lcm/gcd & some NS basic prob.

[thebullfighter]

Q. The positive integer m is a multiple of n. Their sum is 2000. It is known that 3 divides the lcm(m,n) but it does not divide the gcd(m,n). If the number of divisors of m is twice the number of divisors of n, what is n?

[spoiler:c813dcbd79]ans: 500[/spoiler:c813dcbd79]

[capricious]

Let m=kn for some +ve integer k
(k+1)n=2000
lcm(m,n)=m which is a factor of 3 ;m%3=0
gcd(m,n)=n which is not a factor of 3 ;n%3!=0
So it means that k must be a factor of 3( as, m=kn)

No. of divisor's of a number x is doubled if we multiply the number by a prime no. p, which is not itself a divisor of the original no(i.e. x%p!=0). This is because all factors(f1,f2..) will be multiplied by prime no. giving another set of equal no. of factors(p·f1,p·f2,..).And as that prime no. is not a factor of original no. so their will be no replicated values, hence doubled.

as k%3=0 and k is prime => k=3
=>n=2000/(3+1)=500

This question really spun my head, but it was real fun. I hope I'm correct.
You can easily verify : no. of factors of 1500=24; no. of factors of 500=12.