The 50^{text{th}} root of a number x is 12 an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $x^{1/50} = 12 \implies x = 12^{50}$ and $y^{1/50} = 18 \implies y = 18^{50}$.We need to find the remainder of $(12^{50} + 18^{50})$ divided

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(B) Given $x^{1/50} = 12 \implies x = 12^{50}$ and $y^{1/50} = 18 \implies y = 18^{50}$.We need to find the remainder of $(12^{50} + 18^{50})$ divided by $25$.$12^{50} = (12^2)^{25} = 144^{25} = (150 - 6)^{25}$.Using the Binomial Theorem,$(150 - 6)^{25} = \sum_{k=0}^{25} \binom{25}{k} (150)^k (-6)^{25-k} = 25M + (-6)^{25}$.$18^{50} = (18^2)^{25} = 324^{25} = (325 - 1)^{25}$.Using the Binomial Theorem,$(325 - 1)^{25} = \sum_{k=0}^{25} \binom{25}{k} (325)^k (-1)^{25-k} = 25N + (-1)^{25} = 25N - 1$.So,$x + y = 25M + (-6)^{25} + 25N - 1 = 25(M+N) - (6^{25} + 1)$.Now,$6^{25} = (6^2)^{12} \cdot 6 = 36^{12} \cdot 6 = (25 + 11)^{12} \cdot 6$.$(25 + 11)^{12} = 25P + 11^{12} = 25P + (121)^6 = 25P + (125 - 4)^6 = 25Q + (-4)^6 = 25Q + 4096$.$4096 = 25 	imes 163 + 21$.Thus,$6^{25} = (25K + 21) 	imes 6 = 150K + 126 = 25(6K + 5) + 1$.Therefore,$x + y = 25(M+N) - (25(6K+5) + 1 + 1) = 25R - 2$.Since the remainder must be positive,we add $25$ to $-2$,giving $25 - 2 = 23$.

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

Similar Questions

The remainder obtained when $5^{124}$ is divided by $124$ is

The expression $2^{4n} - 15n - 1$,where $n \in N$ (the set of natural numbers),is divisible by

The sum of the squares of three consecutive odd numbers increased by $1$ is divisible by-

The remainder when $10^{10} \cdot (10^{10} + 1) \cdot (10^{10} + 2)$ is divided by $6$ is

The remainder when $3^{100} \times 2^{50}$ is divided by $5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module