Why choose instead of or ?

leaves a remainder of modulo , making binomial expansion especially simple since .

JEE MainMathematics

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Binomial Theorem: JEE Main Mathematics Question with Solution

Q: Why can we reduce the exponent modulo 10?

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

The remainder on dividing

5^{99}

11

is :

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $(11\lambda + 1)^n = 11K + 1$ .

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 $3125 = 3124 + 1 = 11(284) + 1 = 11\lambda + 1$ . Expanding $(11\lambda + 1)^{19}$ yields $5$ . Thus, $5^{99} = 625(11\mu + 1) = 625 \times 11\mu + 625$ . Divide $625$ by $11$ : $625 = 11 \times 56 + 9$ . Therefore, $5^{99} = 11k + 9$ .
verify: Using Fermat's Little Theorem: $5^{10} \equiv 1 \pmod{11}$ . Thus $5^{99} = (5^{10})^9 \times 5^9 \equiv 5^9 \pmod{11}$ . Since $5^2 = 25 \equiv 3$ , $5^9 = (5^2)^4 \times 5 \equiv 3^4 \times 5 = 81 \times 5 \equiv 4 \times 5 = 20 \equiv 9 \pmod{11}$ . Both routes yield 9.

Shortcut:

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

given: Expression $5^{99} \pmod{11}$ , with prime modulus $p = 11$ .

goal: Find $5^{99} \pmod{11}$ .

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

execute: $99 = 10 \times 9 + 9$ , so $5^{99} \equiv (5^{10})^9 \times 5^9 \equiv 1^9 \times 5^9 \equiv 5^9 \pmod{11}$ . Since $5^2 = 25 \equiv 3 \pmod{11}$ , $5^5$ .

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

Binomial Theorem: JEE Main Mathematics Question with Solution

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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