list of first finite differences is 6,10,14,18,22,…. As you can see it’s not a repetition of the

same number. Now let’s calculate the list of second finite differences: 4,4,4,4,…. Whenever

the list of second finite differences is a repetition of the same number it means that the original

list of numbers can be produced by a formula of the form 2an bn c , where again n

represents the position in the list. Here’s why:

original a b c 4 2a b c 9 3a b c 16 4a b c 25 5a b c first finite differences 3a b 5a b 7a b 9a b

second finite differences 2a 2a 2a

In fact, the repeated number in the second finite differences will always be twice the coefficient

of 2n . In our example, it must be that 2a , 3a b c , and 4 2 9a b c . Substituting the

value of a into the next two equations leads to the system 1

2 1

b c

b c

. Subtracting the first

equation from the second equation leads to 0b , so the values are 2a , 0b , and 1c . So

a formula for the list of numbers 3,9,19,33,51,73,… is 22 1n . Similar results hold for higher

order differences which are repetitions of the same number. You may use the method of finite

differences to solve the following two problems.

What’s the Diff?

54. Find a formula for each of the following lists of numbers:

a) 2,5,8,11,…

b) 3 5 72 2 2,2, ,3, ,

c) 4,7,12,19,28,

d) 3,7,13,21,31,43

e) 1, 1 2 , 1 2 3 , 1 2 3 4 , 1 2 3 4 5 ,

f) 1, 1 3 , 1 3 5 , 1 3 5 7 , 1 3 5 7 9 ,

g) 1,15,53,127,249,…

h) 1, 1 4 , 1 4 9 , 1 4 9 16 , 1 4 9 16 25 , 1 4 9 16 25 36 ,

You Want Me To Cut The Pizza Into How Many Slices?

55. A chord is a line segment joining two points on a circle. Here, n is the number of chords.

Associated with each number of chords, n, is the maximum number of regions that the circle is

divided into by the n chords. For the first five numbers of chords, the list of the maximum

number of regions is 2,4,7,11,16 . Use the method of finite differences to find a formula for the

maximum number of regions that a circle can be divided into using n chords, and use the

formula to predict the maximum number of regions a circle can be divided into using 60 chords.

Now What Was That Last Digit?

56. Peggy is writing the numbers 1 to 9,999.

a) If she stops to rest after she has written a total of 150 digits. What is the last digit that

she wrote?

b) If she stops to rest after she has written a total of 1,000 digits. What is the last digit that

she wrote?

{Hint:

Quantity Total number of digits

1 digit numbers(1-9) 9 9

2 digit numbers(10-99) 90 180

3 digit numbers(100-999) 900 2,700

4 digit numbers(1,000-9,999) 9000 36,000

}

1n 2n 3n 4n

5n

On The Mark, Off The Mark, Or Bull’s Eye?

57. In the following figure, the curves are concentric circles with the indicated radii. Which

shaded region has the larger area, the inner circle or the outer ring?

Calculate the area of each region and check your visual estimation ability.

I Hate This Problem To The Nth Degree.

58. Use the following properties of exponents to find the exact value of the given expressions.

n m n mx x x , n n nxy x y ,

mn nmx x

a) 90,000 90,000

89,999 90,000

3 6

2 9

b)

100,000 110,000

100,000 5000

2 3

6 9

We use a base 10 number system. A number written with the digits abcd actually represents the

number 3 210 10 10a b c d , where the digits a, b, c, and d can be any of the numbers 0, 1,

2, 3, 4, 5, 6, 7, 8, or 9. If we want to use a base of 2, then abcd would represent the number 3 22 2 2a b c d , where the digits a, b, c, and d can be the numbers 0 or 1.

3 5

4

I’m Thinking Of A Number From 1 To 31.

59. A person is asked to think of a whole number from 1 to 31. The person is shown the

following 5 cards and asked to identify the cards that contain the number he is thinking of.

1 3 5 7

9 11 13 15

17 19 21 23

25 27 29 31

2 3 6 7

10 11 14 15

18 19 22 23

26 27 30 31

4 5 6 7

12 13 14 15

20 21 22 23

28 29 30 31

8 9 10 11

12 13 14 15

24 25 26 27

28 29 30 31

16 17 18 19

20 21 22 23

24 25 26 27

28 29 30 31

By adding the numbers in the upper left corners of the identified cards, you can always

determine the person’s number. Use base 2 numbers to explain how the trick works.

Oh Yeah, Well I’m thinking Of A Number From 1 to 63.

60. See if you can extend the previous trick to work for whole numbers from 1 to 63 using 6

cards instead of 5. Complete the numbers on the 6 cards in order for the trick to work.

1 3 5 7 9 11 13 15

17 19 21 23 25 27 29 31

33 35 37 39 41 43 45 47

49 51 53 55 57 59 61 63

2

4

8

16

32

Call It Like You See It.

61. Consider the following figure:

a) What fraction of the large square is shaded ?

b) What fraction of the large square is shaded ?

c) What fraction of the large square is shaded ?

Let’s Not Go Overboard.

62. This problem in one form or another has been around for nearly 2,000 years – if not longer.

Josephus, a Jewish historian of the first century A. D., mentions it. In all its forms the

problem seems to have a violent streak. Here’s our version of it: One-hundred sailors are

arranged around the edge of their ship. They hold in order the numbers from 1 to 100.

Starting the count with number 1, every other sailor is pushed overboard into the cold North

Atlantic waters until there is only one left – the survivor.

a) What number do you want to be holding in order to be the survivor?

b) How many times will the survivor be skipped during this process?

c) Find the number of the last sailor to be pushed overboard.

d) Find the number of the second to last sailor to be pushed overboard.

{Hint: Consider the problem for smaller numbers of sailors and work up: For example

with 10 sailors you have

The survivor, 5, is skipped three times. The last sailor to be pushed overboard is

9, and the next to last sailor pushed overboard is 1.}

1 2

3

4

5

6 7

8

9

10

first

elimination

1

3

5 7

9

second

elimination

1

5

9

third

elimination

5

the

survivor

It’s Just Gotta Be 6.

63. a) Show that for any integer n, 6 divides 3n n .

{Hint: 3 2 1 1 1n n n n n n n }

b) Show that for any integer n, 6 divides 5n n .

Even So, It’s Odd.

64. Show that for every positive integer n, 2 3 8n n is even.

{Hint: n is either even or odd, so take it from here.}

Logarithms Are Very Overrated.

65. If 3 4a , 4 5b , 5 6c , 6 7d , 7 8e , and 8 9f , then what is the value of the product

abcdef ?

{Hint: From 3 4a and 4 5b , you can conclude that 3 5ab . So just keep going.}

Real-valued Function, What’s Your Function?

66. a) Suppose that f is a real-valued function with 20142 3x

f x f x for all 0x . Find

2f . {Hint: Plug in 2 and 1007, and solve for 2f .}

b) Suppose that f is a real-valued function with 21

23

x

f xf x

x for all 0x . Find

2f .

An Odd Product.

67. Find the value of the following product:3 5 7 9 4021

1 1 3 5 4017

.

{Hint: What factors can you cancel out?}

An Even Sum.

68. Find the one’s digit of the following sum: 20109 81 729 9 .

{Hint: Start small, and look for a pattern.}

Don’t Put All Of Your Eggs In One Basket!

69. If eggs are taken from a basket two at a time, then one egg remains in the basket. If eggs are

taken three at a time from the same basket, then two eggs remain in the basket. If eggs are

taken four at a time from the same basket, then three eggs remain in the basket. If eggs are

taken five at a time from the same basket, then four eggs remain in the basket. If eggs are

taken six at a time from the same basket, then five eggs remain in the basket. If eggs are

taken seven at a time from the same basket, then no eggs remain in the basket. What is the

fewest possible number of eggs in the basket?

{Hint: If N is the number of eggs in the basket, then 1N must be a multiple of 2, 1N

must be a multiple of 3, 1N must be a multiple of 4, 1N must be a multiple of 5, and

1N must be a multiple of 6. So 1N must be a multiple of the LCM of 2, 3, 4, 5, and 6.

Also, N must be a multiple of 7.}

Lucky 13, I Repeat, Lucky 13. Lucky 7, I Repeat….

70. a) Show that every six-digit number of the form abc,abc (for example 281,281 or 435,435)

is divisible by 13.

b) Show that every six-digit number of the form abc,abc (for example 281,281 or 435,435)

is divisible by 7.

c) Show that every six-digit number of the form abc,abc (for example 281,281 or 435,435)

is divisible by 11.

Just your average airplane flight.

71. An airplane travels between two cities. The first half of the distance between the two cities

is travelled at an average speed of 600 mph. The second half of the distance is travelled at

an average speed of 900 mph. Find the average speed of the airplane for the entire trip.

{Hint: distance travelled

average speedtime of travel

.}

Sometimes Reduction Can Get In The Way Of Induction.

72. Observe that

2

2 2

2 2 2

2 2 2 2

1 31

2 4

1 1 41 1

2 3 6

1 1 1 51 1 1

2 3 4 8

1 1 1 1 61 1 1 1

2 3 4 5 10

.

Use inductive reasoning to make a conjecture about the value of

2 2 2 2

1 1 1 11 1 1 1

2 3 4 n

.

Use your conjecture to find the value of 2 2 2 2

1 1 1 11 1 1 1

2 3 4 1,000,000

.

A Whole Lotta Eights; A Whole Lotta…

73. a) What is the smallest whole number that when multiplied by 9 gives a number whose

digits are all 8’s?

b) What is the smallest whole number that when multiplied by 9 gives a number whose


c) What is the smallest whole number that when multiplied by 9 gives a number whose


It’s All In The Ones.

74. Find all the integer solutions of the equation 2 5 27x y , or show that there aren’t any.

{Hint: What are the possible ones digits of 2x ? What are the possible ones digits of 5y ?}

Middleaged At 40?

75. The counting numbers are written in the pattern below. Find the middle number of the 40th

row.

1

2 3 4

5 6 7 8 9

10 11 12 13 14 15 16

Stick puzzles involve rearranging, removing, or adding sticks in order to accomplish the

requirements of a problem. In the following problems, you might want to use the following

suggestions:

This arrangement of three sticks can be used to represent a square root:

A stick representing an over bar can be used to multiply a Roman numeral by 1,000:

represents 5,000

Just Stick It.

76.

a) Move one stick to make a true equation.

b) Remove two sticks to make a true equation.

c) Move one stick to make a square.

d) Move one stick to make a true equation.

e) Move one stick to make a true equation.

Stick It, Again.

77.

a) Move one stick to make a true equation.

b) Add one stick to make the fraction(1/6) equivalent to one.

c) Move two sticks to make a true equation.

d) Move one stick to make a true equation.

Shifty Four.

78. The leftmost digit of a 6 digit number is 4. If it’s shifted to be the rightmost digit, the new

number is one-fourth of the original number. What’s the original number?

{Hint: If the original number is 4x abcde , then 10 4 4 4x abcde , and if you subtract the

right number from 10 4x , you’ll have the new number.}

You’re A Real Square, Man.

79. A man born in the year 2x died, on his 87th birthday, in the year 2

1x . In what year was

he born?

Now There’s An Odd Sum.

80. The sum of ten positive odd numbers is 20. Find the largest number which can be used as

an addend in this sum.

Zero Is My Hero.

81. If a and b are two unequal numbers, and ax bx , then find the exact numerical value of

3x

a b .

It’s As Easy As 123… . 82. If the digit 9 is written to the right of a certain number, that number is increased by

111,111,111. Find the number.

{Hint: If the original number is x , then 10 9x is the new number.}

The Truth, The Hole Truth, And… .

83. How many cubic centimeters of dirt are in a hole in the ground that is 1 ft. long by 1 ft. wide

by 1 ft. deep?

Slide Your Way Into An Answer.

84. What fractional part of the figure is shaded(assuming that line segments that appear parallel

actually are, all angles are right angles, and the vertical line segments are equally spaced.)?

See How Everything Lines Up.

85. Given the following incomplete distance chart for 4 points in a plane, find the distance from

A to B.

Just Go All The Way Around.

86. The following figure consists of six congruent squares, and it has an area of 294. Find the

perimeter of the figure.

Sister Act.

87. Three sisters leave home on the same day. One returns every 5 days, another returns every

4 days, and the third returns every 3 days. How many days until all three sisters meet at

home again for the first time?

A B C D

A 0 ? 21 9

B ? 0 5 7

C 21 5 0 12

D 9 7 12 0

What An Intersecting Little Problem.

88. Find the area of the following shaded region formed by the two perpendicular intersecting

rectangles.

Don’t Text And Drive.

89. On a stretch of road 75 miles long, two trucks approach each other. Truck A is traveling at

55 miles per hour, and Truck B is traveling at 80 miles per hour. What is the distance

between the two trucks in miles one minute before their head-on collision?

The People Under The Stairs.

90. Three rectangles are connected as in the figure. The first rectangle is 2 by 1; the second

rectangle is 4 by 2; the third rectangle is 8 by 4. A line is drawn from a vertex of the

smallest rectangle to a vertex of the largest rectangle. Find the area of the shaded region.

2 4 8

4

1 2

I Go Cuckoo For Coconuts.

91. Five sailors were stranded on a desert island, and their only food was coconuts. One day

they gathered all the coconuts on the island together, and the next day they would divide

them evenly. The first sailor woke up early and gave one coconut to a monkey and hid his

fifth of the remaining coconuts. Then the second sailor woke up and gave one coconut to

the same monkey and hid his fifth of the remaining coconuts. The third, fourth, and fifth

sailors all did the same. Upon arising the next day, one coconut was given to the monkey,

and the remaining coconuts were divided equally among the five sailors. What is the

smallest starting number of coconuts possible?

N is the starting number of coconuts, and R is the remaining coconuts after the five sailors

have done their secret removals.

Or start with a smaller problem: If one sailor gives a coconut to the monkey, takes one-fifth

of the remaining coconuts, and then another coconut is given to the monkey and the rest are

divided among the five sailors then, the number of remaining coconuts would have to be a

multiple of 5: 45

1 1 5N k . So 5 5 1

14

kN

, which means that 5 1k must be a

multiple of 4, and 16 is the smallest possible multiple of 4 that works. This means that the

smallest number of coconuts in this case would be 21.

Sailor Remaining coconuts

1 45

1N

2 4 45 5

1 1N

3 4 4 45 5 5

1 1 1N

4 4 4 4 45 5 5 5

1 1 1 1N

5 4 4 4 4 45 5 5 5 5

1 1 1 1 1N

Sailor Remaining coconuts

1 5 5 5 5 625 3694 4 4 4 256 64

1 1 1 1R R

2 5 5 5 125 614 4 4 64 16

1 1 1R R

3 5 5 25 94 4 16 4

1 1R R

4 54

1R

5 R

If one sailor gives a coconut to the monkey, takes one-fifth of the remaining coconuts, and

another sailor gives a coconut to the monkey, takes one-fifth of the remaining coconuts, and

another coconut is given to the monkey, and the rest are divided among the five sailors then,

the number of remaining coconuts would have to be a multiple of 5:

4 45 5

1 1 1 5N k . So 125 61

16

kN

, which means that 125 61k must be a

multiple of 16, and 1936 is the smallest possible multiple of 16 that works. This means that

the smallest number of coconuts in this case would be 121. Keep going!

The Euclidean algorithm can be used to find the greatest common factor of two whole numbers.

It can also be used to express the greatest common factor as a combination of the two whole

numbers. Here’s an example:

Let’s find the gcf of 738 and 621 using the Euclidean algorithm, and express it as a combination

of 738 and 621. 738 1 621 117

621 5 117 36

117 3 36 9

36 4 9

So the gcf of 738 and 621 is 9. Now work backwards through the equations.

9 117 3 36 117 3 621 5 117 16 117 3 621 16 738 621 3 621

16 738 19 621.

So 16 738 19 621 9 .

The Greatest Common Combination.

92. Find the greatest common factor of the following pairs of numbers, and express it as a

combination of the two numbers.

a) 24 and 54 b) 72 and 160 c) 5291 and 11951

The greatest common factor of 7 and 9 is 1. The equation 7 5 9 4 1 , expresses 1 as a

combination of the numbers 7 and 9. The equation also gives a way of solving the following

problem:

You have an unlimited supply of water, a drain, a large container, and two jugs which can hold

7 gallons and 9 gallons, respectively. How can you manage to end up with exactly 1 gallon of

water in the large container?

The equation tells you how to do it. Add four of the 9 gallon containers of water into the large

container, and then you remove 5 of the 7 gallon containers of water from the large container.

You’ll be left with 1 gallon of water in the large container.

Forget About Your Whiskey, Can You Hold Your Water?

93. Solve the previous water problem, except this time you have:

a) 36 gallon and 60 gallon jugs, and you want 12 gallons in the large container.

b) 40 gallon and 70 gallon jugs, and you want 10 gallons in the large container.

c) 450 gallon and 750 gallon jugs, and you want 150 gallons in the large container.

Who Said Holes Have To Be Round?

94. Find the area of the square hole in the middle of the square.

{Hint: Find the areas of the four right triangles, and subtract it from the area of the large

square.}

6

3

Trigonometry Is Overrated.

95. What is the height of this rectangle containing a 3-4-5 right triangle?

{Hint: The area of the rectangle is twice the area of the right triangle.}

What’s The Difference?

96. Consider the two overlapping rectangles below:

What is the difference between the areas of the non-overlapping regions of the rectangles?

{Hint: The area of the first non-overlapping region is 80 xy . Find the area of the second

non-overlapping region, and subtract it from the first.}

5

3 4

8 feet

10 feet

5 feet

7 feet

x

y

I Don’t Give A Square’s S!

97. A right triangle with leg measurements of a and b has an inscribed square with side

measurement s as shown in the figure. Find the value of s.

{Hint: The areas of the square and two little right triangles must equal the area of the big

right triangle.}

Naughty Scotty.

98. Lucius, Sarah, Brandon, and Scott take Algebra together. On their first exam, Lucius got a

94, Sarah got a 91, Brandon got a 95, and Scott’s score was strictly between 82 and 88. If

the average of all four of their scores is a whole number, what was Scott’s score on the first

exam?

Mission: Impossible/Possible Perimeters.

99. A rectangle has the following sides. Find all possible numerical values of the perimeter of

the rectangle.

b

a

s

s

s s

2x y

4 8y

20 16x

The Arc Of Triangle.

100. In right triangle ABC with legs of 5 and 12, arcs of circles are drawn, one with center A and

radius 12, the other with center B and radius 5. What is the length of MN?

The Smallest Of The Bigs.

101. a) For x, y, and z real numbers with 1x y z , find the smallest value of M you can so

that xy yz xz M .

Hint: Square both sides of 1x y z , and solve for xy yz xz .

b) Using the symmetry of the problem, let x y z , to find a smaller value of M.

Mary, Mary, Not So Contrary.

102. Lottie and Lucy Hill are both 90 years old. Mary Jones, on the other hand, is half again as

old as she was when she was half again as old as she was when she lacked 5 years of being

half as old as she is now. How old is Mary?

Hint: 3 3 12 2 2

5M is half again as old as she was when she was half again as old as she

was when she lacked 5 years of being half as old as she is now, and M is her age now.

The Age Of Man.

103. A man lived one-sixth of his life in childhood, one-twelfth in youth, and one-seventh as a

bachelor. Five years after his marriage, a son was born who died four years before his

father at half his father’s final age. What was the man’s final age?

Hint: 1 1 1 16 12 7 2

5M M M M is the age of the man up to the death of his son, and this is

4 years less than the man’s final age.

A

B

C

N

M

How Long John?

104. John was three times as old as his sister 2 years ago, and five times as old 2 years before

that. In how many years will John be twice as old as his sister?

Which One Are You Rooting For?

105. Which number is larger, the 10,000,000th root of 10 or the 3,000,000th root of 2?

Hint: Raise both numbers to the 30,000,000th power to decide.

Crazy 8.

106. Show that the difference of the squares of two odd numbers is divisible by 8.

Hint: Suppose the two odd numbers are 2 1x n and 2 1y m .

Then 2 22 2 2 22 1 2 1 4 4 1 4 4 1 4 1x y n m n n m m n m n m .

If you can show that 1n m n m must be divisible by 2, then you’re done.

n m n m 1n m 1n m n m

even even even odd ?

odd odd even odd ?

even odd odd even ?

odd even odd even ?

With Friends Like You… .

107. A group of friends decide to pool their money to buy a wedding gift. They find that if

each of them pays $8, they will have $3 more than the price of the gift, and if they each

pay $7, they will have $4 less than the price of the gift. What’s the size of the group of

friends?

Sometimes You Got To Kiss A Lot Of Frogs.

108. There are some princes and frogs in a fairy-tale. As a group, they have 35 heads and 94

feet. How many princes, and how many frogs are there?

Zeroing In On An Answer.

109. Identify a pattern in the following table, and use it to determine the number of zeros that

the number 10011 1 ends in.

111 1 11 1 10 211 1 121 1 12 0 311 1 1,331 1 1,33 0 411 1 14,641 1 14,64 0 511 1 161,051 1 161,05 0 611 1 1,771,561 1 1,771,56 0 711 1 19,487,171 1 19,487,17 0 811 1 214,358,881 1 214,358,88 0 911 1 2,357,947,691 1 2,357,947,69 0 1011 1 25,937,424,601 1 25,937,424,6 00 1111 1 285,311,670,611 1 285,311,670,610 1211 1 3,138,428,376,721 1 3,138,428,376,72 0 1311 1 34,522,712,143,931 1 34,522,712,143,93 0 1411 1 379,749,833,583,241 1 379,749,833,583,24 0 1511 1 4,177,248,169,415,651 1 4,177,248,169,415,65 0 1611 1 45,949,729,863,572,161 1 45,949,729,863,572,16 0 1711 1 505,447,028,499,293,771 1 505,447,028,499,293,77 0 1811 1 5,559,917,313,492,231,481 1 5,559,917,313,492,231,48 0 1911 1 61,159,090,448,414,546,291 1 61,159,090,448,414,546,29 0 2011 1 672,749,994,932,560,009,201 1 672,749,994,932,560,009,2 00

Find X + Y + Z PDQ.

110. If 2x y z , 1y z x , and 4z x y , then find the value of x y z .

{Hint: Add the three equations together.}

Don’t Bank On The Old Switcheroo.

111. An absentminded bank teller switches the dollars and cents when he cashed a check for

Mr. Spencer, giving him dollars instead of cents, and cents instead of dollars. After buying

a five cent newspaper, Mr. Spencer discovered he had left exactly twice as much as his

original check. What was the original amount of the check?

{Hint: If D = original number of dollars and C = original number of cents, then the

original amount of the check in cents is 100X D C and the incorrect amount of

the check in cents is 100Y C D . This leads to the equation

100 5 2 100C D D C , which can be rearranged into 98 199 5C D or

199 5

98

DC

. We need to find D so that C is a whole number of at most 2-digits.}

D

199 5

98

DC

1 2.081632653

2 4.112244898

3 6.142857143

4 8.173469388

5 10.20408163

6 12.23469388

7 14.26530612

8 16.29591837

9 18.32653061

49 99.55102041

Where There’s A Will.

112. A father in his will left all his money to his children in the following manner:

$1000 to the first born and 1/10 of what then remains, then

$2000 to the second born and 1/10 of what then remains, then

$3000 to the third born and 1/10 of what then remains, and so on.

When this was done each child had the same amount.

a) How much money did the father leave in his will?

b) How many children were there?

{Hint: If S is the amount in the will, then the first born gets 1

1,000 1,00010

S , and the

second born gets 1 9

2,000 1,000 2,00010 10

S

. Set them equal to each other, and

solve for S.}

Don’t Shoot The Messenger.

113. a) Two armies are advancing toward each other, each at 1 mph. When the two armies are

10 miles apart, a messenger leaves the first army and races toward the second army at 9

mph. Upon reaching the second army, the messenger immediately turns around and

races back to the first army at 9 mph. How many miles apart are the two armies when

the messenger returns to the first army?

{Hint: It takes the messenger 1 hour reach the second army, at which time the two

armies are 8 miles apart.}

b) Solve the same problem, except this time the messenger travels at 4 miles per hour.

problem solving(due by sep. 11)

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