1 For some integer m, every even integer is of the form (A) m (B) m + 1 (C) - Maths - Real Numbers

Question

1 For some integer m, every even integer is of the form
(A) m (B) m + 1
(C) 2m (D) 2m + 1
2 For some integer q, every odd integer is of the form
(A) q (B) q + 1
(C) 2q (D) 2q + 1
3
3 n
2
– 1 is divisible by 8, if n is
(A) an integer (B) a natural number
(C) an odd integer (D) an even integer
4 If the HCF of 65 and 117 is expressible in the form 65m
– 117, then the value of m is
(A) 4 (B) 2
(C) 1 (D) 3
5 The largest number which divides 70 and 125, leaving
remainders 5 and 8,
respectively, is
(A) 13 (B) 65
(C) 875 (D) 1750
6 If two positive integers a and b are written as
a = x
3
y
2
and b = xy
3
; x, y are prime numbers, then HCF (a, b) is
(A) xy (B) xy
2
(C) x
3
y
3
(D) x
2
y
2
7 If two positive integers p and q can be expressed as
p = ab
2
and q = a
3
b; a, b being prime numbers, then LCM (p, q) is
(A) ab (B) a
2
b
2
(C) a
3
b
2
(D) a
3
b
3
8 The product of a non-zero rational and an irrational number
is
(A) always irrational (B) always rational
(C) rational or irrational (D) one
9 The least number that is divisible by all the numbers from 1
to 10 (both inclusive) is
(A) 10 (B) 100 (C) 504 (D) 2520
10 The decimal expansion of the rational number
14587
1250
will terminate after:
(A) one decimal place (B) two decimal places
(C) three decimal places (D) four decimal places

Ajanta Trivedi · Accepted Answer

any number can be written in the form of 2q or 2q+1.

if number is even i.e. it is divisible by 2. it is of the form 2q.

if number is odd it is of the form 2q+1.

therefore

for query 1 option (c) is correct.

for query 2 option (d) is correct.

3.

$n^{2} - 1$ is divisible by 8, if n is an odd integer.

let n = 2q+1. where q is an integer.

$n^{2} - 1 = (2 q + 1)^{2} - 1 = 4 q^{2} + 4 q + 1 - 1 = 4 q^{2} + 4 q = 4 q (q + 1)$

now two case arises;

case I : q is an even number.

therefore 4q is divisible by 8.

thus 4q(q+1) i.e. $n^{2} - 1$ is divisible by 8.

case II: q is an odd integer.

therefore q+1 is an even number.

and 4q(q+1) is divisible by 8.

4.

let us find the HCF of 65 and 117,

117 = 1*65 + 52

65 = 1*52 + 13

52 = 4*13 +0

therefore HCF of 65 and 117 is 13.

65m-117 = 13

65m = 117+13 = 130

m =130/65 =2

thus option (b) is correct.

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