rnrn show that n2 - 1 is divisible by 8 if n is an odd positive integer rn show that - Maths - Real Numbers

Question

rnrn
show that n2 - 1 is divisible by 8 if n is an odd
positive integerrn
show that the square of any positive integer is of the form 4q
0r 4q+1 for the integer qrn
if the H C F of 210 and 55 is expressible in the form 210*5+55y
, then find yrn
using 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 numbersrn
Find the greatest number that will divides 445 , 572 and 699
leaving remainders 4 , 5 and 6 respectivelyrn
Express 0 57 in the form of p/qrn
Find the H C F of the smallest composite number and the
smallest prime number
the decimal expansion of the rational number
43/24*53 will terminate after how many places of
decimal

Ajanta Trivedi · Accepted Answer

1 if n is an odd positive integer, let n = 2m+1 , where m is a positive integer $n^{2} - 1 = (2 m + 1)^{2} - 1 = 4 m^{2} + 4 m + 1 - 1 = 4 m^{2} + 4 m = 4 m (m + 1)$ 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 $4 m (m + 1)$ 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 $n^{2} = 4 m^{2} = 4 q w h e r e q = m^{2}$ case II: if n = (2m+1) $n^{2} = (2 m + 1)^{2} = 4 m^{2} + 4 m + 1 = 4 (m^{2} + m) + 1 = 4 q + 1 w h e r e q = m^{2} + m$ 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: $5 = 210 * 5 + 55 y 5 = 1050 + 55 y 55 y = 5 - 1050 55 y = - 1045 y = - \frac{1045}{55} = - 19$ 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!!