JEE MainMathematics

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Finding Remainder of Large Powers Modulo a Prime

Q: Why choose 5^5 instead of 5^2 or 5^3?

5^5 = 3125 leaves a remainder of +1 modulo 11, making binomial expansion especially simple since (11λ + 1)^n = 11K + 1.

Q: Why can we reduce the exponent modulo 10?

Because for any prime p, the multiplicative group modulo p has order p-1 = 10, so powers repeat every 10.

The remainder on dividing 5^99 by 11 is:

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

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Why choose

5^{5}

instead of

5^{2}

5^{3}

$5^{5} = 3125$ leaves a remainder of $+ 1$ modulo $11$ , making binomial expansion especially simple since $(11 λ + 1)^{n} = 11 K + 1$ .

Why can we reduce the exponent modulo 10?

Because for any prime $p$ , the multiplicative group modulo $p$ has order $p - 1 = 10$ , so powers repeat every 10.

Answer

The remainder when dividing $5^{99}$ by $11$ is $9$ .

Why each option works or fails

Step-by-step solution

Given: We are given the expression $5^{99}$ and the divisor $11$ .
Goal: Find the remainder when $5^{99}$ is divided by $11$ .
Approach: Express $5^{99}$ by factoring out powers such that a power of $5$ is close to a multiple of $11$ , specifically $5^{5} = 3125 = 3124 + 1$ where $3124 = 11 \times 284$ . Then expand using the Binomial Theorem.
Execute: Split the power: $5^{99} = 5^{4} \times 5^{95} = 625 \times (5^{5})^{19} = 625 \times (3125)^{19}$ . Notice $3125 = 3124 + 1 = 11 (284) + 1 = 11 λ + 1$ . Expanding $5^{99}$ yields $11 μ + 1$ . Thus, $5^{99} = 625 (11 μ + 1) = 625 \times 11 μ + 625$ . Divide $625$ by $11$ : $625 = 11 \times 56 + 9$ . Therefore, $5^{99} = 11 k + 9$ .
Verify: Using Fermat's Little Theorem: $5^{10} \equiv 1 (m o d 11)$ . Thus $5^{99} = (5^{10})^{9} \times 5^{9} \equiv 5^{9} (m o d 11)$ . Since $5^{2} = 25 \equiv 3$ , $5^{9} = (5^{2})^{4} \times 5 \equiv 3^{4} \times 5 = 81 \times 5 \equiv 4 \times 5 = 20 \equiv 9 (m o d 11)$ . Both routes yield 9.

Shortcut:

When to use it: Fastest in competitive exams using modular arithmetic and Fermat's Little Theorem.

Given: Expression $5^{99} (m o d 11)$ , with prime modulus $p = 11$ .

Goal: Find $5^{99} (m o d 11)$ .

Approach: By Fermat's Little Theorem, $a^{p - 1} \equiv 1 (m o d p)$ for $\gcd (a, p) = 1$ . Here, $5^{10} \equiv 1 (m o d 11)$ . Reduce the exponent modulo $10$ .

Execute: $99 = 10 \times 9 + 9$ , so $5^{99} \equiv (5^{10})^{9} \times 5^{9} \equiv 1^{9} \times 5^{9} \equiv 5^{9} (m o d 11)$ . Since $5^{2} = 25 \equiv 3 (m o d 11)$ , $(11 λ + 1)^{n} = 11 K + 1$ .

Verify: Alternatively, $5 \equiv - 6 (m o d 11)$ , $5^{5} = 3125 = 11 \times 284 + 1 \equiv 1 (m o d 11)$ . Then $5^{99} = 5^{4} \times (5^{5})^{19} \equiv 625 \times 1^{19} \equiv 9 (m o d 11)$ . Consistently 9.

Finding Remainder of Large Powers Modulo a Prime

What feels right?

The remainder when dividing $5^{99}$ by $11$ is $9$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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