IPM PREP | APRIL 2026
Last Updated: April 2026
• Number Theory appears in every IPMAT Indore and JIPMAT paper — 5-8 questions
• HCF, LCM, remainders, and factors are the highest-frequency topics
• Correct approach can solve most problems in under 60 seconds
• Short-answer section in IPMAT Indore makes Number Theory even more critical
IPMAT Number Theory — Why It Matters
Number Theory is one of the most predictable and scoring areas in the IPMAT Quantitative Aptitude section. In both IPMAT Indore and JIPMAT, IPMAT number theory practice questions on HCF, LCM, divisibility rules, remainders, and factors appear regularly — contributing 5-8 questions per paper. Mastering these topics gives you quick, high-confidence points in the exam.
Unlike Data Interpretation or Geometry, Number Theory questions are largely formula-driven. With the right shortcuts and sufficient practice, you can solve most Number Theory questions in 45-90 seconds — making this a time-efficient scoring area.
Topic 1: HCF and LCM — Methods and Shortcuts
Key Formulas
• HCF (Highest Common Factor) = Product of common prime factors with lowest powers
• LCM (Lowest Common Multiple) = Product of all prime factors with highest powers
• HCF × LCM = Product of two numbers (for exactly TWO numbers only)
• HCF of fractions = HCF of numerators / LCM of denominators
• LCM of fractions = LCM of numerators / HCF of denominators
Solved Examples
Example 1: Find the HCF and LCM of 36 and 48.
Prime factorisation: 36 = 2² × 3² | 48 = 2⁴ × 3
HCF = 2² × 3¹ = 12 | LCM = 2⁴ × 3² = 144
Verification: 12 × 144 = 1728 = 36 × 48 ✓
Example 2: The LCM of two numbers is 120 and their HCF is 8. One number is 24. Find the other.
HCF × LCM = Product of numbers → 8 × 120 = 960
Other number = 960 / 24 = 40
Example 3: Three bells ring at intervals of 12, 15, and 20 minutes. They ring together at 9 AM. When will they ring together again?
LCM(12, 15, 20) = 60 minutes → They ring together at 10 AM.
Example 4: Find the largest number that divides 98, 160, and 200 leaving remainder 2 in each case.
Subtract remainders: 96, 158, 198. HCF(96, 158, 198) = 2. Answer = 2.
Topic 2: Divisibility Rules (2 through 13)
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit is 0, 2, 4, 6, or 8 | 246 is divisible by 2 |
| 3 | Sum of digits is divisible by 3 | 123: 1+2+3=6 ✓ |
| 4 | Last two digits form a number divisible by 4 | 1236: 36÷4=9 ✓ |
| 5 | Last digit is 0 or 5 | 345 ÷ 5 ✓ |
| 6 | Divisible by both 2 and 3 | 126: even + digit sum 9 ✓ |
| 7 | Double last digit, subtract from remaining; check divisibility by 7 | 161: 16 − 2 = 14 ✓ |
| 8 | Last three digits divisible by 8 | 1736: 736÷8=92 ✓ |
| 9 | Sum of digits divisible by 9 | 729: 7+2+9=18 ✓ |
| 10 | Last digit is 0 | 340 ÷ 10 ✓ |
| 11 | Alternating sum of digits = 0 or multiple of 11 | 121: 1−2+1=0 ✓ |
| 12 | Divisible by both 3 and 4 | 144 ✓ |
| 13 | Add 4× last digit to remaining; check if divisible by 13 | 117: 11 + 4×7=39=3×13 ✓ |
Solved Examples — Divisibility
Example 5: What is the smallest number greater than 100 that is divisible by both 6 and 9?
LCM(6, 9) = 18. Smallest multiple of 18 greater than 100 = 108.
Example 6: How many numbers between 1 and 200 are divisible by both 4 and 6?
LCM(4, 6) = 12. Numbers: 12, 24, 36… 192. Count = 192/12 = 16.
Topic 3: Remainders — Theorem and Shortcuts
• Remainder of (a + b) ÷ n = [Remainder(a÷n) + Remainder(b÷n)] ÷ n
• Remainder of (a × b) ÷ n = [Remainder(a÷n) × Remainder(b÷n)] ÷ n
• Cyclicity: Powers of 2 have unit digit cycle 2,4,8,6 (period 4)
• Cyclicity: Powers of 3 have unit digit cycle 3,9,7,1 (period 4)
• Cyclicity: Powers of 7 have unit digit cycle 7,9,3,1 (period 4)
• Fermat’s Little Theorem: a^(p-1) ≡ 1 (mod p) when p is prime and HCF(a,p)=1
Solved Examples — Remainders
Example 7: What is the remainder when 2^100 is divided by 7?
Powers of 2 mod 7: 2, 4, 1, 2, 4, 1… (cycle of 3)
100 = 33×3 + 1 → Remainder = 2¹ mod 7 = 2
Example 8: What is the unit digit of 7^53?
Cycle of unit digits of powers of 7: 7, 9, 3, 1, 7, 9… (period 4)
53 = 4×13 + 1 → Unit digit = 7
Example 9: A number when divided by 5 gives remainder 3. When divided by 7, gives remainder 5. Find the smallest such number.
Numbers with remainder 3 when divided by 5: 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, 63, 68…
Check which gives remainder 5 when divided by 7: 68 ÷ 7 = 9 remainder 5 ✓ → Answer: 68
Topic 4: Number of Factors
If N = p^a × q^b × r^c… then Number of factors = (a+1)(b+1)(c+1)…
Examples:
• 360 = 2³ × 3² × 5¹ → Factors = (3+1)(2+1)(1+1) = 4×3×2 = 24
• 100 = 2² × 5² → Factors = (2+1)(2+1) = 9
• Perfect square numbers always have an odd number of factors
Solved Examples — Factors
Example 10: How many factors does 720 have?
720 = 2⁴ × 3² × 5¹ → Factors = (4+1)(2+1)(1+1) = 5×3×2 = 30
Example 11: Find the number of even factors of 360.
360 = 2³ × 3² × 5¹. For even factors, at least one power of 2 must be included.
Even factors = (3)(2+1)(1+1) = 3×3×2 = 18 (power of 2 can be 1, 2, or 3)
Example 12: What is the smallest number with exactly 12 factors?
12 = 12×1 or 6×2 or 4×3 or 3×2×2 or 2×2×3
Try p^11 = 2^11 = 2048; p^5 × q = 2^5 × 3 = 96; p^3 × q^2 = 2^3 × 3^2 = 72; p^2 × q × r = 2^2 × 3 × 5 = 60 → Smallest = 60
Topic 5: Number Properties and Shortcuts
Unit Digit Patterns
| Base | Cycle of Unit Digits | Period |
|---|---|---|
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 4 | 4, 6 | 2 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
| 9 | 9, 1 | 2 |
| 0, 1, 5, 6 | Same digit always | 1 |
Solved Examples — Unit Digits
Example 13: Find the unit digit of 3^99 × 7^100.
3^99: 99 = 4×24 + 3, unit digit of 3³ = 7
7^100: 100 = 4×25, unit digit of 7^4 ends in 1
Unit digit of product = 7 × 1 = 7
Example 14: Sum of first 50 natural numbers = n(n+1)/2 = 50×51/2 = 1275
Example 15: Is 437 prime? Check divisibility by primes up to √437 ≈ 20.9 (2,3,5,7,11,13,17,19). 437 = 19 × 23 → NOT prime.
Common Mistakes to Avoid in IPMAT Number Theory
- Using HCF×LCM = product formula for more than two numbers (it only works for exactly two numbers)
- Not checking if a “remainder” question is asking for Chinese Remainder Theorem setup
- Forgetting the cyclicity period when computing unit digits of large powers
- Confusing “number of factors” with “sum of factors” (different formulas)
- Not simplifying fractions before applying HCF/LCM rules
Practice Quiz — IPMAT Number Theory
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Frequently Asked Questions
How many questions appear from Number Theory in IPMAT?
Number Theory typically contributes 5-8 questions in the IPMAT Quantitative Aptitude section across both IPMAT Indore and JIPMAT. The most frequently tested topics are HCF and LCM (word problems), remainders (cyclicity and Chinese Remainder Theorem), divisibility rules, and number of factors. Given the predictable nature of these questions, Number Theory is considered one of the highest-return areas to prepare.
What is the fastest way to find HCF of two large numbers in IPMAT?
The Euclidean Algorithm is the fastest method. Divide the larger number by the smaller, then divide the smaller by the remainder, and keep repeating until the remainder is 0. The last non-zero remainder is the HCF. For example: HCF(48, 72) — 72 = 1×48 + 24, then 48 = 2×24 + 0, so HCF = 24. This takes under 30 seconds for most IPMAT-level numbers.
How do I find unit digits of large powers quickly in IPMAT?
Use cyclicity rules. Find the unit digit of the base and identify its cycle period. For bases ending in 2, 3, 7, 8 — the cycle length is 4. Divide the power by 4 and use the remainder to identify which position in the cycle applies. If remainder is 0, use the 4th position. For example, unit digit of 7^35: 35 mod 4 = 3, so use 3rd position in cycle (7, 9, 3, 1) = 3. Answer: 3.
Are Number Theory questions in IPMAT MCQ or short answer?
In IPMAT Indore, Number Theory questions appear in both the MCQ section (Section 1 — 40 questions) and the Short Answer section (Section 3 — 20 questions where you type the exact answer). JIPMAT has only MCQs. For the short answer section, precision matters — there is no partial credit and no negative marking for short answer questions in IPMAT Indore.
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