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Binomial Theorem: JEE Main Mathematics Question with Solution

Q: Why can we split modulo 35 into modulo 5 and modulo 7?

Because 5 and 7 are coprime, knowing the remainder modulo 5 and modulo 7 uniquely determines the remainder modulo 35 by the Chinese Remainder Theorem.

The remainder when

(2023)^{2023}

is divided by 35 is

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Step-by-step solutionView

Correct answer

The remainder when $(2023)^{2023}$ is divided by 35 is 7.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

We need to find the remainder when $N = (2023)^{2023}$ is divided by $35$ .

02goal

Find an integer $R \in \{0, 1, \dots, 34\}$ such that $N \equiv R \pmod{35}$ .

03approach

Since $35 = 5 \times 7$ and $\gcd(5, 7) = 1$ , evaluate $N \pmod 7$ and $N \pmod 5$ separately, then combine the modular conditions.

04execute

Check divisibility of $2023$ by $7$ : $2023 = 7 \times 289$ . Thus, $2023 \equiv 0 \pmod 7$ , which implies: $N = (2023)^{2023} \equiv 0 \pmod 7$ .

05execute

Now compute $N \pmod 5$ : $2023 \equiv 3 \equiv -2 \pmod 5$ . So $N = (2023)^{2023} \equiv (-2)^{2023} \pmod 5$ . $(-2)^{2023} = -2 \cdot ((-2)^2)^{1011} = -2 \cdot (4)^{1011} \equiv -2 \cdot (-1)^{1011} = -2 \cdot (-1) = 2 \pmod 5$ .

06execute

Find $R \in \{0, 1, \dots, 34\}$ satisfying $R \equiv 0 \pmod 7$ and $R \equiv 2 \pmod 5$ . The multiples of $7$ in the range $[0, 34]$ are $0, 7, 14, 21, 28$ . Testing modulo $5$ : $0 \equiv 0 \pmod 5$ $7 \equiv 2 \pmod 5$ Hence, $R = 7$ .

✓verify

Check $7 \pmod 7 = 0$ and $7 \pmod 5 = 2$ . Both match the derived remainders uniquely in the range $0 \le R < 35$ .

Hints that build this answer step by step

Since 35 factors into the coprime pair $5 \times 7$ , what is the best strategy to compute $2023^{2023} \pmod{35}$ ?
Compute the remainder modulo 5 and modulo 7 separately, then combine them using the Chinese Remainder Theorem.
What is the remainder of $2023^{2023}$ modulo 7?
0
What is the remainder of $2023^{2023}$ modulo 5?
2
Which integer between 0 and 34 satisfies $x \equiv 0 \pmod 7$ and $x \equiv 2 \pmod 5$ ?
7

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we split modulo 35 into modulo 5 and modulo 7?

Because 5 and 7 are coprime, knowing the remainder modulo 5 and modulo 7 uniquely determines the remainder modulo 35 by the Chinese Remainder Theorem.

Answer

The remainder when $(2023)^{2023}$ is divided by 35 is 7.

Why each option works or fails

Step-by-step solution

given: We need to find the remainder when $N = (2023)^{2023}$ is divided by $35$ .
goal: Find an integer $R \in \{0, 1, \dots, 34\}$ such that $N \equiv R \pmod{35}$ .
approach: Since $35 = 5 \times 7$ and $\gcd(5, 7) = 1$ , evaluate $N \pmod 7$ and $N \pmod 5$ separately, then combine the modular conditions.
execute: Check divisibility of $2023$ by $7$ : $2023 = 7 \times 289$ . Thus, $2023 \equiv 0 \pmod 7$ , which implies: $N = (2023)^{2023} \equiv 0 \pmod 7$ .
execute: Now compute $N \pmod 5$ : $2023 \equiv 3 \equiv -2 \pmod 5$ . So $N = (2023)^{2023} \equiv (-2)^{2023} \pmod 5$ . $(-2)^{2023} = -2 \cdot ((-2)^2)^{1011} = -2 \cdot (4)^{1011} \equiv -2 \cdot (-1)^{1011} = -2 \cdot (-1) = 2 \pmod 5$ .
execute: Find $R \in \{0, 1, \dots, 34\}$ satisfying $R \equiv 0 \pmod 7$ and $R \equiv 2 \pmod 5$ . The multiples of $7$ in the range $[0, 34]$ are $0, 7, 14, 21, 28$ . Testing modulo $5$ : $0 \equiv 0 \pmod 5$ $7 \equiv 2 \pmod 5$ Hence, $R = 7$ .
verify: Check $7 \pmod 7 = 0$ and $7 \pmod 5 = 2$ . Both match the derived remainders uniquely in the range $0 \le R < 35$ .

Binomial Theorem: JEE Main Mathematics Question with Solution

What feels right?

The remainder when $(2023)^{2023}$ is divided by 35 is 7.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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