IPMAT Indore 2026 · Quantitative Ability · Question 3
IPMAT Indore 2026 previous year question on Remainder Theorem, with the answer and a step-by-step solution.
The remainder when {7}^{103}+{7}^{101} is divided by 9 is ____
Answer & solution
Answer: 2
Step 1 — factor out the smaller power.
{7}^{103}+{7}^{101}={7}^{101}\left({7}^{2}+1 \right)={7}^{101}\times 50
Step 2 — find the cycle of 7 modulo 9.
{7}^{1}\equiv 7,{7}^{2}\equiv 49\equiv 4,{7}^{3}\equiv 28\equiv 1 (\mod 9)
The powers repeat every 3 steps.
Step 3 — reduce the exponent. 101=3\times 33+2, so
{7}^{101}\equiv{7}^{2}\equiv 4 (\mod 9)
Step 4 — handle the other factor. 50=45+5\equiv 5 (\mod 9).
Step 5 — multiply.
{7}^{101}\times 50\equiv 4\times 5=20\equiv 2 (\mod 9)
Answer: 2
Shortcut. Factoring first is what makes this quick — {7}^{101}\left({7}^{2}+1 \right) turns two large powers into one. And {7}^{3}\equiv 1 (\mod 9) is worth memorising: it collapses any power of 7 to one of three values.
More Remainder Theorem questions
- Q2The remainder when 1!+2!+3!+cdots +95! is divided by 15 is _____IPMAT Indore 2023 · Quantitative Ability · Remainder Theorem
- Q40The remainder when 11¹⁰¹¹+1011¹¹ is divided by 9 isIPMAT Indore 2025 · Quantitative Ability · Remainder Theorem
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