previous years school summative-exam ...cbsenotes.com/notes/files/sa-1.pdfq36 -given that lcm(91,...

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PREVIOUS YEARS SCHOOL SUMMATIVE-EXAM QUESTIONS ( REAL NUMBERS)

Q1-H.C.F. of two consecutive even numbers is : (A) 0 (B) 1 (C) 4 (D) 2

Q2-If the HCF of 85 and 153 is expressible in the form 85n - 153, then value of n is :

(A) 3 (B) 2 (C) 4 (D) 1

Q3-If the HCF of 65 and 117 is expressible in the form 65 m-117, then the value of m

is : (A) 4 (B) 2 (C) 3 (D) 1

Q4-Rational number

, q 0 will be terminating decimal if the prime factorisation of

q is of the form. (m and n are non negative integers : (A) 2m x3n (B) 2mx 5n (C) 3m x 5n (D) 3m x7n

Q5-For any two positive integers a and b, there exist unique integers q and r such that

a=bq+r, 0 If b =4 then which is not the value of r ?

(A) 1 (B) 2 (C) 3 (D) 4

Q6-The decimal expansion of

will terminate after how many places of decimal ?

(A) 1 (B) 2 (C) 3 (D) 4 Q7-Which of the following rational numbers has non terminating and repeating decimal expansion

(A)

(B)

(C)

(D)


will terminate after how many places of decimal ?

(A) 1 (B) 2 (C) 3 (D) 4


will terminate after how many places of decimals ?

(A) 2 (B) 1 (C) 3 (D) will not terminate Q10-For some integer ‘m’ every odd integer is of the form :

(A) m (B) m+1 (C) 2 m (D) 2 m+1

Q11-If two positive integers A and B can be expressed as A= ab2 and B = a3b, where a,

b, are prime numbers, then LCM (A, B) is : (A) ab (B) a2b2 (C) a3b2 (D) a4b3

2

Q12-Decimal expansion of

is :

(A) 0.115 (B) 1.15 (C) 11.5 (D) 0.0115 Q13-Which of the following numbers has terminating decimal expansion ?

(A)

(B)

(C)

(D)

Q14-Which of the following rational numbers have a terminating decimal expansion?

(A) )

(B) )

(C) )

(D)


will terminate after how many places of

decimals ? (A) 2 (B) 3 (C) 4 (D) 5

Q16-Given that HCF (2520, 6600) = 40, LCM (2520, 6600) = 252 X k, then the value of k

is : (A) 1650 (B) 1600 (C) 165 (D) 1625

Q17-From the following, the rational number whose decimal expansion is terminating is

(A)

(B)

(C)

(D)


the rational number will terminate after :

(A) one decimal place (B) two decimal places (C) three decimal places (D) more than 3 decimal places



(A) one decimal place (B) two decimal places (C) three decimal places (D) more than 3 decimal places

Q20-n2 -1 is divisible by 8, if n is

(A) an integer (B) a natural number (C) an odd integer (D) an even integer

Q21-If d=LCM(36, 198), then the value of d is :

(A) 396 (B) 198 (C) 36 (D) 1

Q22-A rational number between and is : (A) 1.4 (B) 1.75 (C) 1.45 (D) 1.8

Q23-If d=HCF(48, 72), the value of d is :

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(A) 24 (B) 48 (C) 12 (D) 72

Q24-Given that LCM(91, 26)=182, then HCF(91, 26) is :

(A) 13 (B) 26 (C) 7 (D) 9 Q25-H.C.F. of two co–prime numbers is : (A) 1 (B) 2 (C) 0 (D) 3 Q26-The product of the HCF and LCM of the smallest prime number and the smallest composite number is : (A) 2 (B) 4 (C) 6 (D) 8 Q27-The HCF and LCM of two positive integers are ‘h’ and ‘l’ respectively, if one integer is a, then other will be :

(A)

(B)

(C)

(D)

Q28-1192-1112 is :

(A) Prime number (B) Composite number (C) An odd prime number (D) An odd composite number

Q29-If HCF (a, 8) = 4, LCM (a, 8) = 24, then a is :

(A) 8 (B) 10 (C) 12 (D) 14



(A) 3 places (B) 4 places (C) 5 places (D) 1 place Q31-Which of the following is rational ?

(A) (B) (C) (D)

Q32-If m=dn+r, where m, n are positive integers and d and r are integers, then n is

the H.C.F of (m, n) if.

(A) r =1 (B) 0 < r 1 (C) r=0 (D) r is a real number

Q33-A pair of irrational numbers whose product is a rational number is :

(A) (B) (C) (D) Q34-The decimal expansion of is : (A) terminating (B) non-terminating and non-recurring (C) non-terminating and recurring (D) doesn’t exist

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Q35-If the HCF of 55 and 99 is expressible in the form 55 m - 99, then the value of m

is : (A) 4 (B) 2 (C) 1 (D) 3

Q36-Given that LCM(91, 26)=182, then HCF(91, 26) is :

(A) 13 (B) 26 (C) 7 (D) 9


the rational number will terminate after

(A) one decimal place (B) two decimal places (C) three decimal places (D) four decimal place

Q38-For given positive integers ‘a’ and ‘b’ if a=bq+r ; 0 r < b then :

(A) HCF (a, b)=HCF (b, r) (B) HCF (a, b)=HCF (a, r) (C) HCF (a, q)=HCF (a, r)

(D) HCF (b, q)=HCF (b, r)

Q39-Prove that 7 - 2 2 is irrational.

Let 7 - 2 2 is a rational number

7 - 2 2

7 -

=

=

Irrational = rational Which is not possible

So that 7 - 2 2 is irrational.

Q40-Prove that is irrational.

Let is a rational number

Squaring both sides

6 +2 +2 =

2 =

- 8

=


So that is irrational

5

Q41-Prove that is irrational


Squaring both sides

3 +2 +2 =

2 =

- 5

=



Q42-Prove that is irrational.

Let is a rational number then there exist p and q such that

p and q are co-prime

2

5

P2= 5q2 5|p2 5|p ------------------------------------------------------(1) P=5c for some integer c P2|25c2

25c2 =5q2 q2=5c2 5|q2 5|q -------------------------------------------------------(2) So that 5 is a common factor of p and q , but p and q are co-prime i.e. HCF(p,q)=1 This means that our supposition is wrong

is irrational number

Q43-Prove that

is an irrational.

Let

is a rational number


So that

is irrational

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Q44-Prove that

is an irrational

Let



So that

is irrational

Q45-Prove that

-

is irrational.

Let

-


Squaring both sides

= 9 +20 - 12

12 = 29 -

=


so that

-

is irrational number



Squaring both sides

9 +5 - 6 =

-6 = -(14 -

)

=




7

Squaring both sides

25 +3 - 10 =

-10 = -(28 -

)

=


Q48-Prove that 13 + 25 2 is irrational.

Let 13+25 2 is a rational number

13+25 2

- 13

=

=

irrational = rational Which is not possible

So that 13 + 25 2 is irrational.

Q49-Prove that 5 + 2 3 is irrational.


5+2 3

- 5

=

=


So that 5+2 3 is irrational.

Q50-Prove that 15 + 17 3 is irrational


5+17 3

- 15

=

=

8


So that 5+17 3 is irrational.

Q51-Prove that (3 + 2 5)2 is irrational

Let (3 + 2 5)2 is rational number

(3 + 2 5)2 =

9 +20 +12 =

12 =

- 29

=

=


So that (3 + 2 5)2 is irrational.

Q52-Prove that

is irrational

Let

is rational number

=

=

=

( rationalization the Dr)

=

=

rational =ir rational Which is not possible

So that

is irrational.


Let 6+ 2 is a rational number

+ 2

- 6

=


So that 6+ 2 is irrational.

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Q54-Prove that

is irrational

Let

is rational number


So that

is irrational

Q55-Prove that

is irrational.

Let

is rational number

=

2 - =

2-

=

=

rational =irrational Which is not possible

So that

is irrational

Q56-Prove that 2 - 5 3 is irrational.

Let 2 - 5 3 is a rational number

2-5 3

=

=


So that 2 - 5 3 is irrational

Q57-Prove that 2 3 - 7 is irrational

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Let 2 3 - 7 is a rational number

2 3- 7

+7

=

=


So that 2 3 - 7 is irrational



Squaring both sides

6 +2 +2 =

2 = (

- 8)

=

=




Let 12 3 - 41 is a rational number

12 3- 41

+41

=

=


So that 12 3 - 41 is irrational



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Squaring both sides

3 +5 +2 =

2 = (

- 8)

=

=




Let 3+ 2 is a rational number

3+ 2

- 3

=


So that 3+ 2 is irrational.

Q63-Prove that

is irrational

Let

is rational number

X

2+

=

rational = irrational Which is not possible

So that

is irrational

Q64-If x is rational and is irrational, then prove that (x+ )is irrational.

Let x+ is a rational number

x+

12

Squaring both sides

x +y +2 =

2 = (

– x -y)

=

=


So that x+ is irrational Q65-Using fundamental theorem of arithmetic, find the HCF of 26, 51 and 91. 26= 2 x13 51= 3 x 17 91= 7 x13 HCF = 1

Q66- Find the HCF and LCM of 306 and 54. Verify that HCF x LCM = Product of the

two numbers. 54= 2 x3 x3 x3 306= 2 x 3 x 3 x 17 HCF= 2 x 32= 18 LCM= 33 x 2 x 17= 918

HCF x LCM = 1st no. x 2nd no.

18 x918 = 306 x54 16524 = 16524 Q67-Find the LCM and HCF of 15, 18, 45 by the prime factorization method. 15= 3 x5 18 = 2 x 3 x 3 45 = 3 x 3 x 5 HCF= 3 LCM= 2 x 32 x 5= 90

Q68-If d is the HCF of 45 and 27, find x, y satisfying d = 27x+ 45y

45 = 27 x 1 + 18 -------------------(1) 27 = 18 x 1 + 9 --------------------(2) 18 = 9 x2 + 0 ---------------------(3) HCF = 9 From (2) 9 = 27 – 18 x1 From (1) 18 = 45 – 27 x1 9 = 27 –( 45- 27 x 1) 9 = 27 x2 – 45

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= 27 x2 + 45 (-1) = 27 x + 45y X = 2, y = -1 Q69- Use Euclid’s division algorithm to find the HCF of 10224 and 9648. 10224 = 9648 x 1 + 576 9648 = 576 x 16 + 432 576 = 432 x1 + 144 432= 144 x 3 + 0 HCF = 144 Q70-Find the LCM of 72, 80 and 120 using the fundamental theorem of arithmetic. 72 = 2 x 2x 2 x 3x3 80= 2x2x2x2x5 120= 2x2x2x3x5 LCM= 24 x 32x 5 = 720 Q71-Find the HCF of 867 and 255, using Euclid’s division algorithm. 867= 255 x 3 + 102 255 = 102 x2 +51 102 = 51 x 2 + 0 HCF = 51 Q72-The HCF and LCM of two numbers are 9 and 90 respectively. If one number is 18, find the other. Let 2nd number = p

HCF x LCM = 1st no. x 2nd no.

9 x 190 = 18 x p P = 45 Q73- Find the LCM and HCF of 120 and 144 by using fundamental theorem of Arithmetic. 120= 2 x 2 x 2 x 3 x 5 144= 2x2x2x2x3x3 HCF= 23 x 3 = 24 LCM = 24 x 32 x 5= 720 Q74-Using fundamental theorem of arithmetic, find the HCF of 26, 51 and 91. 26= 2 x13 51= 3 x 17 91= 7 x13 HCF = 1

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Q75-Write the denominator of

in the form of 2m 5n, where m, n are non-negative

integers. Also write its decimal expansion without actual division. Dr = 2 x 54 its decimal expansion up to 4 decimal places because highest power in Dr is 4 Q76-An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ? 616 = 2x2x2x7x11 32= 2x2x2x2x2 HCF= 2x2x2 = 8 No. of columns = 8 Q77-An army contingent of 1000 members is to march behind an army band of 56 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ? 1000= 2x2x2x5x5x5 56=2x2x2x7 HCF= 2x2x2=8 No. of columns = 8 Q78-There are 156, 208 and 260 students in Groups A, B, C respectively. Buses are to be hired to take them for a field trip. Find the minimum number of buses to be hired if the same number of students should be accommodated in each bus. HCF of 156, 208 and 260 is 52 So that 52 students will be accommodate in each bus Total students = 156 +208 + 260 = 624 No. of buses = 624 ÷ 52 = 12 buses

Q79- Explain why 11x13x15x17+17 is a Composite Number.

Since ( 11 x 13 x 15 + 1) x 17 It has more than two factors so it is composite number

Q80-Explain why 5x4x3x2x1+3 is a composite number

Since ( 5 x 4 x 2 x 1+1) x 3 It has more than two factors so it is composite number

Q81-Is 7 x 6 x 5 x 4x 3 x 2 x 1+5 a composite number ? Justify your answer.

Since ( 7 x6x4x3x2x1+1) x5 It has more than two factors so it is composite number

Q82-Is 7X11x13+11 a composite number ? Justify your answer.

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Since (7x13+1) x11 It has more than two factors so it is composite number

Q83-Is 75x32x5+3 a composite number ? Justify your answer.

Since (75x 3x5+1) x 3 It has more than two factors so it is composite number Q84-Show that 4n can never end with the digit zero for any natural number n. 4n =(22)n , only prime factor of 4n is 2 5 does not occur in the prime factorization of 4n So that 4n does not end is zero Q85-Show that 8n cannot end with the digit zero for any natural number n. 8n =(23)n , only prime factor of 8n is 2 5 does not occur in the prime factorization of 8n So that 8n does not end is zero Q86-Check whether 6n can end with the digit 0 for any natural number n ? 6n =(2 x 3)n , prime factor of 6n is 2 and 3 5 does not occur in the prime factorization of 6n So that 6n does not end is zero Q87-Check whether 15n can end with the digit 0 for any natural number n ? 15n =(3 x 5)n , prime factor of 15n is 3 and 5 2 does not occur in the prime factorization of 15n So that 15n does not end is zero

Q88-Show that square of any positive integer is either of the form 3m or (3m +1) for

some integer m. Let a be any +ve integer By Euclid’s division lemma a=bq + r where b = 3 a= 3q +r 0 r <3 a= 3q when r= 0 a= 3q+1 , if r= 1 a= 3q +2 if r = 2 “ a “ be any +ve integer in the form of 3q, 3q+1 and 3q + 2 for some integer q

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Case I- a = 3q a2 = ( 3q)2 = 3(3q2) a2 = 3(3q2) = 3m where m = 3q2 Case II- a = 3q +1 a2 = ( 3q+1)2 = 9q2+6q + 1 a2 = 3(3q2+ 2q) + 1 = 3m+1 where m = 3q2+ 2q Case III - a = 3q +2 a2 = ( 3q+2)2 = 9q2+12q + 4 = 9q2+12q +3 + 1 a2 = 3(3q2+ 4q + 1) + 1 = 3m+1 where m = 3q2+ 4q+1 Q89-Use Euclid’s division lemma to show that cube of any positive integer is either

of form 9q, 9q+1, or 9q +8 for some integer q.

Let a be any +ve integer By Euclid’s division lemma a=bm + r where b = 3 a= 3m +r 0 r <3 a= 3m when r= 0 a= 3m+1 , if r= 1 a= 3m +2 if r = 2 “ a “ be any +ve integer in the form of 3m, 3m+1 and 3m + 2 for some integer m Case I- a = 3m a3 = ( 3m)3 = 9(3m3) = 9q where q = 3m3 Case II- a = 3m +1 a3 = ( 3m+1)3 = 27m3+27m2 + 9m+ 1 a3 = 9(3m3+3m2+m) + 1 = 9q+1 where q = 3m3+3m2+m Case III - a = 3m +2 a3 = ( 3m+2)3 = 27m3+ 54m2+ 36m+ 8 a3 = 9(3m3+ 6m2 +4m ) + 8 = 9q+8 where q = 3m3+ 6m2 +4m

So that cube of any positive integer is either of form 9q, 9q+1, or 9q +8 for some integer

q.

Q90-Show that, any positive integer is of the form 3q, 3q+1 or 3q+2, where q is some

integer. Let a be any +ve integer By Euclid’s division lemma

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a=bq + r where b = 3 a= 3q +r 0 r <3 a= 3q when r= 0 a= 3q+1 , if r= 1 a= 3q +2 if r = 2 “ a “ be any +ve integer in the form of 3q, 3q+1 and 3q + 2 for some integer q

Q91-Show that any positive odd integer is of the form 4q+1 or 4q+3 where q is a

positive integer. Let a be any +ve integer By Euclid’s division lemma a=bq + r where b = 4 a= 4q +r 0 r <4 a= 4q when r= 0 even for any +ve integer q = 1,2,3 --- ( a= 4, 8,12-----) a= 4q+1 , if r= 1 odd for any +ve integer q = 1,2,3------( a= 5,9,13-----) a= 4q +2 if r = 2 even for any +ve integer q = 1,2,3 --- ( a= 6, 10,14 -----) a= 4q +3 if r = 3 odd for any +ve integer q = 1,2,3 --- ( a= 7, 11,15-----) any +ve odd integer in the form of 4q+1 and 4q + 3 for some integer q Q92-Show that every positive even integer is of the form 2q and that every positive

odd integer is of the form 2q+1, where q is some integer.

Let a be any +ve integer By Euclid’s division lemma a=bq + r where b = 2 a= 2q +r 0 r <2 a= 2q when r= 0 even for any +ve integer q = 1,2,3 --- ( a= 2, 4,6-----) a= 2q+1 , if r= 1 odd for any +ve integer q = 1,2,3------( a= 3,5, 7 -----) so that every even + ve integer in the form of 2q and odd + ve integer in the form of 2q+1

Q93-Show that the square of any positive integer cannot be of the form 5q+2 or 5q+3

for any integer q. Let a be any +ve integer By Euclid’s division lemma a=bm + r where b = 5 a= 5m +r 0 r <5 a= 5m when r= 0 a= 5m+1 , if r= 1 a= 5m +2 if r = 2

18

a= 5m+3 , if r= 3 a= 5m +4 if r = 4 “ a “ be any +ve integer in the form of 5m, 5m+1 , 5m + 2 , 5m+3 ,5m+ 4 for some integer m Case I- a = 5m a2 = ( 5m)2 = 25m2 a2 = 5(5m2) = 5q , where q = 5m2 Case II- a = 5m +1 a2 = ( 5m+1)2 = 25m2+10m + 1 a2 = 5(5m2+ 10m) + 1 = 5q+1 , where q = 5m2+ 10m Case III - a = 5m +2 a2 = ( 5m+2)2 = 25m2+10m + 4 a2 = 5(5m2+ 2m) + 4 = 5q+4 , where q = 5m2+ 2m Case IV- a = 5m +3 a2 = ( 5m+3)2 = 25m2+30m + 9 a2 = 25m2+30m +5 + 4 a2 = 5(5m2+ 6m +1) + 4 = 5q+4 , where q = 5m2+ 6m+1 Case V- a = 5m +4 a2 = ( 5m+4)2 = 25m2+40m + 16 a2 = 25m2+40m +15 + 1 a2 = 5(5m2+ 8m +3) + 1 = 5q+1 , where q = 5m2+ 8m+3 So that square of any +ve integer cannot be in the form of 5q+2 and 5q+3

Q94-Show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5 where q

is some integer. Let a be any +ve integer By Euclid’s division lemma a=bq + r where b = 6 a= 6q +r 0 r <6 a= 6q when r= 0 even for any +ve integer q = 1,2,3 --- ( a= 6, 12,18-----) a= 6q+1 , if r= 1 odd for any +ve integer q = 1,2,3------( a= 7,13,19-----) a= 6q +2 if r = 2 even for any +ve integer q = 1,2,3 --- ( a= 8, 14,20 -----) a= 6q +3 if r = 3 odd for any +ve integer q = 1,2,3 --- ( a= 9, 15,21-----) a= 6q +4 if r = 4 even for any +ve integer q = 1,2,3 --- ( a= 10, 16,22 -----) a= 6q +5 if r = 5 odd for any +ve integer q = 1,2,3 --- ( a= 11, 17,23-----) any +ve odd integer in the form of 6q+1 or 6q + 3 or 6q+5 for some integer q

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Q95-Show that any positive odd integer is of the form 8q + 1 or 8q + 3 or 8q + 5 or 8q

+ 7,

Let a be any +ve integer By Euclid’s division lemma a=bq + r where b = 8 a= 8q +r 0 r <8 a= 8q when r= 0 even for any +ve integer q = 1,2,3 --- ( a= 8,16,24-----) a= 8q+1 , if r= 1 odd for any +ve integer q = 1,2,3------( a= 9,17,25-----) a= 8q +2 if r = 2 even for any +ve integer q = 1,2,3 --- ( a= 10,18,26 -----) a= 8q +3 if r = 3 odd for any +ve integer q = 1,2,3 --- ( a= 11,19,27-----) a= 8q +4 if r = 4 even for any +ve integer q = 1,2,3 --- ( a= 12,20,28 -----) a= 8q +5 if r = 5 odd for any +ve integer q = 1,2,3 --- ( a= 13,21,29-----) a= 8q +6 if r = 6 even for any +ve integer q = 1,2,3 --- ( a= 14,22,30 -----) a= 8q +7 if r = 7 odd for any +ve integer q = 1,2,3 --- ( a= 15,23,31-----) any +ve odd integer in the form of 8q+1 or 8q + 3 or 8q+5 or 8q+7 for some integer q Q96-Use Euclid’s division lemma to show that the square of any positive integer is

either of the form 5m, 5m+1 or 5m+4 for some integer m.

Let a be any +ve integer By Euclid’s division lemma a=bq + r where b = 5 a= 5q +r 0 r <5 a= 5q when r= 0 a= 5q+1 , if r= 1 a= 5q +2 if r = 2 a= 5q+3 , if r= 3 a= 5q +4 if r = 4 “ a “ be any +ve integer in the form of 5q, 5q+1 , 5q + 2 , 5q+3 ,5q+ 4 for some integer q Case I- a = 5q a2 = ( 5q)2 = 25q2 a2 = 5(5q2) = 5m , where m = 5q2 Case II- a = 5q +1 a2 = ( 5q+1)2 = 25q2+10q + 1 a2 = 5(5q2+ 10q) + 1 = 5m+1 , where m = 5q2+ 10q Case III - a = 5q +2 a2 = ( 5q+2)2 = 25q2+10q + 4 a2 = 5(5q2+ 2q) + 4

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= 5m+4 , where m = 5q2+ 2q Case IV- a = 5q +3 a2 = ( 5q+3)2 = 25q2+30q + 9 a2 = 25q2+30q +5 + 4 a2 = 5(5q2+ 6q +1) + 4 = 5m+4 , where m = 5q2+ 6q+1 Case V- a = 5q +4 a2 = ( 5q+4)2 = 25q2+40q + 16 a2 = 25q2+40q +15 + 1 a2 = 5(5q2+ 8q +3) + 1 = 5m+1 , where m = 5q2+ 8q+3

So that the square of any positive integer is either of the form 5m, 5m+1 or 5m+4 for

some integer m.

Q97-Show that the square of any positive odd integer is of the form 8 m+1, for some integer

m. Let a be any +ve integer By Euclid’s division lemma a=bq + r where b = 8 a= 8q +r 0 r <8 a= 8q when r= 0 even for any +ve integer q = 1,2,3 --- ( a= 8,16,24-----) a= 8q+1 , if r= 1 odd for any +ve integer q = 1,2,3------( a= 9,17,25-----) a= 8q +2 if r = 2 even for any +ve integer q = 1,2,3 --- ( a= 10,18,26 -----) a= 8q +3 if r = 3 odd for any +ve integer q = 1,2,3 --- ( a= 11,19,27-----) a= 8q +4 if r = 4 even for any +ve integer q = 1,2,3 --- ( a= 12,20,28 -----) a= 8q +5 if r = 5 odd for any +ve integer q = 1,2,3 --- ( a= 13,21,29-----) a= 8q +6 if r = 6 even for any +ve integer q = 1,2,3 --- ( a= 14,22,30 -----) a= 8q +7 if r = 7 odd for any +ve integer q = 1,2,3 --- ( a= 15,23,31-----) “ a “ be any +ve odd integer in the form of 8q+1 ,8q+3 and 8q+5 for some integer q Case I- a = 8q +1 a2 = ( 8q+1)2 = 64q2+16q + 1 a2 = 8(8q2+ 2q) + 1 = 8m+1 , where m = 8q2+ 2q Case II - a = 8q +3 a2 = ( 8q+3)2 = 64q2+48q + 9 = 64q2+48q + 8 + 1 a2 = 8(8q2+ 6q + 1) + 1 = 8m+1 , where m = 8q2+ 6q + 1 Case III - a = 8q +5 a2 = ( 8q+5)2 = 64q2+80q + 25 = 64q2+80q + 24 + 1

21

a2 = 8(8q2+ 10q + 3) + 1 = 8m+1 , where m = 8q2+ 10q + 3

So that the square of any positive odd integer is of the form 8 m+1, for some integer m.

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