What is the residue of 9^2010, modulo 17?

How many zeroes does 10! end in, when written in base 11?

Let $n$ be the integer such that $0 \le n < 31$ and $3n \equiv 1 \pmod{31}$. What is $\left(2^n\right)^3 - 2 \pmod{31}$? Express your answer as an integer from $0$ to $30$, inclusive.

What is the sum of all positive integer solutions less than or equal to $20$ to the congruence $13(3x-2)\equiv 26\pmod 8$?

Creeperhissboom May 27, 2018

#1**+1 **

**9^2010 mod 17 = 13**

10! base 10 =3,628,800 =205940a in base 11. So, there are no trailing zeros in base 11.

Sorry, can't read the other 2.

Guest May 27, 2018

#2**+2 **

What is the residue of \(9^{2010}\), modulo 17?

We can see the pattern:

\(9^1\equiv9\pmod{17}\\ 9^2\equiv13\pmod{17}\\ 9^3\equiv15\pmod{17}\\ 9^4\equiv16\pmod{17}\\ 9^5\equiv8\pmod{17}\\ 9^6\equiv4\pmod{17}\\ 9^7\equiv2\pmod{17}\\ 9^8\equiv1\pmod{17}\\\)

The residues then repeat after that.

Therefore, since \(2010÷8=251\ R\ 2\), the residue is 13.

I hope this helped,

Gavin

GYanggg May 27, 2018

#3**+2 **

Let n be the integer such that \(0\le n < 31\ \text{and}\ 3n \equiv 1 \pmod{31}.\) What is \(\left(2^n\right)^3 - 2 \pmod{31}\)? Express your answer as an integer from 0 to 30, inclusive.

\(3n\equiv1\pmod{31}\\ 3n\cdot21\equiv1\cdot21\pmod{31}\\ 63n\equiv21\pmod{31}\\ 62n+n\equiv21\pmod{31}\\ \boxed{n\equiv21\pmod{31}} \)

If \(0\le n < 31\ \text{and}\ 3n \equiv 1 \pmod{31},\) n = 21

You can finish this on your own.

What is the sum of all positive integer solutions less than or equal to 20 to the congruence \(13(3x-2)\equiv 26\pmod 8\)?

Note: you usually can not divide in linear congruences, but since 13 is relatively prime to 8, you can.

\(13(3x-2)\equiv 26\pmod 8\\ 3x-2\equiv2\pmod8\\ 3x\equiv4\pmod8\\ 3x\cdot3\equiv4\cdot3\pmod8\\ 9x\equiv12\pmod8\\ 8x+x\equiv12\pmod8\\ x\equiv12\pmod8 \)

I hope this helped,

Gavin

GYanggg May 27, 2018