Prove that there exists a multiple of $2016$, such that the multiple has only $4$ and $6$ as its digits. Any ideas how to go about proving this? I tried it with pigeonhole-principle, i tried to find a similar demonstration like this one : Existance of multiple of $n$ with only 0 and 1 as it's digits , but i couldn't.
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Use the pigeonhole principle to find a multiple of $63$ that is a repunit in base 100.
Multiply that number by $64$.
Note that $63\cdot 64=2\cdot 2016$.
hmakholm left over Monica
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@AndreasCaranti: Mostly because it actually works. :-) – hmakholm left over Monica Jan 15 '17 at 16:17
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Exactly! Otherwise there could have been zeroes. 10101010101010101 – Andreas Caranti Jan 15 '17 at 16:19
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Thank you very much! I didn't think about that! :-) – Deni Jan 15 '17 at 16:27