2021/044) Prove that $13^{99} - 19^{93}$ is divisible by 162

we have $162= 2 * 81 =2 * 3^4$

So we need to show that the given expression is divisible by 2 and 81

Divisible by 2 is simple and both terms are odd so the difference is even so divisible by 2

Now let us find the value of expression mod 81

To find the same let us evaluate each term mod 81

$13^{99}= (12+1)^{99} = \sum_{n=0}^{99}{99 \choose n}12^n 1^{99-n}$

for n = 4 to 99 there is power of $12^n$ so each of the terms is divisible by $3^4$ or 81

so we have $13^{99} \equiv 1 + 99 * 12 + \frac{99 *98}{2} * 12^2 +   \frac{99 *98 * 96 }{6} * 12^3 \pmod {81}cdots(1)$

out of above the 3rd term onwards divisible by 81( as it contains 99 so one 9 is there so we require another 9 that comes from $12^2$ so taking mod 81 all the terms from 3rd term onwards is zero

So

$13^{99} \equiv 1 + 99 * 12 =  1+ (81 * 18) * 12 \equiv 1 + 18 * 12 \equiv 217 \equiv 55 \pmod {81}$

Now

$19^{93}= (18+1)^{93} = \sum_{n=0}^{93}{93 \choose n}18^n 1^{93-n}= 1 + 18 * 92 + \sum_{n=2}^{93}{93 \choose n}18^n 1^{93-n}$

except 1st 2 terms all are divisible by 81 so we have

$19^{93}\equiv 1 + 18 * 93 \equiv 1 + 18 * 12 \equiv 217 \equiv 55  \pmod {81}\cdots(2)$

from (1) and (2)

$13^{99} - 19 ^{93} \equiv 0 \pmod {81}$

as $13^{99} - 19 ^{93} \equiv 0 \pmod {2}$

from above 2 we get $13^{99} - 19 ^{93} \equiv 0 \pmod {162}$