$a,b,c,d,e$ are different non-zero digits.We define $N(a,b,c,d,e)$ as the number of elements of the set $\{ \overline{abcde},\overline{bcdea},\overline{cdeab},\overline{deabc},\overline{eabcd}\}$ which are divisible by $41$.How many values of $N(a,b,c,d,e)$ are possible?
Divisibility of elements of $\{ \overline{abcde},\overline{bcdea},\overline{cdeab},\overline{deabc},\overline{eabcd}\}$ by $41$
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elementary-number-theory
divisibility
decimal-expansion
1 Answers
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Note that the next number in the list is obtained from the preceding by first multiplying it by $10$ and then subtracting a multiple of $100000-1=41\cdot 2439$. As $10$ is coprime to $41$, this means that the nxt number is divisible by $41$ iff the preceding one is. Hence either all or none of the numbers are divisible by $41$.