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A 5 digit number ABCDE is divided by 2,3,4,5,6 and gives A,B,C,D,E as the remainders respectively.

Can you find the only possible number?

This was asked in my math exam, I think it would make a good puzzle.

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  • $\begingroup$ To avoid confusion, A,B,C,D,E are not necessarily distinct digits. And since they're remainders mod 2,3,4,5,6, (and easily shown to be non-zero), there aren't many possibilities for them at all. $\endgroup$
    – smci
    Feb 27 at 0:58

2 Answers 2

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The answer is

$11311$.

Solution:

$A$ is nonzero and a remainder upon division by $2$. Thus $A=1$ and $E$ is odd.
Now suppose $E=5$. Then $D=0$ (by divisibility by $5$) and $C=1$ (by divisibility by $4$), but no $B$ work. Thus $E \neq 5$.
Then the remainder when divided by $5$ is simply $E$, since $E < 5$. So $D=E$. Since $E$ is odd, there are two cases.
If $D=E=1$, then $\overline{ABCDE} = \overline{1BC11}$, so $C=3$ by dividing by $4$. Only $B=1$ works, so the solution is $\boxed{11311}$.
If $D=E=3$, then $\overline{ABCDE} = \overline{1BC33}$, so $C=1$ by dividing by $4$. No $B \in \{0, 1, 2\}$ work, so there are no solutions here. We are done.

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I got the same solution as @flame, but through a slightly different path.

$A$ must be $1$, as that is the only nonzero remainder$\bmod 2$. $E$, in turn, is one of $1,3,5$.
If $E=5$, then $B$ must be $2$, since $5=2 \bmod 3$. However, by the nature of our base ten number system, $D=0$ and $C=1$, but $1+2+1+0+5 =9 = 0 \bmod 3$, which does not equal $2$.
If $E=3$, then $B$ must be $0$, since $3=0 \bmod 3$. However, as above, $D=3$ and $C=1$, but $1+0+1+3+3 = 7 = 1 \bmod 3$, which does not equal $0$.
If $E=1$, then $B$ must be $1$, since $1 = 1 \bmod 3$. However, as above, $D=1$ and $C=3$, and $1+1+3+1+1 = 7 = 1 \bmod 3$, so the answer is $11311$!

Just for fun, I considered the cases where we let $A = 0$:

If $A=0$, then $E$ is one of $0,2,4$.
If $E=4$, then $B$ must be $1$. However, $D=4$ and $C=0$, but $0+1+0+4+4 = 9 =0 \bmod 3$, which does not equal $1$.
If $E=2$, then $B$ must be $2$. However, $D=2$ and $C=2$, and $0+2+2+2+2 = 8 = 2 \bmod 3$, so $02222$ works.
If $E=0$, then $B$ must be $0$. However, $D=0$ and $C=0$, and $0+0+0+0+0 = 0 = 0 \bmod 3$, so $00000$ works. Which of course it does, every digit's zero.

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  • $\begingroup$ I liked your A=0 part, I could have tweaked the question a bit to fit in "02222". $\endgroup$
    – I'm Nobody
    Feb 27 at 5:51

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