Question

$50^{\text {th }}$ root of a number $x$ is 12 and $50^{\text {th }}$ root of another number $y$ is 18 Then the remainder obtained on dividing $(x+y)$ by 25 is_____

Answer 1

Correct Answer: 23

$x+y=12^{50}+18^{50}=(150-6{)}^{25}+(325-1{)}^{25}$
$=25K-\left(6^{25}+1\right)=25K-\left((5+1{)}^{25}+1\right)$
$=25K_1-2$
Remainder $=23$
So , the correct answer is 23.

Answer 2

Given:
\(x = 12^{50}, \quad y = 18^{50}\)
We want to calculate \(x+y \mod 25\)
Calculate \(12^{50} \mod 25\):
\(12^2 = 144 \equiv 19 \mod 25\)
\(12^4 = (12^2)^2 \equiv 19^2 = 361 \equiv 11 \mod 25\)
\(12^8 = (12^4)^2 \equiv 11^2 = 121 \equiv 21 \mod 25\)
\(12^{16} = (12^8)^2 \equiv 21^2 = 441 \equiv 16 \mod 25\)
\(12^{32} = (12^{16})^2 \equiv 16^2 = 256 \equiv 6 \mod 25\)
\(12^{50} = 12^{32} \cdot 12^{16} \cdot 12^2 \equiv 6 \cdot 16 \cdot 19 \mod 25\)
\(6 \cdot 16 = 96 \equiv 21 \mod 25\)
\(21 \cdot 19 = 399 \equiv 24 \mod 25\)
So, \(12^{50} \equiv 24 \mod 25\)

Calculate \(18^{50} \mod 25\):
\(18^2 = 324 \equiv 24 \mod 25\)
\(18^4 = (18^2)^2 \equiv 24^2 = 576 \equiv 1 \mod 25\)
\(18^8 = (18^4)^2 \equiv 1^2 = 1 \mod 25\)
\(18^{16} = (18^8)^2 \equiv 1^2 = 1 \mod 25\)
\(18^{32} = (18^{16})^2 \equiv 1^2 = 1 \mod 25\)
\(18^{50} = 18^{32} \cdot 18^{16} \cdot 18^2 \equiv 1 \cdot 1 \cdot 24 \mod 25\)
\(1 \cdot 24 = 24 \mod 25\)
Therefore, \(x + y \equiv 24 + 24 \equiv 48 \equiv 23 \mod 25\)
So, the remainder obtained when x+y is divided by 25 is 23.

So, the answer is 23