Question

Find the last digit of the product $7^{23}\times 8^{13}$.

Answer 1

Both 7 and 8 exhibit a cyclicity of 4 The last digits are
$7^1=7$ $8^1=8$
$7^2=9$ $8^2=4$
$7^3=3$ $8^3=2$
$7^4=1$ $8^4=6$
$7^5=7$ $8^5=8$
The cycle would repeat itself for higher powers $7^{23}$ ends with the same last digit as $7^3$ i.e.$3$.
$8^{13}$ ends with the same last digit as $8^1$ i.e. $8$.

Hence, the product of the two numbers would end with the same last digit as that of $3\times 8$ i.e. $4$.