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Determine the unit’s digit (one’s digit) of the sum of the expression: \[ (1!)^3 + (2!)^3 + (3!)^3 + \cdots + (13!)^3 + (14!)^3 + (15!)^3 \]

${15! (15!+1) \div 2}^2$

Your solution is not correct; look, it asks for the last digit. It's actually a modular arithmetic exercise...

remember that every integer from $5!$.......... is divisible by 10.so just calculate the first four.

sorry Moon bhaia, i forgot to correct it....

It's like this-
As for every integer $n>5,10\mid n!$ so the last digit (suppose $a$) will be-
$a\equiv 1!+2!+3!+4!\equiv 1+2+6+4(mod 10)$
$a\equiv 3(mod 10)$
Or,$a=3$(as $a<10$)

shouldn't the solution be like this?
\[(1!)^3+(2!)^3+(3!)^3+(4!)^3=14049\equiv 9(mod10)\]
so the last digit should be 9.
am i correct?

Yes the answer is correct, but I could not do such multiplications in the hall, so here is another solution frm me

$(1!)^3\equiv 1(mod 10)$
$(2!)^3= 2^3\equiv 8(mod 10)$
$(3!)^3=6^3\equiv 6(mod 10)$( 6 last digit amon number er jekono power last digit 6)
$(4!)^3=24^3\equiv 4(mod 10)$( 4 last digit hole odd power e last digit hoy 4, even power e last digit hoy 6)
after these $(n!)^3\equiv 0(mod 10)$(last digit 0)
so we find the last digit = $19\equiv 9(mod10)$
the answer is $9$