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This problem was on a math competition on a no calculator section. How can it be solved without a calculator? The problem is to find the remainder of : $\frac{10^9+23^9}{11}$.

Bill Dubuque
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Python_coder
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    What level competition was this? Are you sure you weren't asked simply to show that this was an integer? – lulu Apr 29 '22 at 23:28
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    Ah, the remainder... that makes a huge difference... :) – paul garrett Apr 29 '22 at 23:35
  • Normally, the term remainder refers to the smallest non-negative integer that is congruent to the original number, based on the pertinent modulus. It is often helpful, when (for example) working with a modulus like $11$, to consider the remainders (AKA congruence classes) of $S = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}.$ If $a \equiv b\pmod{c}$, then $a^n \equiv b^n \pmod{c}.$ Ask yourself, what element $(a)$ from the set $S$ is such that $10 \equiv a\pmod{11}$? Similarly, what element element $b$ from the set $S$ is such that $23 \equiv b\pmod{11}$? – user2661923 Apr 29 '22 at 23:37
  • $10^9=(11-1)^9$, $23^9=(22+1)^9$. – lhf Apr 29 '22 at 23:38
  • $!\bmod 11!:,\ \color{#c00}{10\equiv -1},, \color{#0a0}{23\equiv 1}\Rightarrow \color{#c00}{10}^9 +\color{#0a0}{23}^9\equiv \color{#c00}{-1}^9+\color{#0a0}1^9\equiv -1+1\equiv 0,$ by basic congruence arithmetic rules in the linked dupe. – Bill Dubuque Apr 30 '22 at 01:52

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Modular arithmetic is certainly one option, but for those who may not be too well versed in it, another option is simply using the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$: $$10^9+23^9=(10^3)^3+(23^3)^3=(10^6-10^3\times23^3+23^6)(10^3+23^3)$$ $$=(10^6-10^3\times23^3+23^6)(10^2-10\times23+23^2)(10+23)$$ $$=33(10^6-10^3\times23^3+23^6)(10^2-10\times23+23^2)$$ Since the number has a factor of 33 in it, dividing it by 11 will give a 0 remainder.

person
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You can use modular arithmetic. Essentially instead of evaluating the exponents directly, you can discard most of the information at each step. $$10^2\equiv 100\equiv 1+9\cdot11\equiv 1\pmod{11}$$ Thus, you can see that $$10^9\equiv 10(10^8)\equiv 10(10^2)^4\equiv 10(1^4)\equiv 10 \pmod{11}$$ And since $$23=1+2\cdot 11$$ $$23^9\equiv 1^9\equiv 1\pmod{11}$$ And so the remainder of the division is $10^9+23^9\pmod{11}=0$

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