Question

(ii) (a) Describe briefly, in each case, the integers a for which (a) g.c.d(a, 3) = 3; (B) g.c.d(a, 6) = 3; (7) g.c.d(a, –15) = –15. (b) Let m be a fixed integer. Describe briefly the integers a, and any restrictions on m, for which g.c.d(a,m) = m.

Answer 1

G.c.d : let $a\epsilon Z,b\epsilon Z$ such that $a\neq 0,b\neq 0$ .

the greatest common divisor(gcd) of two numbers a and b is the positive integer d such that

i) d/a and d/b

ii) if c/a and c/b then c/d.

II) $\alpha$ ) g.c.d.(a,3) = 3

i.e. 3 divides a and 3 divides 3.

and 3 can divide a iff a is multiple of 3.

so , a is multiple of 3.

$\beta$)

g.c.d.(a,6) = 3

g.c.d divides a and 6.

i.e. 3 divides a and 3 divides 6.

so a can be 3 or multiple of 3 such that gcd(a,2)=1.

so , a is multiple of 3 with condition g.c.d(a,2)=1.

$\gamma )$ g.c.d (a,-15) = -15

g.c.d can not be negative. so there is not exist any value of a for which g.c.d can become a negative number.

b) let m be a fixed integer and a is an integer and

g.c.d(a,m) = m

since g.c.d is always positive so m is also positive. i.e. m>0

and a is multiple of m. it can be write as

a = k m, where k is any integer,

a either positive or negative but multiple of m.