# rnshow that n2 - 1 is divisible by 8 if n is an odd positive integer.rnshow that the square of any positive integer is of the form 4q 0r 4q+1 for the integer q.rnif the H.C.F. of 210 and 55 is expressible in the form 210*5+55y , then find yrnusing prime factorisation method , find the H.C.F. and L.C.M. of 72 , 126 and 168. Also show that H.C.F.*L.C.M. not equals to the prodouct of the three numbers.rnFind the greatest number that will divides 445 , 572 and 699 leaving remainders 4 , 5 and 6 respectively.rnExpress 0.57 in the form of p/q.rnFind the H.C.F of the smallest composite number and the smallest prime number.rnthe decimal expansion of the rational number 43/24*53 will terminate after how many places of decimal.

1.

if n is an odd positive integer, let n = 2m+1 , where m is a positive integer.

now 2 cases will arise:

case I:

m is an odd integer, therefore (m+1) is an even integer.

(m+1) is divisible by 2

is divisible by 8.

case II:

m is an even integer, m is divisible by 2.

m(m+1) is divisible by 2

4m(m+1) is divisible by 8.

2.

since any integer can be written in the form of 2m or 2m+1. (by Euclid's division Lemma)

case I: if n = 2m

case II: if n = (2m+1)

thus square of any integer is of the form 4q or 4q+1.

3.

to find the HCF of  210 and 55 we will find it by Euclid's division Lemma

210 = 55*3 + 45

since remainder is not zero.

again applying the Euclid's division Lemma

55 = 45*1+10

45 = 10*4 + 5

10 = 5*2 +0

thus the HCF of 210 and 55 is 5.

given:

thus y = -19.

hope this helps you.

due to constraint of time, we are answering only three questions here.

hope you can try the remaining questions again, if you have still any difficulty, do get back to us.

cheers!!

• 8

let, n = 7, where n is an odd positive integer.

then, n2 - 1 = 2 - 1 = 49 - 1 = 48

48 is divisible by 8

Hence, n2 - 1 is divisible by 8

• 3
What are you looking for?