JEE Mathematics 2023 Question with Solution

Given: Find the remainder when $(2023)^{2023}$ is divided by $35$.

Find: The remainder modulo $35$.

Using modular arithmetic,

$$2023 = 2030 - 7 \implies 2023 \equiv -7 \pmod{35}$$

Thus,

$$(2023)^{2023} \equiv (-7)^{2023} \pmod{35}$$

Also,

$$(-7)^{2023} = -7 \cdot 7^{2022} = -7 \cdot (7^2)^{1011}$$

Now,

$$7^2 = 49 = 50 - 1$$

So,

$$-7 \cdot (7^2)^{1011} = -7(50-1)^{1011}$$

Using binomial expansion,

$$(50-1)^{1011} = {}^{1011}C_0 50^{1011} - {}^{1011}C_1 50^{1010} + \cdots + {}^{1011}C_{1011}$$

Hence this can be written as

$$(50-1)^{1011} = 5\lambda - 1$$

for some integer $\lambda$. Therefore,

$$-7(50-1)^{1011} = -7(5\lambda - 1) = -35\lambda + 7$$

So the number is of the form $35N + 7$ for some integer $N$.

Therefore, when $(2023)^{2023}$ is divided by $35$, the remainder is $7$.

Given: $2023 = 35K - 7$ for some integer $K$.

Find: The remainder of $(2023)^{2023}$ upon division by $35$.

Write

$$(2023)^{2023} = (35K - 7)^{2023}$$

Expanding by the binomial theorem,

$$(35K - 7)^{2023} = {}^{2023}C_0 (35K)^{2023}(-7)^0 + {}^{2023}C_1 (35K)^{2022}(-7) + \cdots + {}^{2023}C_{2022}(35K)(-7)^{2022} + (-7)^{2023}$$

Every term except the last contains a factor of $35$, so all those terms are divisible by $35$. Hence,

$$(2023)^{2023} = 35N - 7^{2023}$$

for some integer $N$.

Now,

$$-7^{2023} = -7(49)^{1011} = -7(50-1)^{1011}$$

and from the expansion of $(50-1)^{1011}$, all terms except the constant term are multiples of $5$. Thus,

$$(50-1)^{1011} = 5\lambda - 1$$

for some integer $\lambda$. Therefore,

$$-7^{2023} = -7(5\lambda - 1) = -35\lambda + 7$$

So,

$$(2023)^{2023} = 35M + 7$$

for some integer $M$.

Therefore, the required remainder is $7$.