Question

Find the remainder when $7^{21}+7^{22}+7^{23}+7^{24}$ is divided by 25:
$\left(a\right)0$
$\left(b\right)2$
$\left(c\right)4$
$\left(d\right)6$

Answer 1

Hint: We need to find the remainder obtained when we divide the given sum by 25. For this, we will not calculate the actual value of $7^{21}, 7^{22}$ and so on. We would rather see the remainder obtained when we solve the small powers of 7 and see the remainder. And then we will do multiplication in the power to obtain the remainder for the bigger exponent powers of 7.

Complete step by step answer:
Observe that when we divide 7 by 25, we obtain 7 as the remainder.
If we divide $7^2$ by 25 i.e. dividing 49 by 25, we obtain 24 as the remainder.
Next, if we divide $7^3$ by 25 i.e. 343 by 25, we obtain 18 as the remainder.
Further, if we divide $7^4$ by 25 we get 1 as the remainder.
After this, the cycle will continue the same way i.e. $7^5$ divided by 25 will again give 7 and so on.
So, we have the powers $7^{21}$ which when divided by 25 will obtain 7 as the remainder.
Similarly, $7^{22}$ when divided by 25 will obtain 24 as the remainder.
And, $7^{23}$ which when divided by 25 will obtain 18 as the remainder.
Finally, $7^{24}$ which when divided by 25 will obtain 1 as the remainder.
If we sum up these remainders, we get $7+24+18+1=50$ which when divided by 25 gives 0 as the remainder. Hence the remainder obtained when we divide $7^{21}+7^{22}+7^{23}+7^{24}$ by 25 is 0.

So, the correct answer is “Option a”.

Note: We can also write the expression $7^{21}+7^{22}+7^{23}+7^{24}$ as $7^{21}\left(1+7+7^{2}+7^{3}\right)=7^{21}\left(400\right)$. We know that 400 is divisible by 25. So, the remainder obtained will be 0. So from this trick also, we get the same answer.