# Thread: fastest way to solve:

1. ## fastest way to solve:

Given $N = 10^{22} + 1$. If ‘N’ is divided by 11, then
Col A: The remainder
Col B: 2

This is a gre math Q. Would the fastest way to solve be to recognize that for every even power of 10, there is a remainder of 1 when product is divided by 11? Then the +1 on the end would make remainder 2. Any other ways to tackle this problem?

2. Originally Posted by sfspitfire23
Given $N = 10^{22} + 1$. If ‘N’ is divided by 11, then
Col A: The remainder
Col B: 2

This is a gre math Q. Would the fastest way to solve be to recognize that for every even power of 10, there is a remainder of 1 when product is divided by 11? Then the +1 on the end would make remainder 2. Any other ways to tackle this problem?
Fastest way is to do $10^{22}+1 \equiv (-1)^{22} + 1 \equiv 1 + 1 \equiv 2 \pmod{11}$. Are you familiar with this notation?

3. I am familiar with modular arithmetic...however, I don't understand how you got 10^22 to be (-1)^22

4. 10 is one less than 11. Since $11\equiv 0$ (mod 11), $10\equiv -1$ (mod 11).

Or, to put it another way, 10+ 1= 11: $10+ 1\equiv 11\equiv 0$ (mod 11).

In modulo 11, $10^{22}\equiv (-1)^{22}\equiv 1$ (mod 11).

5. Ah right. Now, could I then change the +1 to -10 as 1 is 10 less than 11?

(-1)^22+-10 is equivalent to 1-10 = -9 is equivalent to 2 (mod 11)? And thus, the remainder is 2? Can we conclude that from method?

6. Originally Posted by sfspitfire23
Ah right. Now, could I then change the +1 to -10 as 1 is 10 less than 11?

(-1)^22+-10 is equivalent to 1-10 = -9 is equivalent to 2 (mod 11)? And thus, the remainder is 2? Can we conclude that from method?
You could do so, but it adds an extra step.