A pair of integers( m, n) satisfy m × n = to gcd (m, n) +1cm(m,n) then m+n is
Answers
Given :- A pair of integers( m, n) satisfy m × n = to gcd (m, n) + lcm(m,n) then m+n is ?
Solution :-
we know that, for two number a and b
- a * b = LCM of a and b * GCD of a and b .
- LCM is always divisible by HCF .
so, for 2 number m and n , given that,
→ m * n = gcd(m,n) + lcm(m,n)
and , we know that,
→ m * n = gcd(m,n) * lcm(m,n)
comparing value of m * n , we get,
→ gcd(m,n) + lcm(m,n) = gcd(m,n) * lcm(m,n)
now, since lcm is multiple of gcd(m,n) or we can say
- lcm(m,n) = M * gcd(m,n) { where M is any natural number. }
putting this value in both sides we get,
→ gcd(m,n) + M * gcd(m,n) = gcd(m,n) * M * gcd(m,n)
→ gcd(m,n)[1 + M] = gcd(m,n) * M * gcd(m,n)
→ [1 + M] = M * gcd(m,n)
→ [1 + M]/M = gcd(m,n)
given that, m and n both are integers . then, M must be equal to 1 .
→ 2 = gcd(m,n)
therefore,
→ lcm(m,n) = M * gcd(m,n)
→ lcm(m,n) = 1 * 2
→ lcm(m,n) = 2
as,
→ lcm(m,n) = gcd(m,n)
→ m = n = 2 .
hence,
→ m + n = 2 + 2 = 4 (Ans.)
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