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

Question

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.

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
@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
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 or as here 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

score 1 · Answer 1 · answered Jan 24 '23 at 01:58

1

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

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

answered Jan 24 '23 at 01:58

IntegralLover

382

note that $2^4\equiv1\pmod5$, so $2^{12}\equiv1\pmod5$ – J. W. Tanner Jan 24 '23 at 02:00
Yep there are too many ways. I used $2^{12}=4096$ too frequently :D – IntegralLover Jan 24 '23 at 02:01
Easier: $,12^{12} = (145-1)^6\overset{\rm BT} \equiv (-1)^6\equiv 1\pmod{! 5}.,$ It's easier to use mod arithmetic (e.g. Congruence Power Rule) rather than BT = Binomial Theorem. – Bill Dubuque Jan 24 '23 at 02:13

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

1 Answers1