# Problem Of The Week #405 Feb 20th, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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Let $n$ be the smallest positive integer such that $n$ is divisible by 20, $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$?

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### anemone

##### MHB POTW Director
Staff member
Congratulations to the following members for their correct solution!

1. castor28
2. kaliprasad
3. MegaMoh

Solution from castor28 :
If $p$ is a prime that divides $n$ with exponent $\alpha_p$, $2\alpha_p$ must be divisible by $3$, and $3\alpha_p$ must be divisible by $2$; therefore $\alpha_p$ must be divisible by $6$.

If $p$ is other than $2$ or $5$, the smallest acceptable multiple of $6$ is $0$: the only primes dividing $n$ are $2$ and $5$.

On the other hand, since $20 \mid n$, we must have $\alpha_2 \ge 2$ and $\alpha_5 >= 1$, this gives $\alpha_2=\alpha_5=6$.

We have therefore $n=2^6\cdot5^6 = 10^6$, a number of $\bf 7$ digits.

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