Impossible Sigma

Number Theory Level 2

∑ k = 1 100 k × k ! \large \displaystyle\sum _{ k=1 }^{ 100 } k\times k!

Find the last digit of the number above.


The answer is 9.

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Lee Care Gene by Lee Care Gene , 20, Malaysia

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3 solutions

Kay Xspre
Jan 31, 2016

For each k k k ( k ! ) = ( ( k + 1 ) − 1 ) k ! = ( k + 1 ) ! − k ! k(k!) = ((k+1)-1)k! = (k+1)!-k! The sum above is then 101 ! − 1 101!-1 . Since every k ≥ 5 k≥5 would make k ! k! ends with zero, subtracting with 1 gives the unit digit 9 9

17 Helpful 0 Interesting 0 Brilliant 0 Confused
Ashwin K
Feb 1, 2016

This question has a simpler approach: 5! and above taking into consideration is meaningless as all the numbers has a '0' as last digit. Hence we need first 4 terms which are, 1+4+18+96 which has last digit as '9'.

8 Helpful 0 Interesting 0 Brilliant 0 Confused

EXACTLY. last digit of x! is 0 for any x>4

Nicuşoe Linguearu - 5 years, 4 months ago

From 5! and up we know the last digit is always zero, so you just sum k from 1 to 4 so you get the last digit.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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