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.
Answer
Yes of course even if in this case is very simply as you noticed since
$$1!+2!+....+99!= 1!+2!+3!+4!+10\cdot N = 3 +10\cdot M$$
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