When it is said to find the last digit of a number and the number is given in $a^b$ or $a^{b^c}$ format it is easy to find using either basic congruency or Fermat's little theorem or Euler's phi-function. But in the exam question is of this type $$1!+2!+….+99!$$ and I did it in this way
$1!=1\equiv 1$ $mod$ $10$
$2!=2 \equiv 2$ $mod$ $10$
$3!=6 \equiv 6$ $mod$ $10$
$4!=24 \equiv 4$ $mod$ $10$, and
$n!=n.(n-1).(n-2)….4.3.2.1 \equiv 0$ $mod$ $10$ for $n \geq5$
Hence adding up all the factorials and the remainders we get
$1!+2!+3!+4!+n! \equiv 1+2+6+4 \equiv 13 \equiv 3$ $mod$ $10$
Hence, $$1!+2!+….+99! \equiv 3\mod10$$
My question is should I use $\mod10$ here because no other conditions are given to solve it. The remainder we find that is the digit in unitary place of a number, I think. Any help is appreciated.
Best Answer
Yes, you are doing fine. To find the last digit, use $\mod{10}$
In general, to find the last $k$ digits, use $\mod{10^k}$.