When $4^{101} + 6^{101}$ is divided by $25$ , the remainder is:

$20$
$10$
$5$
$0$

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Solution

The correct option is B
$10$

Step 1: Use binomial expansion formula

$4^{101} = {(5 - 1)}^{101} [∵ {(x - 1)}^{n} = {}^{n}{C_{0}} x^{n} - {}^{n}{C_{1}} x^{n - 1} + . . . - {}^{n}{C_{n}} x^{0}] = {}^{101}C_{0} 5^{101} - {}^{101}C_{1} 5^{100} + . . . - {}^{101}C_{101} 5^{0}$
Now,
$6^{101} = {(5 + 1)}^{101} [∵ {(x + 1)}^{n} = {}^{n}{C_{0}} x^{n} + {}^{n}{C_{1}} x^{n - 1} + . . . + {}^{n}{C_{n}} x^{0}] = {}^{101}C_{0} 5^{101} + {}^{101}C_{1} 5^{100} + . . . + {}^{101}C_{101} 5^{0}$

Step 2: Find remainder when $4^{101} + 6^{101}$ is divided by $25$

$4^{101} + 6^{101} = 2 [{}^{101}C_{0} 5^{101} + {}^{101}C_{2} 5^{99} + . . . + {}^{101}C_{98} 5^{3} + {}^{101}C_{100} 5^{1}] [A f t e r s u b s t r a c t i n g l i k e t e r m s] = 2 \times 5^{2} [{}^{101}C_{0} 5^{99} + {}^{101}C_{2} 5^{97} + . . . + {}^{101}C_{98} 5^{1}] + 2 \times {}^{101}C_{100} 5^{1} = 25 [2 ({}^{101}C_{0} 5^{99} + {}^{101}C_{2} 5^{97} + . . . + {}^{101}C_{98} 5^{1})] + 2 \times 101 \times 5 = 25 K + 1010$
Where, $k = ({}^{101}C_{0} 5^{99} + {}^{101}C_{2} 5^{97} + . . . + {}^{101}C_{98} 5^{1})$
Again,
$4^{101} + 6^{101} = 25 K + 25 \times 40 + 10 = 25 M + 10$
Where, $M = K + 40$
Therefore, the remainder will be $10$ .
The correct answer is option (B).

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