Congruence problems

• May 10th 2009, 12:51 PM
Currybullen
Congruence problems
I got an assignment from my math teacher to solve some congruence problems, thing is, I have a really hard time understanding how to solve them (Worried). So, help from you guys would be greatly appreciated.

Here are problems:

1.

Prove that 3246(mod 11) = 3246(mod 3) + 3246(mod 5)

2.

Calculate and simplify:

1^3 + 2^3 + ... + 13^2 + 14^2 (mod 7)

Any = should be read as the congruence symbol, don't know how to type that one (Nerd).
• May 10th 2009, 05:26 PM
Soroban
Hello, Currybullen!

Quote:

2. Calculate and simplify: . $\displaystyle 1^3 + 2^3 + \hdots + 13^2 + 14^2 \text{ (mod 7)}$
We have:

. . $\displaystyle \begin{array}{ccc}1^3 &\equiv& 1^3 \\ 2^3 &\equiv&2^3 \\ 3^3 &\equiv&3^3 \\ 4^3 &\equiv& (\text{-}3)^3 \\ 5^3 &\equiv&(\text{-}2)^3 \\ 6^3 &\equiv& (\text{-}1)^3 \\ 7^3 &\equiv & 0^3 \end{array}\;\;\text{(mod 7)}$

$\displaystyle \text{Total} \;\;\:\equiv\quad0\:\;\;\text{(mod 7)}$

We find that the same is true for: .$\displaystyle 8^3 + 9^3 + \hdots + 14^3$

Therefore: .$\displaystyle 1^3 + 2^3 + 3^3 + \hdots + 14^3 \;\equiv\;0\;\text{ (mod 7)}$

• May 11th 2009, 10:13 AM
Currybullen
Thanks I think I grasp that one now. Anyone know about the first one? I'm not even sure it's formulated in the right way, our teacher can be a bit odd at some times.
• May 11th 2009, 10:34 AM
Media_Man
Well, $\displaystyle 3246 \equiv 1 (\bmod 11)$, $\displaystyle 3246 \equiv 0 (\bmod 3)$, $\displaystyle 3246 \equiv 1 (\bmod 5)$. $\displaystyle 1=0+1$. QED.