# modular arithmetics - remainder

I need to find the remainder of ${1011}^{10}+{10}^{11}$ when divided by $101$.

According to this website it is 55 but I fail to see how.

For 1011, we can write it as $1$ because $1011=10*101+1$ so ${1011}^{10}$ is $1$.

For ${10}^{11}$, I wrote the $11$ like $8+2+1$ or $2^3+2^1+2^0$

Then multiplying the results $10\cdot100\cdot1$ and taking the remainder of 101 I got 91. $$1000=9\cdot101+91$$ So overall I have 92 for the answer, but the website says 55. Was my method or understaning wrong or is there is a problem or limitation on large numbers in that calculator?

Also is there any website or mehtod to check myself with this kind of big numbers?

• You did fine. I would do the $10^{11}$ part as follows: $$10^2=100\equiv-1\pmod{101},$$ so $$10^{11}=(10^2)^5\cdot10\equiv(-1)^5\cdot10=-10\equiv91\pmod{101}.$$ Looks like that website doesn't deliver. Nov 9 '16 at 20:38
• Even simpler than what I did, thanks a lot for showing me this way and for the answer. Nov 9 '16 at 20:41

I don't think it's the web-site's fault. The input on that page was 1.1156078e30. You're not giving it the correct number $1011^{10} + 10^{11}$, which is 1115607835569227940375059334601, just the first $8$ decimal digits. All the digits matter here!

• I took whatever google gave as a result, with e+, so the mistake was indeed on my side, what did you use to get the full number? Nov 9 '16 at 20:49
• Example how to calculate full number codepad.org/5JsuDZ6z Nov 9 '16 at 20:52
• Oh codepad.org is indeed a nice website, thanks a lot. Nov 9 '16 at 20:55

$$1011^{10}+10^{11}\equiv (101\times 10+1)^{10}+10\times 100^5\equiv(0\times10 +1)^{10}+10\times(-1)^5\equiv1-10\equiv92\mod101$$

• sorry, -9 mod 101 is 92... Nov 9 '16 at 20:47
• @kotomord Fair point, I was correcting it... Sorry it is hard to write from my smarphone.
– user378947
Nov 9 '16 at 20:47