How to prove that if a $5$-digit number is divisible by $41$,then all the numbers generated from it by cyclic shift are also divisible by $41$
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$\begingroup$Say $n = \mathbf{abcde}$ is a 5-digit number. Then $10n = \mathbf{abcde}0$ and $10n-99999\cdot\mathbf{a} = \mathbf{bcdea}$, which is a cycled version of $n$. $99999 = 2439 \cdot 41$. If $n$ is divisible by $41$, then $10n$ and $99999\cdot\mathbf{a}$ are too, so $10n-99999\cdot\mathbf{a}$ is divisible by $41$.
$\endgroup$ $\begingroup$Check out this link for divisibility by most primes.
For your convenience, what the website says about divisibility for 41 is this: Test for divisibility by 41. Subtract four times the last digit from the remaining leading truncated number. If the result is divisible by 41, then so was the first number. Apply this rule over and over again as necessary. Example: 30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0, remainder is zero and so 30873 is also divisible by 41.
$\endgroup$ $\begingroup$That is true,in the similar way you may see these answers too which are given as 8*41*271,7*41*271,6*41*271,5*41*271,4*41*271,3*41*271,2*41*271 & 1*41*271.
$\endgroup$ $\begingroup$The reason is that $10^5 \equiv 1 \bmod 41$:$$ 0 \equiv 10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0 \bmod 41 \implies\\ 0 \equiv 10(10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0) \equiv 10^4 a_3 + 10^3 a_2 + 10^2 a_1 + 10 a_0 + a_4 \bmod 41 $$
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