The five digit integer ABCDE, where each letter represents a digit, not necessary distinct, is divided by the numbers $2$,$3$,$4$,$5$, and $6$. The remainders are A, B, C, D, and E respectively. What is the integer ABCDE?
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Does "five digit number" mean $A>0$? – Thomas Andrews Dec 14 '14 at 03:22
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If $A=0$ is allowed, I suppose $A=B=C=D=E=0$ is a trivial solution. – Thomas Andrews Dec 14 '14 at 03:26
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Notice $A=1$ ,there are $3$ possible values for $E$ since $E$ is odd and between $0$ and $5$. once $E$ has been fixed we can easily obtain $D$. once $D$ has been obtained we have the last two digits so we can obtain $C$. Also notice from $E$ you can obtain $B$ immediately.
Thus there are three cases(three possible values of $E$), they are:
$11311$ This one works
$10133$ This one doesn't work since the number is not a multiple of three
$12105$ This one doesn't work since the number is a multiple of three
Asinomás
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I obtained these numbers by writing $E$ and then deducing $D$, then deducing $C$, and then deducing $B$, until the number was complete. – Asinomás Dec 14 '14 at 03:48
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$A=1$, because it won't be a five digit number if $A=0$ and those are the only remainders when dividing by $2$. $C$ and $E$ are then odd, $C=1,3$ and $E=1,3,5$. $B\lt 3$ and $D\lt 5$ We are down to $90$ possibilities, within range of brute force on a contest.
Ross Millikan
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Of course, if you know $E$, you know $B$. $(E,B)=(1,1), (3,0),$ or $(5,2)$. – Thomas Andrews Dec 14 '14 at 03:31
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And $D$ is also entirely determined by $E$ for different reasons, since modulo $5$, none of the other digits matter. Down to $6$. So $(E,B,D)=(1,1,1),(3,0,3),$ or $(5,2,0)$. – Thomas Andrews Dec 14 '14 at 03:34