tcy - lcm hcf.ppt

7/28/2019 TCY - LCM HCF.ppt

1/12

App l ications o f LCM & HCF

Greatest Number that will leave no remainder when divides a, b and c

Required number = HCF of a, b and c

A shopkeeper has three cakes of weight 18 kg, 45 kg and 36 kg. If hewants to make these cakes into pieces of equal weight withoutwastage, what is the maximum possible weight of each piece?

(1) 1 (2) 9 (3) 12 (4) 13

Example


2/12

The greatest number that will divide a, b & c leavingremainderx, y & z respectively

Required number = HCF of{(a x), (b y), (c z)}

What is the greatest number that divide 20, 50, and 40leaving 2, 5 and 4 as remainder?

Here a x=

b y=

c

z=Required Number =


Example


3/12

To find the greatest number that will divide x, y and z leaving thesame remainderr in each case.

Required number = HCF of (x r), (y r) and (z r)

Find the greatest number which will divide 369, 449, 689,5009 and 729 so as to leave the remainder 9 in each case

(A) 2 (B) 49 (C) 35 (D) 40

Required Number is

HCF of {(369-9), (449-9), (689-9), (5009 9), (729 9)}= HCF of {360, 440, 680, 5000, 720}

= 40


Example


4/12

To find the greatest number that will divide x, y and zleaving the same remainder in each case.

Required number = HCF of |x y|, |y z| and |z x|

What is the greatest number that will divide 1305, 4665 and

6905 leaving in each case the same remainder? Also

calculate the remainder.

(1) 1210, 158 (2) 1120, 158

(3) 1120, 185 (4) 1210, 185


Example


5/12

Solut ion

Here we have

|4665 1305|, |6905 4665|, |6905 1305|

= 3360, 2240, 5600

= 1120 x 3, 1120 x 2, 1120 x 5Required number = HCF of (3360, 2240, 5600) = 1120

To calculate Remainder

1305 = 1120 x 1 + 185

So the Remainder = 185


6/12

Least number which is exactly divisible by a, b, c

Required number = LCM of x, y and z

Sequence = n x LCM

(Where n is a natural Number)

Find the greatest number of five digits which is divisible by

32, 36, 40, 42 and 48

(a) 999720 (b) 90702 (C) 90720 (d) 90730


Example


7/12

Step II

Find greater number of 5 digit which multiple of 10080

9279

10080 99999

907209

Greatest number = 90720

LCM of (32, 36, 40, 42, 48) = 10080

Solut ion

Step I


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To find least number which when divided by a, b, c leavesr as a remainder

Required number= LCM of (a, b, c) + r

What is the smallest sum which a person can have such that

when he distributed @ Rs. 2.5 or Rs 20 or Rs 12 or Rs. 7.5 per

person in a group, he is always left with Rs. 2.00?

(A) Rs. 62 (B) Rs. 80 (C) Rs. 90 (D) Rs. 100

App l icat ions o f LCM & HCF

Example


9/12

According to the problem the person must have the money

equal to the LCM of2.5, 20, 12 and 7.5 and the remainder

money always left.

Solut ion


10/12

Least number which when divided by a, b, c leaves x, y, zas remainder, such that a x = b y = c z = k (say)

Required number = LCM of (a, b, c) k

What is the greatest number of 4 digits that when dividedby the numbers 6, 9, 12, 17 leaves 5, 8, 11, 16 asremainders respectively?

(a) 9791 (b) 9793 (c) 9895 (d) 9497

App l icat ions o f LCM & HCF

Example


11/12

Step 1 Here,

6 5 = 9 8 = 12 11= 17 16 = 1 = K (let)

Step 2 LCM of 6, 9, 12, 17 = 17 x 9 x 4 = 17 x 36 = 612

Step 3 Find Greatest 4 digit number which is multiple of 612

612 9999

612

3879

3672

207

16

Greatest Number= 9999 207 = 9792

Step 4 Required number = (LCM) n - K

= 9792 - 1 = 9791

Solut ion


12/12

Thank you !!

tcy - lcm hcf.ppt

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