IPMAT Indore 2025 · Quantitative Ability · Question 37
IPMAT Indore 2025 previous year question on Divisibility Rules, with the answer and a step-by-step solution.
- A{3}^{20}+4
- B{3}^{30}+2
- C{3}^{20}+8
- D{3}^{30}+8
Answer & solution
Answer: Option D
Step 1 — substitute to make it readable. Let
t={3}^{10}
Then the divisor is t+2, and the options become {t}^{2}+4, {t}^{3}+2, {t}^{2}+8, {t}^{3}+8.
Step 2 — recall the standard factorisation.
{t}^{3}+8={t}^{3}+{2}^{3}=\left( t+2 \right)\left({t}^{2}-2t+4 \right)
That is an exact factorisation, so t+2 divides {t}^{3}+8 with no remainder.
Step 3 — check the others fail. Using the Remainder Theorem (substitute t=-2):
Expression | Value at t=-2 | Divisible? |
|---|---|---|
{t}^{2}+4 | 4+4=8 | No |
{t}^{3}+2 | -8+2=-6 | No |
{t}^{2}+8 | 4+8=12 | No |
{t}^{3}+8 | -8+8=0 | Yes |
Since {3}^{30}+8={t}^{3}+8, that is the answer.
Answer: Option 4 — {3}^{30}+8
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