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 Mar 5, 2012
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Given a number that does not end in 0 or 5, prove that it has a multiple that consists of only 6's and 7's.
For instance 12 has the multiple 7776.
For instance 12 has the multiple 7776.
Given a number that does not end in 0 or 5, prove that it has a multiple that consists of only 6's and 7's.
For instance 12 has the multiple 7776.
It is instrumental that it has n digits, so I think that should be mentioned beforehand rather than as an afterthought. There are smaller numbers. It's just that the proof by induction won't work on them.If a number is divisible by $2^n$ then either it divisible by $2^{n+1}$ or remainder when divided by $2^{n+1}$ is $2^n$
if it is divisible by $2^{n+1}$ then add $6*10^{n+1}$ to it to make is divisible by$ 2^{n+1}$ as both terms are divisible by $2^{n+1}$
if it is not divisible by $2^{n+1}$ that is remainder is $2^n$ then add $7*10^{n+1}$ to it to make is divisible by$ 2^{n+1}$ as both terms leave a remainder $2^n$ and sum zero divisible by $2^{n+1}$
so in both cases we have a number consisting of 7 and 6 (n digits) which is divisible by $2^n$
I don't think the case of $C*2^n$ is covered where C is a composite number.if it is even number that is $p*2^n$ where p is a prime

Thank you kaliprasad for a correct solution!
Just a nitpick:
It is instrumental that it has n digits, so I think that should be mentioned beforehand rather than as an afterthought. There are smaller numbers. It's just that the proof by induction won't work on them.
I don't think the case of $C*2^n$ is covered where C is a composite number.
I'm going to sue you for deliberately trying to hurt my brain...Given a number that does not end in 0 or 5, prove that it has a multiple that consists of only 6's and 7's.
For instance 12 has the multiple 7776.
Easily fixed. Just get your brain into gear lol.I'm going to sue you for deliberately trying to hurt my brain...
Dan
Our solution is very much like yours.I would like to have a solution from OP/Satya.