# find the remainder

• May 5th 2009, 07:15 PM
find the remainder
find the remainder when-

http://latex.codecogs.com/gif.latex?...10}(n^{2}+n)n!

is divided by 10.

No idea. (Thinking)

i just know the sum of n.n!=(n+1)!-1
• May 5th 2009, 07:25 PM
TheEmptySet
Quote:

find the remainder when-

http://latex.codecogs.com/gif.latex?...10}(n^{2}+n)n!

is divided by 10.

No idea. (Thinking)

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

Notice for $\displaystyle n \ge 5$ 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 $\displaystyle (2)1+(6)2+12(6)=86$ so the remainder is 6
• May 5th 2009, 07:43 PM
Quote:

Originally Posted by TheEmptySet
Notice for $\displaystyle n \ge 5$ 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 $\displaystyle (2)1+(6)2+12(6)=86$ so the remainder is 6

yes for the fourth term too we have 16+4=20 (hence, divisible by 10)

thank you. (Happy)