The 50th root of a number x is 12 and 50th root of another number y is 18.

Question

The 50th root of a number $x$ is 12 and 50th root of another number $y$ is 18. Then the remainder obtained on dividing $(x+y)$ by 25 is \_\_\_\_\_\_\_.

Answer: 23

Accepted Answer

$x=12^{50},\ y=18^{50}$.

$$x+y=12^{50}+18^{50}=(150-6)^{25}+(325-1)^{25}$$

Wait, more directly: $12^{50}=(150-6)^{25}$... using the source's approach: $12^{50}+18^{50}=25K-\left(6^{25}+1\right)=25K-\left((5+1)^{25}+1\right)$

Expanding $(5+1)^{25}=25K_2+1$ (mod 25, only constant term $1$ survives), so

$$x+y=25K_1-\big[(25K_2+1)+1\big]=25K-2$$

Remainder = $25-2=23$

Solutions · 1