Question

The last digit of $583!+7^{291}$ is:

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

Answer: (D)

$3$

The last digit of $583!$ would be $0$ as there is $10$ in the multiplication.

Now, for $7^{291}$, consider the last digits of smaller powers of $7$ which can simply be found by multiplication of the previous power's last digit with 7.

$7^1⟹7$

$7^2\left(49\right)⟹9$

$7^3\left(343\right)⟹3$

$7^2\left(2401\right)⟹1$

$7^5\left(16807\right)⟹7$

and so on the cycle continues for every set of 4 powers.

Therefore, $\frac{291}{4}$ leaves a remainder of $3$ and hence, the last digit of $7^{291}$ is same as that for $7^3$ i.e. $3$.