what is the remainder of the summation when divided by ten

**swanz** · July 24th 2010, 02:08 AM

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.

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

i personally do not what to do with the n!
im lost

**Dinkydoe** · July 24th 2010, 02:33 AM

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, $n!\equiv 0$ mod 10 for all $n\geq 5$

**Defunkt** · July 24th 2010, 02:35 AM

Start by noting that for any $n \geq 5, ~ n! \equiv 0 ~ (mod 10)$ .

Ah, Dinkydoe got to it before I did...

!

**swanz** · July 24th 2010, 02:39 AM

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

**swanz** · July 24th 2010, 02:43 AM

sorry ....i got it :L ...
i just thought there wud be a method going about this :L
thanx for the insight Defunkt and Dinkydoe

**Dinkydoe** · July 24th 2010, 02:49 AM

Do you actually know that insight has the following implication?

$\sum_{n=1}^{10}(n^2+3n+1)n!$ has the same remainder as $\sum_{n=1}^{4}(n^2+3n+1)n!$ when divided by 10

**swanz** · July 24th 2010, 02:51 AM

yup got that bit ....dunno wat hppned b4... brain freezed i guess :L
thanx a bunch Dinkydoe

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