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Euler 4: Supplementary Number – needed explanation
So I’ve been looking at the problem Euler 4. I stumbled on this:
Find the number of composite numbers that, when multiplied together, equal $7^n$.
I have no idea how to solve this.
Can someone help me understand the logic here?
A:
Hint:
$$n=\frac{\log(7)}{\log(7)-1}$$
This is a remainder of division of $7^{\log(7)}$ by $7-1$.
By cross product we have $7\cdot7\cdot7=343$, so $7^{\log(7)}\equiv343\mod7$. Can you finish the problem now?
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