JEEOlympiad
Let N be the smallest positive integer such that N/2 is a perfect…
Question
Let N be the smallest positive integer such that N/2 is a perfect square, N/3 is a perfect cube, and N/5 is a perfect fifth power. Find the number of positive divisors of N.
✓ Verified answer: 1232checked by our engine — not a guess
Step-by-step solution
N must be of the form 2^a·3^b·5^c.
The three conditions translate into congruences on each exponent: N/2 a perfect square requires a−1, b, c all even; N/3 a perfect cube requires a, b−1, c all ≡0 (mod 3); N/5 a perfect fifth power requires a, b, c−1 all ≡0 (mod 5).
For exponent a: a≡1 (mod 2), a≡0 (mod 3), a≡0 (mod 5) ⇒ smallest a = 15.
For b: b≡0 (mod 2), b≡1 (mod 3), b≡0 (mod 5) ⇒ smallest b = 10.
For c: c≡0 (mod 2), c≡0 (mod 3), c≡1 (mod 5) ⇒ smallest c = 6.
So N = 2^15·3^10·5^6, and the number of divisors is (15+1)(10+1)(6+1) = 16·11·7 = 1232.
Final answer1232
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