+166 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 $13(3x-2)\equiv 26\pmod 8$?
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.
+985 What is the residue of 92010, modulo 17?
We can see the pattern:
91≡9(mod17)92≡13(mod17)93≡15(mod17)94≡16(mod17)95≡8(mod17)96≡4(mod17)97≡2(mod17)98≡1(mod17)
The residues then repeat after that.
Therefore, since 2010÷8=251 R 2, the residue is 13.
I hope this helped,
Gavin
+985 Let n be the integer such that 0≤n<31 and 3n≡1(mod31). What is (2n)3−2(mod31)? Express your answer as an integer from 0 to 30, inclusive.
3n≡1(mod31)3n⋅21≡1⋅21(mod31)63n≡21(mod31)62n+n≡21(mod31)n≡21(mod31)
If 0≤n<31 and 3n≡1(mod31), 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)≡26(mod8)?
Note: you usually can not divide in linear congruences, but since 13 is relatively prime to 8, you can.
13(3x−2)≡26(mod8)3x−2≡2(mod8)3x≡4(mod8)3x⋅3≡4⋅3(mod8)9x≡12(mod8)8x+x≡12(mod8)x≡12(mod8)
I hope this helped,
Gavin
