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In my class we were given a problem, and I am struggling to find the answer for it. Here's the problem:

(5000001 * 1468162901)^11 is congruent to x mod 33.

I first broke the problem into two numbers, where I did 5000001^11 mod 33 and 1468162901^11 mod 33. I broke the 11 using binary, making it 5000001^1, 5000001^2, and 5000001^8. Set them to mod 33 and solved and got 6, 3, 9. I multiplied them together, and set it with mod 33: 162 mod 33 and got 30. I did the same process for 1468162901 and got 17. I multiplied those together and did mod: 510 mod 33 = 15. The online grader says I am getting it wrong but I have no idea where I went wrong. If I am getting my process wrong then please let me know.

GrilledCatEggs
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  • Welcome to Math SE. I have only checked part of what you did. So far, while I also got that $5000001^1\equiv 6\pmod{33}$, $5000001^2\equiv 6^2\equiv 3\pmod{33}$, but I then got $5000001^8\equiv (5000001^2)^4\equiv 3^4\equiv 81\equiv 15\pmod{33}$, not $9$ as you did (where it seems you calculated $5000001^4\pmod{33}$ instead). Also, note there are faster and easier ways to get the answer, such as those explained in Modular exponentiation by hand ($a^b\bmod c$), as well as some of the various questions linked to it. – John Omielan Oct 19 '24 at 02:44
  • Casting $11$'s shows $\bmod 11!:\ x\equiv (6\cdot 7)^{11}\equiv 9^{11}\equiv \color{#c00}9,$ by Fermat. Casting $3$'s shows $,x\equiv (0\cdot\square)^{11}\equiv 0\equiv \color{#c00}9,,$ so $,x\equiv \color{#c00}9\pmod{3\cdot 11},$ by CCRT. $\ \ $ – Bill Dubuque Oct 19 '24 at 03:24
  • See the linked dupe for a similar example of using CRT and casting out $11$'s and $9$'s. $\ \ $ – Bill Dubuque Oct 19 '24 at 03:36
  • e.g. casting $11$'s by computing the alternating digit sum yields $$,1468162901 \equiv -1+4-6+8-1+6-2+9-0+1\equiv 7!!\pmod{!11}\qquad$$ There are many web calculators you can use to check your modular arithmetic, e.g. Wolfram Alpha. This site is not meant to be used to perform arithmetic checks. $\ \ $ – Bill Dubuque Oct 19 '24 at 03:46

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