JEE Mathematics 2023 Question with Solution

Given: We need the remainder when $19^{200} + 23^{200}$ is divided by $49$.

Find: The remainder modulo $49$.

Write

$$19^{200} + 23^{200} = (21 - 2)^{200} + (21 + 2)^{200}$$

Using binomial expansion, odd-powered terms cancel and even-powered terms remain:

$$(21 + 2)^{200} + (21 - 2)^{200} = 2\left[\binom{200}{0}21^{200} + \binom{200}{2}21^{198}2^2 + \cdots + \binom{200}{198}21^2 2^{198} + 2^{200}\right]$$

Every term inside the bracket except $2^{200}$ contains $21^2 = 441 = 49 \times 9$, so those terms are divisible by $49$. Hence,

$$(21 + 2)^{200} + (21 - 2)^{200} = 2[49I_1 + 2^{200}] = 49I_1 + 2^{201}$$

for some integer $I_1$. Now,

$$2^{201} = (8)^{67} = (1 + 7)^{67}$$

Again by binomial expansion,

$$(1+7)^{67} = 49I_2 + \binom{67}{0} + \binom{67}{1}\cdot 7 = 49I_2 + 1 + 469 = 49I_2 + 470$$

So,

$$2^{201} = 49I_2 + 470 = 49I_2 + 49 \times 9 + 29$$

Therefore the remainder on division by $49$ is $29$.

Given: $19 = 21-2$ and $23 = 21+2$.

Find: $19^{200} + 23^{200} \pmod{49}$.

Use symmetry:

$$(21+2)^{200} + (21-2)^{200}$$

All odd-power terms of $2$ cancel. Every surviving term except the last contains a factor $21^2$, and since $21^2 = 441$ is divisible by $49$, all such terms vanish modulo $49$. Thus,

$$19^{200} + 23^{200} \equiv 2^{201} \pmod{49}$$

Now,

$$2^{201} = 8^{67} = (1+7)^{67}$$

Modulo $49$, only the first two terms matter:

$$(1+7)^{67} \equiv 1 + 67\cdot 7 = 470 \equiv 29 \pmod{49}$$

Therefore, the remainder is $29$.