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7-digit numbers are formed by the digits 1, 2, 3, 4, 5, 6, 7. In each number, no digit is repeated. Prove that among all these numbers, there is no number, which is a multiple of another number.

My Attempt:

Total no. of ways of arranging it is 7!= 5040. So now we cannot check cases.

Suppose $x$ and $y$ are 2 nos. formed from the given digits with $x|y$ Then $y= 6x,5x,4x,3x,2x$.

But this leads nowhere...

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    $\begingroup$ Try it with 1,2,7,8. There is a classic recreational math problem hidden here. $\endgroup$ Jul 24 at 1:49
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Hint. Each of the $7!$ numbers has the same remainder when divided by $9$ which is $1$ (because $1+2+3+4+5+6+7=28=9\cdot 3+1$). Now if we take one of them, say $x$, then what are the remainders of $6x$, $5x$, $4x$, $3x$, $2x$ when they are divided by $9$?

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  • $\begingroup$ Thanks for the solution...🙏🏻🙏🏻🙏🏻 $\endgroup$ Jul 23 at 18:27

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