# Thread:

1. ## find the remainder

find the remainder when-

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

is divided by 10.

No idea.

i just know the sum of n.n!=(n+1)!-1

2. Originally Posted by adhyeta
find the remainder when-

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

is divided by 10.

No idea.

i just know the sum of n.n!=(n+1)!-1
Notice for $n \ge 5$ each term is divisable by 10 so they have a remainder of zero(why)

Now we only need to worry about the first 4 terms

We also don't need to worry about the 4 term (why)

so we get $(2)1+(6)2+12(6)=86$ so the remainder is 6

3. Originally Posted by TheEmptySet
Notice for $n \ge 5$ each term is divisable by 10 so they have a remainder of zero(why)

Now we only need to worry about the first 4 terms

We also don't need to worry about the 4 term (why)

so we get $(2)1+(6)2+12(6)=86$ so the remainder is 6
yes for the fourth term too we have 16+4=20 (hence, divisible by 10)

thank you.
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