Question

The remainder when $4^{87}+6^{87}$ is divided by $25$, is?

• A. $19$
• B. $20$
• C. $21$
• D. $22$

Answer 1

The correct option is A $19$

We know that Remainder of a number from $25$ will be the same as the remainder of the last two digits of the number from $25$.

$4^{87}+6^{87}$

Remainder of $\frac{4^{87}+6^{87}}{25}$

$=Lasttwodigitsof{4}^{87}+Lasttwodigitsof{6}^{87}$

$=Lasttwodigitsof[{\left(4^5\right)}^{17}\times 4^2]+Lasttwodigitsof[{\left(6^5\right)}^{17}\times 6^2]$

$=Lasttwodigitsof[{\left(1024\right)}^{17}\times 16]+Lasttwodigitsof[{\left(7776\right)}^{17}\times 36]$

$=(24\times 16)+(76\times 36)$

$=84+36$

$=120\left(mod100\right)$

$=20$