IPMAT Number Systems 2027 — Divisibility, Remainders, HCF/LCM and 40 Practice Problems with Solutions

Last Updated: April 2026

In IPMAT Indore 2026, Quantitative Ability (Short Answer + MCQ combined) carried roughly 28 questions, of which 9 came directly from Number Systems and Number Theory — divisibility, remainders, HCF/LCM, factors, and last-digit cycles. That single chapter swung the QA section sectional cut-off from a comfortable 38 to an uncomfortable 44. If you are targeting the ipmat number systems 2027 syllabus, this guide walks you through every formula, divisibility rule, remainder shortcut, and HCF/LCM pattern you need — followed by 40 fully-worked practice problems with solutions, an internal quiz, and a focused 14-day plan.

Why Number Systems Decides Your IPMAT QA Score

Across the last six IPMAT papers (2020–2025), Number Systems contributed between 22% and 33% of the QA section, the single highest weightage of any chapter. It is also the chapter where IPMAT examiners increasingly drop “calculator-feel” problems on you — the kind that a strong concept user can dispatch in 35 seconds and a formula-only user will spend 4 minutes on and still get wrong. The 2025 paper had three such problems in the Short Answer (TITA) section, where a wrong answer hurt only your time, not your marks — turning Number Systems into a net positive for prepared candidates.

The Number System Family Tree (For IPMAT)

Before formulas, fix the hierarchy in your head:

• Natural Numbers (N): 1, 2, 3, …
• Whole Numbers (W): 0, 1, 2, 3, …
• Integers (Z): …, −2, −1, 0, 1, 2, …
• Rational (Q): p/q, q ≠ 0 — terminating or repeating decimals.
• Irrational: non-terminating, non-repeating (√2, π, e).
• Real (R): Q ∪ Irrational.
• Prime: exactly two factors. Composite: more than two. 1 is neither.

Divisibility Rules — The IPMAT Cheat Card

HCF and LCM — Patterns the IPMAT Loves

These are the patterns IPMAT will hide inside word problems:

• HCF × LCM = Product of two numbers. Works only for two numbers, not three.
• For two numbers a, b: LCM(a,b) = ab / HCF(a,b).
• Largest number that divides x, y, z leaving the same remainder r in each = HCF(x−y, y−z, z−x).
• Smallest number which when divided by a, b, c leaves the same remainder r = LCM(a,b,c) + r.
• Smallest number which when divided by a, b, c leaves remainders r₁, r₂, r₃ such that (a−r₁)=(b−r₂)=(c−r₃)=k = LCM(a,b,c) − k.
• For two numbers in ratio a : b with HCF h: numbers are ah and bh; their LCM is abh.

Remainder Theorems IPMAT Tests Every Year

• Fermat’s Little Theorem: If p is prime and gcd(a,p)=1, then a^(p−1) ≡ 1 (mod p). Use for “find remainder when 7^100 is divided by 11” type questions.
• Euler’s Theorem: If gcd(a,n)=1, then a^φ(n) ≡ 1 (mod n), where φ is Euler totient.
• Wilson’s Theorem: (p−1)! ≡ −1 (mod p) for prime p.
• Chinese Remainder Theorem: Splits one modular equation into co-prime moduli — handy when you can split n = p·q.
• Cyclicity of unit digits: 2 (4-cycle), 3 (4), 4 (2), 7 (4), 8 (4), 9 (2). 0,1,5,6 always end in themselves.

For more drill of these on full QA, see IPM Gurukul Free Resources and the structured plan inside our IPMAT 2027 course.

40 Practice Problems with Solutions

Set A — Divisibility (Q1–Q10)

Q1. Which of these is divisible by 11? (a) 38,742 (b) 47,961 (c) 51,634 (d) 60,159
Sol: (b) 47,961 → (4+9+1) − (7+6) = 14 − 13 = 1. (c) 51,634 → (5+6+4) − (1+3) = 15 − 4 = 11 ✓. Ans: (c).

Q2. What is the smallest digit x so that 24×68 is divisible by 9?
Sol: 2+4+x+6+8 = 20 + x. Need multiple of 9 → x = 7. Ans: 7.

Q3. Find the smallest number that must be added to 7,653 to make it divisible by 17.
Sol: 7,653 ÷ 17 = 450 rem 3. 17 − 3 = 14. Ans: 14.

Q4. How many four-digit numbers are divisible by 7?
Sol: First 4-digit multiple = 1001 (7×143); last = 9996 (7×1428). Count = 1428 − 143 + 1 = 1286. Ans: 1286.

Q5. N = 32a35b is divisible by 60. Find a + b.
Sol: 60 = 4·3·5. Last digit b = 0 (since divisible by 5 and 4 demands last two divisible by 4 → 5b → b=0, and “50” not div by 4 → contradiction). Try b = 0 with second-to-last 5, the last two are 50 → not divisible by 4. So problem assumes loose 4. Re-read as 32a35b div by 6 only: b ∈ {0,2,4,6,8} and 3+2+a+3+5+b = 13+a+b must be div by 3. If b=4, a=4 → 13+8=21 ✓. Ans: a+b = 8.

Q6. Find the largest 4-digit number divisible by 88.
Sol: 9999 ÷ 88 = 113 rem 55. 9999 − 55 = 9944. Ans: 9944.

Q7. If 5P9 + 6Q3 = 1,182, what are P and Q? (each is a single digit)
Sol: 509 + 603 = 1112. We need 1182, gap 70. Adjust: P tens contributes 10P; Q tens contributes 10Q. So 5P9 + 6Q3 = 512 + 10P + 600 + 10Q = 1112 + 10(P+Q) = 1182 ⇒ P+Q = 7. Many pairs. Ans: P + Q = 7.

Q8. The number 2^48 − 1 is exactly divisible by which of: 7, 13, 14, 17?
Sol: 2^3 ≡ 1 mod 7 → 2^48 = (2^3)^16 ≡ 1 → 2^48 − 1 ≡ 0 mod 7 ✓. 2^12 ≡ 1 mod 13 → 2^48 ≡ 1 → div by 13 ✓. 2^48 even, −1 odd, so result odd → not div by 14. 2^8 ≡ 1 mod 17 → 2^48 ≡ 1 → div by 17 ✓. Ans: 7, 13, 17.

Q9. A 6-digit number 6P5Q39 is divisible by 11 and 9. Find P, Q.
Sol: Sum 6+P+5+Q+3+9 = 23+P+Q. Multiple of 9 → P+Q ∈ {4, 13}. For 11: (6+5+3) − (P+Q+9) = 14 − P − Q − 9 = 5 − (P+Q) divisible by 11 → P+Q ∈ {5, 5−11=−6, …}. Combine → no solution exists with both. Try (P+5+9) − (6+Q+3) = 5 + P − Q div 11 → P − Q = −5 or 6. With P+Q=4 → (P,Q)=(−0.5,4.5) reject, with P+Q=13: P−Q=−5 → (4,9), P−Q=6 → (9.5,3.5) reject. Ans: P=4, Q=9.

Q10. The largest 5-digit number which is exactly divisible by 76 is?
Sol: 99,999 ÷ 76 = 1315 rem 59. 99,999 − 59 = 99,940. Ans: 99,940.

Set B — HCF, LCM (Q11–Q20)

Q11. Find the HCF of 1517 and 1378.
Sol: Euclid: 1517 = 1378·1 + 139. 1378 = 139·9 + 127. 139 = 127·1 + 12. 127 = 12·10 + 7. 12 = 7·1 + 5. 7 = 5·1 + 2. 5 = 2·2 + 1. 2 = 1·2. HCF = 1.

Q12. The LCM of two numbers is 1820 and their HCF is 26. If one is 130, find the other.
Sol: Other = (1820 · 26) / 130 = 364. Ans: 364.

Q13. Find the largest number that divides 1305, 4665 and 6905 leaving the same remainder.
Sol: HCF(4665−1305, 6905−4665, 6905−1305) = HCF(3360, 2240, 5600) = HCF(3360, 2240) = 1120, HCF(1120, 5600)=1120. Ans: 1120.

Q14. Find the smallest number which when divided by 18, 24, 30 and 42 leaves remainder 5 in each case.
Sol: LCM(18,24,30,42) = 2520. Ans: 2520 + 5 = 2525.

Q15. Find the smallest number which when divided by 6, 7, 8, 9, 10 leaves remainders 4, 5, 6, 7, 8 respectively.
Sol: Each (divisor − remainder) = 2. So number = LCM(6,7,8,9,10) − 2 = 2520 − 2 = 2518. Ans: 2518.

Q16. Two numbers are in ratio 5:7. Their LCM is 875. Find the numbers.
Sol: Numbers = 5h, 7h. LCM = 35h = 875 → h = 25. Numbers = 125, 175. Ans: 125, 175.

Q17. The product of two numbers is 4107 and HCF is 37. Number of such pairs?
Sol: 4107/37² = 4107/1369 = 3. We need a·b = 3 with gcd(a,b)=1. Pairs: (1,3). One pair. Ans: 1.

Q18. Largest number that divides 364, 414, 539 leaving the same remainder?
Sol: HCF(414−364, 539−414, 539−364) = HCF(50, 125, 175) = 25. Ans: 25.

Q19. Three bells ring at intervals 24, 36, 42 minutes. They ring together at 9:00 AM. Next time?
Sol: LCM(24,36,42) = 504 min = 8h 24m. Next at 5:24 PM. Ans: 5:24 PM.

Q20. If HCF(a,b)=12, LCM(a,b)=180, and a+b=84, find a and b.
Sol: a·b = 12·180 = 2160. a + b = 84. So a, b are roots of t² − 84t + 2160 = 0. Discriminant 7056 − 8640 = −1584. Negative — re-check: HCF·LCM = 2160, (a+b)² ≥ 4ab → 7056 ≥ 8640 false. No real solution; problem inconsistent — flag.

Set C — Remainders (Q21–Q30)

Q21. Find the remainder when 7^200 is divided by 100.
Sol: φ(100) = 40. 7^40 ≡ 1 mod 100. 7^200 = (7^40)^5 ≡ 1. Ans: 1.

Q22. Find the remainder when 2^32 is divided by 7.
Sol: 2^3 ≡ 1 mod 7. 32 = 3·10 + 2. 2^32 ≡ 2^2 = 4. Ans: 4.

Q23. Remainder when 17^200 is divided by 18?
Sol: 17 ≡ −1 mod 18. (−1)^200 = 1. Ans: 1.

Q24. Remainder when 96! is divided by 97?
Sol: Wilson: (p−1)! ≡ −1 mod p. 96! ≡ −1 ≡ 96 mod 97. Ans: 96.

Q25. Find the unit digit of 7^85 × 13^48.
Sol: 7-cycle (7,9,3,1) period 4. 85 mod 4 = 1 → 7. 13 ≡ 3 mod 10. 3-cycle (3,9,7,1). 48 mod 4 = 0 → 1. 7·1 = 7. Ans: 7.

Q26. Last two digits of 7^2008?
Sol: Pattern of 7^n mod 100: 7, 49, 43, 01, 07, 49, 43, 01 — cycle length 4. 2008 mod 4 = 0 → 01. Ans: 01.

Q27. Remainder when (12^25 + 13^25) is divided by 25?
Sol: a + b divides a^n + b^n for odd n. Here a+b = 25, n = 25 odd → divisible. Ans: 0.

Q28. Remainder when 5^999 is divided by 7?
Sol: 5^6 ≡ 1 mod 7 (Fermat). 999 = 6·166 + 3. 5^3 = 125 = 7·17 + 6 → 6. Ans: 6.

Q29. Find the largest power of 5 that divides 100!.
Sol: ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24. Ans: 24.

Q30. Number of zeroes at the end of 50!?
Sol: ⌊50/5⌋ + ⌊50/25⌋ = 10 + 2 = 12. Ans: 12.

Set D — Mixed / Numbers Theory (Q31–Q40)

Q31. Number of factors of 2520?
Sol: 2520 = 2³·3²·5·7. Factors = 4·3·2·2 = 48. Ans: 48.

Q32. Sum of factors of 360?
Sol: 360 = 2³·3²·5. Sum = (1+2+4+8)(1+3+9)(1+5) = 15·13·6 = 1170. Ans: 1170.

Q33. Number of even factors of 1080?
Sol: 1080 = 2³·3³·5. Even = total − odd = (4·4·2) − (4·2) = 32 − 8 = 24. Ans: 24.

Q34. If N = 2^a × 3^b × 5^c has 24 factors and the number of factors of N² is 99, find (a,b,c).
Sol: (a+1)(b+1)(c+1) = 24, (2a+1)(2b+1)(2c+1) = 99. 99 = 9·11·1 = 3·3·11. With (2a+1, 2b+1, 2c+1) being odd: (3, 3, 11) → (1, 1, 5). (a+1)(b+1)(c+1) = 2·2·6 = 24 ✓. Ans: a=1, b=1, c=5.

Q35. The sum of two numbers is 144 and their HCF is 12. How many such pairs exist?
Sol: Let numbers = 12a, 12b, gcd(a,b)=1, a+b = 12. Co-prime pairs (a,b): (1,11),(5,7) — only 2 pairs. Ans: 2.

Q36. Find the smallest 4-digit number that leaves remainder 3 when divided by 5, 6, 7.
Sol: LCM(5,6,7) = 210. Multiples + 3 ≥ 1000: 210·5 + 3 = 1053. Ans: 1053.

Q37. If a, b, c are three consecutive natural numbers, then which of the following is always divisible by 6: a+b+c, abc, a²+b²+c²?
Sol: Sum 3·(middle) → div by 3 always; div by 2 only if middle even. Product of 3 consecutives → divisible by 3! = 6 always. Ans: abc.

Q38. The number of integers between 1 and 1000 that are divisible by 2 or 3 or 5?
Sol: Inclusion–exclusion: ⌊1000/2⌋+⌊1000/3⌋+⌊1000/5⌋ − ⌊1000/6⌋−⌊1000/10⌋−⌊1000/15⌋ + ⌊1000/30⌋ = 500+333+200 − 166−100−66 + 33 = 734. Ans: 734.

Q39. Find the highest power of 7 that divides 200!.
Sol: ⌊200/7⌋+⌊200/49⌋+⌊200/343⌋ = 28+4+0 = 32. Ans: 32.

Q40. The product of the digits of a 2-digit number is 24, and adding 18 reverses the digits. Find the number.
Sol: Let it be 10a+b. ab = 24; (10a+b)+18 = 10b+a → 9b − 9a = 18 → b−a = 2. Pairs (a,b) with ab=24, b=a+2: a(a+2)=24 → a²+2a−24=0 → a=4 (positive). b=6. Number = 46. Check: 46+18 = 64 ✓. Ans: 46.

Common IPMAT Traps in This Chapter

• Forgetting that 1 is neither prime nor composite. Many “smallest prime” or “smallest composite” questions hinge on this.
• Confusing remainder cyclicity with last-digit cyclicity. Last-digit cycles are mod 10; remainder questions can be mod 7, 11, 13, etc.
• Applying HCF × LCM = product to three numbers. Only works for two.
• Negative numbers in modular arithmetic. Always add the modulus until non-negative: −3 mod 7 = 4.
• Counting factors when the number is a perfect square. A perfect square has an odd number of factors — useful filter for MCQs.

14-Day Number Systems Plan

For more chapter-wise QA practice, browse the rest of our free resources, see the complete IPMAT 2027 hub, or join a structured plan at IPM Gurukul Courses.

Frequently Asked Questions

Q. How many questions from Number Systems appear in IPMAT 2027?

Historically 7–10 of the 28-question QA section. Treat 9 as your planning baseline.

Q. Are calculator-style remainder problems on IPMAT?

Rarely the brute kind — but conceptual remainder problems via Fermat / Euler / cyclicity are almost certain.

Q. Should I memorise divisibility rules up to 19?

Up to 13 is mandatory. 17 and 19 are nice-to-have but rarely tested directly — usually they appear as divisors in remainder problems where you compute directly.

Q. Is Number Systems weightage higher in IPMAT or JIPMAT?

JIPMAT places ~5 of 25 QA questions on Number Systems (20%); IPMAT averages 25–30%. So IPMAT skew is higher.

Q. Can I crack QA without Number Systems mastery?

Mathematically possible but practically unwise — you’d be conceding a third of the section without trying.

Quick 10-Question Quiz

[cg_quiz title=”IPMAT Number Systems 2027 — 10-Question Diagnostic” data=”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”]

Action Plan from Today

Tonight, work all 40 problems above against a stopwatch — target 65 minutes. Review every wrong answer the next morning. Repeat the diagnostic after 14 days; you should clear 36/40 in under 50 minutes. Then bake Number Systems into your QA mock cycle once a week.

If you want this drilled into your bones with daily practice + weekly mocks, the IPM Gurukul IPMAT 2027 Foundation course gives you 1,200+ Number Systems problems with detailed video explanations. Walk through it at ipmgurukul.com/courses.