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The sum of the cubes of two numbers is 7163. What are these numbers?

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Find the smallest numbers a, b, and c such that: a^2 = 2b^3 = 3c^5 What can you say about other solutions to this problem?

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Factorial Fun

How many divisors does factorial n (n!) have?


Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Ping sent this solution from Thailand.

(1) If $5^m=4^m 2^n$, then $5 = 2^{2m + n}$ which is impossible as 2 and 5 are prime so there are no positive integer solutions $m$ and $n$ of this equation.

(2) We have $a^m d^n=c^n b^m$. As $a$ and $b$ are coprime, we get $a^m|c^n$. Because $c$ and $d$ are coprime, so $c^n|a^m$. This means $a^m=c^n$. Similarly, $b^m=d^n$.

If $a^m=c^n$ and $b^m=d^n$, then obviously $(a/b)^m=(c/d)^n$.

This implies that $a^m$ and $c^n$ have the same prime factors. Write $a = p_1^{u_1}...p_k^{u_k}$ and $c = p_1^{v_1}...p_k^{v_k}$ and for all $j$ we have $mu_j = nv_j$ so that $${u_1\over v_1} = {u_2\over v_2} = ... = {u_k\over v_k} = {m\over n}.$$ Similarly for $b$ and $d$. This is a very special necessary relationship between $a$ and $c$ and also between $b$ and $d$ so solutions rarely occur to the equation: $$\left({5\over 4}\right)^m = \left({2\over 1}\right)^n.$$ We now show this is a sufficient condition. Conversely suppose $a = p_1^{u_1}...p_k^{u_k}$ and $c = p_1^{v_1}...p_k^{v_k}$ and $${u_1\over v_1} = {u_2\over v_2} = ... = {u_k\over v_k}.$$ We call this common ratio ${n\over m}$ then $u_jm=v_jn$ for all $j$ and $a^m = b^n$. Similarly if corresponding ratios of the powers of the prime factors of $b$ and $d$ are constant and also equal to ${n\over m}$ then $b^m=d^n$ giving $$\left({a\over b}\right)^m = \left({c\over d}\right)^n.$$