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This is a problem from a school textbook that I currently learn math with, this is the chapter on divisibility properties from the number theory.

There is two theorems in the chapter:


  1. $\overline{a_n a_{n-1} a_{n-2}...a_1 a_0} ≡ a_n + a_{n-1} + ... + a_1 + a_0$ (mod. 9) => (mod. 3) as well.


  2. $\overline{a_n a_{n−1} a_{n−2} ... a_1 a_0} ≡ a_0 − a_1 + a_2 − a_3 + ... + ( −1 )^n a_n$ (mod 11).

I understood the proofs of both of the theorems and yet the theorems seem to be inapplicable to the equation below that needs to be proven. Neither can I apply anything learned before in the previous lessons on the number theory.

$\overline{a_na_{n-1}...a_ka_{k-1}...a_0} ≡ \overline{a_{k-1}a_{k-2}...a_0}$ mod($5^k$)

How to approach such problems, how to prove this one in particular ?

Bill Dubuque
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curioushuman
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  • Have you tried the cases $k=1$ and $k=2$? – Dietrich Burde Apr 12 '23 at 19:19
  • $\overline{a_na_{n-1}...a_ka_{k-1}...a_0} = 10^k q + r,,$ for $q = \overline{a_na_{n-1}...a_k},,$ $,r = \overline{a_{k-1}...a_0}$ so the claim is that $\bmod \color{#c00}{5^k}!:\ 10^k q + r\equiv r,,$ true by $10^k = \color{#c00}{5^k} 2^k\equiv 0,$ by linked congruence laws. – Bill Dubuque Apr 12 '23 at 19:40
  • Or we can view it as a reduction of a divisibility test in radix (base) $b=10^k,,$ i.e. an integer $n$ in radix $b$ is congruent to it "units" digits $,= n\bmod b,,$ so the congruence remains true mod any divisor of $b$, just like we can compute $,n\bmod 5$ by first computing $,n\bmod 10 = u = $ units digit in base $10$ then reducing $u$ modul0 $5,,$ see the method of simpler multiples – Bill Dubuque Apr 12 '23 at 19:50
  • @BillDubuque very nice of you. Close my question and give me links to "similar" questions, whereas there were no even radixes and polynomial congruences yet in the textbook that I learn math with – curioushuman Apr 20 '23 at 10:41

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Your two numbers differ by a number ending in $k$ zeroes. Such number is a multiple of $10^k=5^k2^k$, hence it is divisible by $5^k$. Hence they coincide $\textrm{mod }5^k$.

GReyes
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