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How to show that $12^{12} + 9^{9}$ can be divided by $15$ using Binomial Theorem?

I don't know where to start, given the specific method needed. Since both powers are divisible by 3, I know that the sum is divisible by 3. I don't know how to show that the sum is also divisible by 5.

PRD
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  • Do you know Fermat's Little Theorem? – eyeballfrog Jan 24 '23 at 01:47
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    Try expressing $12^9=(10+2)^9$ and $9^{12}=(10-1)^{12}$. You'll see that @JoséCarlosSantos is correct. When divided by $5$, the sum has a remainder of $3$. – Robert Shore Jan 24 '23 at 01:49
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    The sum is *not* divisible by $5$. – José Carlos Santos Jan 24 '23 at 01:50
  • @eyeballfrog, good day. Familiar but I will try to learn how to use it. – PRD Jan 24 '23 at 01:51
  • @RobertShore Thanks for the comment. I just have to expand it and show that every addend is divisible by 5. Am I right? or there is a better way? – PRD Jan 24 '23 at 01:53
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    All but one of the addends will be divisible by $5$. – Robert Shore Jan 24 '23 at 01:54
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    On the other hand, $12^{12}+9^9$ *is* divisible by $5$ and by $15$ – J. W. Tanner Jan 24 '23 at 01:54
  • Thanks for the feedback, ladies, and gentlemen. I am sorry for the error on the exponenents. – PRD Jan 24 '23 at 01:57
  • **Hint** $\, $ Write it as: $\ (\color{#c00}{145}-1)^{\large 6} + (\color{#c00}{10}-1)^{\large 9}$ and note $5$ divides $\color{#c00}{145,\, 10}.\,$ Similarly, more generally we can show that $a$ divides $(ab-1)^{2n}+(ac-1)^{2k+1}\ \ $ – Bill Dubuque Jan 24 '23 at 02:25
  • Use the above hint along with the Binomial Theorem as [here](https://math.stackexchange.com/a/3013406/242) or as [here](https://math.stackexchange.com/a/3838498/242) in the linked dupes. It is better to learn modular arithmetic for problems like this, as mentioned in remarks there. – Bill Dubuque Jan 24 '23 at 02:39

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Clearly it's divisible by 3, it suffices to prove it's divisible by 5. Consider

$$12^{12}+9^9\equiv(10+2)^{12}+(10-1)^9\equiv 2^{12}+(-1)^9=4095\equiv0, \mod 5$$

IntegralLover
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