I got this question in a maths competition. the question was that what was the remainder of the equation given below when divided by 10.

$\displaystyle \sum_{n=1}^{10} (n^2+3n+1)n!$

i personally do not what to do with the n!

im lost (Doh)

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- Jul 24th 2010, 02:08 AMswanzwhat is the remainder of the summation when divided by ten
I got this question in a maths competition. the question was that what was the remainder of the equation given below when divided by 10.

$\displaystyle \sum_{n=1}^{10} (n^2+3n+1)n!$

i personally do not what to do with the n!

im lost (Doh) - Jul 24th 2010, 02:33 AMDinkydoe
Hee swanz,

Do you know how to do modulo-calculations? That is, how to add and multiply residu-classes modulo 10?

This problem is actually quite simple if you can do this and make one simple observation. Namely, $\displaystyle n!\equiv 0$ mod 10 for all $\displaystyle n\geq 5$ - Jul 24th 2010, 02:35 AMDefunkt
Start by noting that for any $\displaystyle n \geq 5, ~ n! \equiv 0 ~ (mod 10)$.

Ah, Dinkydoe got to it before I did... :o! - Jul 24th 2010, 02:39 AMswanz
no i am in kindda in high school still lol

sorry =/ but u can tell me whateva i need to learn to solve this thereby i can learn some of it online,

yes i knw n! mod 10 = 0

yes i did make that observation that n! in mod 10 is 0 for all values of n more than equal to 5...but still lost :L - Jul 24th 2010, 02:43 AMswanz
sorry ....i got it :L ...

i just thought there wud be a method going about this :L

thanx for the insight Defunkt and Dinkydoe - Jul 24th 2010, 02:49 AMDinkydoe
Do you actually know that insight has the following implication?

$\displaystyle \sum_{n=1}^{10}(n^2+3n+1)n!$ has the same remainder as $\displaystyle \sum_{n=1}^{4}(n^2+3n+1)n!$ when divided by 10 - Jul 24th 2010, 02:51 AMswanz
yup got that bit ....dunno wat hppned b4... brain freezed i guess :L

thanx a bunch Dinkydoe