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Let $n$ be a 6-digit number, perfect square and perfect cube. If $n-6$ is not even or a multiple of 3, find $n$.

My try

Playing with the first ten perfect squares and cubes I ended with:

The last digit of $n \in (1,5,9)$

If $n$ last digit is $9$, then the cube ends in $9$, Ex: if $n$ was $729$, the cube is $9^3$ (ends in $9$) and the square ends in $3$ or $7$

If $n$ last digit is 5, then the cube ends in 5 and the square ends in 5

If $n$ last digit is 1, then the cube ends in 1 and the square ends in 1

By brute force I saw that from $47^3$ onwards, the cubes are 6-digit, so I tried some cubes (luckily for me not for long) and $49^3 = 343^2 = 117649$ worked.

So I found $n=117649$ but I want to know what is the elegant or without brute force method to find this number because my method isn't very good, just pure luck maybe.

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Note that the required number is both a square and a cube, so it must be a sixth power. Already $10^6=1000000$ has seven digits and $5^6=15625$ has only five digits, so that leaves us with $6^6,7^6,8^6,9^6$ to test.

Furthermore, we are given that $n-6$ is not even and not a multiple of 3, which implies that $n$ itself is also not even and not a multiple of 3. This eliminates $6^6,8^6$ and $9^6$ immediately, leaving $7^6$ as the only possible answer.

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  • $\begingroup$ Pretty clever, i don't know how i didn't think about that. Thanks $\endgroup$ – Rodrigo Pizarro Feb 15 '18 at 2:29
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If $n$ is both a perfect square and perfect cube then. $n = a^6$

If $n-6$ is neither even nor divisible by $3$, then $n$ is not even nor divisible by $3$ and $a$ is not even or divisible by 3.

$a^6$ is a $6$ digit number

$6<a<10$

$7$ is the only integer in that interval that is not divisible by $2$ or by $3.$

$n = 7^6$

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