Question

If $a$ is the digit in unit place of the sum $1!+2!+\cdots +39!,b$ is the digit in unit place of $(16{)}^{256}$ and $c$ is the digit in unit place of $(3{)}^{400},$ then $a+b+c$ is equal to

• A. $11$
• B. $10$
• C. $9$
• D. $12$

Answer 1

The correct option is B $10$

We know, for $n\ge 5,$
$n!$ ends with $0.$
Now, $1!+2!+3!+4!=33$
$\therefore a=3$

$(16{)}^n$ always ends with $6,\forall n\in N$
So, $b=6$

$3^{400}=(81{)}^{100}=(1+80{)}^{100}$
$=1+{}^{100}{C}_1\cdot 80+{}^{100}{C}_2\cdot {80}^2+\cdots +{}^{100}{C}_{100}\cdot {80}^{100}$
$=1+100k$ for some $k\in N$
$\therefore c=1$
Hence, $a+b+c=3+6+1=10$