Question

Find the last digit of the sum 0 ! + 2 ! + 4 ! + … + 2010 ! + 2012 !

Answer 1

For some whole number $n,$ the ones digit of $n!$ should be zero if atleast one among the positive integers less than or equal to $n$ is a multiple of 5.

Because,

→ that integer, less than $n$ or $n$ itself, can be an even multiple of 5, or multiple of 10, and that's enough to make the ones digit zero.

E.g.: 10 is enough to make ones digit of 10! zero.

→  if that integer is an odd multiple of 5, then its predecessor should be an even number, so the product of these two numbers is a multiple of 10 which makes ones digit of $n!$ zero.

E.g.: There is 5 as well as 4 in 5!, and their product is 20, which is enough to imply that 5! ≡ 0 (mod 10).

For a whole number $n,$ if there exists atleast one positive integer less than or equal to $n$ which is a multiple of 5, then undoubtedly $n\geq5.$

This implies the ones digit of $n!$ must be zero for every positive integer $n\geq5.$

In the sum $0!+2!+4!+\,\dots\,+2010!+2012!,$ the terms $6!,\ 8!,\ 10!,\,\dots\,,\ 2012!$ each end in 0 as they are greater than 5, so they do not make any change in the ones place of the given sum.

So the ones digit of the sum is determined by $0!,\ 2!$ and $4!$ only, and is equal to the ones digit of $0!+2!+4!.$

$\longrightarrow0!=1\equiv1\pmod{10}$

$\longrightarrow2!=2\equiv2\pmod{10}$

$\longrightarrow4!=24\equiv4\pmod{10}$

So,

$\longrightarrow0!+2!+4!\equiv1+2+4=\mathbf{7}\pmod{10}$

Hence the ones digit of the sum is 7.