Factors, Prime Factorisation and Number of Divisors
Factors, Prime Factorisation and Number of Divisors
Once a number is written as a product of prime powers, every factor question (how many, how many odd, how many even, their sum, the multiplier for a perfect square, trailing zeros) becomes one line of arithmetic. NTPC repeats these with new numbers every cycle.
1. Prime Factorisation: the Starting Point
Divide by primes in increasing order (2, 3, 5, 7, 11, ...) until 1 remains. Test primes only up to √N: if no prime up to √N divides N, then N is prime.
1,080 = 2³ × 3³ × 5 and 4,500 = 2² × 3² × 5³ and 9,408 = 2⁶ × 3 × 7²
Prime test example: is 401 prime? √401 ≈ 20, so test 2, 3, 5, 7, 11, 13, 17, 19. None divides 401, so 401 is prime. Numbers that look prime but are not (favourite NTPC options): 91 = 7 × 13, 119 = 7 × 17, 133 = 7 × 19, 143 = 11 × 13, 161 = 7 × 23, 187 = 11 × 17, 209 = 11 × 19, 221 = 13 × 17, 247 = 13 × 19, 253 = 11 × 23, 299 = 13 × 23, 323 = 17 × 19, 391 = 17 × 23.
2. Formula Box for N = aᵖ × bᵠ × cʳ (a, b, c distinct primes)
| What is asked | Formula | Example N = 360 = 2³ × 3² × 5 |
|---|---|---|
| Total number of factors | (p + 1)(q + 1)(r + 1) | 4 × 3 × 2 = 24 |
| Number of odd factors | Drop the power of 2, multiply the rest | 3 × 2 = 6 |
| Number of even factors | Total − odd, or p × (q + 1)(r + 1) when a = 2 and p is the power of 2 | 24 − 6 = 18 |
| Sum of all factors | [(a^(p+1) − 1)/(a − 1)] × [(b^(q+1) − 1)/(b − 1)] × ... | 15 × 13 × 6 = 1,170 |
| Product of all factors | N^(d/2), d = number of factors | 360¹² |
| Ways to write N as product of two factors | d/2 (if N is not a perfect square), (d + 1)/2 (if it is) | 24/2 = 12 |
| Factors that are perfect squares | Count choices of EVEN exponents for each prime | 2: {0, 2}, 3: {0, 2}, 5: {0} gives 4 |
| Factors divisible by a given number k | Fix the minimum exponents k needs, count the rest | Multiples of 6: 2 from {1,2,3}, 3 from {1,2}, 5 from {0,1} gives 12 |
| Total prime factors (with repetition) | p + q + r | 3 + 2 + 1 = 6 |
Factors come in pairs whose product is N. So for 60 the factors 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 pair up as (1, 60), (2, 30), (3, 20), (4, 15), (5, 12), (6, 10). A number has an odd count of factors only when it is a perfect square.
3. Perfect Square and Perfect Cube Makers
To make N a perfect square, every prime exponent must be even; for a cube, a multiple of 3. Least multiplier = product of primes needed to complete the exponents. Least divisor = product of the primes that have to be removed.
2,880 = 2⁶ × 3² × 5¹ → multiply by 5 → 14,400 = 120²
9,408 = 2⁶ × 3¹ × 7² → divide by 3 → 3,136 = 56²
4. Trailing Zeros and Highest Power of a Prime in n!
Each trailing zero needs one pair (2 × 5). In n! there are always more 2s than 5s, so:
Trailing zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...
Highest power of prime p in n! = ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + ...
Example: 250! has 50 + 10 + 2 = 62 zeros. Highest power of 3 in 50! = 16 + 5 + 1 = 22.
✗ Trailing zeros of any product = count of 5s | ✓ Count 2s and 5s separately and take the smaller; in a product such as 5 × 10 × 15 × ... × 500 the 2s can run out first
हिंदी नोट: किसी संख्या के गुणनखंडों की संख्या ज्ञात करने के लिए पहले उसे अभाज्य गुणनखंडों की घातों में लिखिए, फिर हर घात में 1 जोड़कर गुणा कीजिए। विषम गुणनखंड गिनते समय 2 की घात छोड़ दीजिए।
Exam Pointer: Verified NTPC shifts: total factors (23 July 2021), even factors of a prime-power product (14 March 2021), smallest divisor for a perfect cube (31 July 2021) and picking the prime from options (15 March 2021); the 2016 cycle asked the median of factors. The trap options are the count with 1 and N excluded, the distinct-prime count in place of total prime factors, and the count of 5s in place of min(2s, 5s).
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Number of factors and factors with a condition
[PYQ: NTPC CBT-1 23-Jul-2021 Shift-1]
EXAM LEVEL
Q. How many factors does 1,080 have?
1,080 = 2³ × 3³ × 5¹, so the count is (3 + 1)(3 + 1)(1 + 1) = 4 × 4 × 2 = 32.
Answer: 32
EXAMATLAS LEVEL
Q. How many factors of 4,500 are multiples of 15?
4,500 = 2² × 3² × 5³. A multiple of 15 needs at least 3¹ and 5¹. Choices: 2 from {0, 1, 2} gives 3 options, 3 from {1, 2} gives 2 options, 5 from {1, 2, 3} gives 3 options. Count = 3 × 2 × 3 = 18.
Answer: 18
Pattern 2: Odd, even and perfect-square factors
[PYQ: NTPC CBT-1 14-Mar-2021 Shift-1]
EXAM LEVEL
Q. Find the number of odd factors and even factors of 360.
360 = 2³ × 3² × 5. Total = 4 × 3 × 2 = 24. Odd factors ignore the 2: 3 × 2 = 6. Even = 24 − 6 = 18.
Answer: 6 odd and 18 even
EXAMATLAS LEVEL
Q. How many even factors of N = 2⁴ × 3³ × 7² are perfect squares?
A perfect-square factor needs every exponent even, and an even factor needs the exponent of 2 to be at least 1. So 2 can take {2, 4} (2 options), 3 can take {0, 2} (2 options), 7 can take {0, 2} (2 options). Count = 2 × 2 × 2 = 8. Including 2⁰ would wrongly add the odd squares.
Answer: 8
Pattern 3: Least multiplier or divisor for a perfect square or cube
[PYQ: NTPC CBT-1 31-Jul-2021 Shift-2]
EXAM LEVEL
Q. Find the smallest number by which 2,880 must be multiplied to make it a perfect square.
2,880 = 2⁶ × 3² × 5¹. Only 5 has an odd exponent, so multiply by 5. Result 14,400 = 120².
Answer: 5
EXAMATLAS LEVEL
Q. By what least number must 9,408 be divided so that the quotient is a perfect square, and what is the square root of that quotient?
9,408 = 2⁶ × 3¹ × 7². Only 3 has an odd exponent, so divide by 3. Quotient = 3,136 = 2⁶ × 7², square root = 2³ × 7 = 56.
Answer: Divide by 3; square root 56
Pattern 4: Sum, median or product of factors
[PYQ: NTPC 2016 cycle]
EXAM LEVEL
Q. Find the sum of all the factors of 72.
72 = 2³ × 3². Sum = (1 + 2 + 4 + 8)(1 + 3 + 9) = 15 × 13 = 195.
Answer: 195
EXAMATLAS LEVEL
Q. Find the median of all positive factors of 60 and show why it can be found without listing them.
60 = 2² × 3 × 5 has 3 × 2 × 2 = 12 factors, an even count, so the median is the average of the 6th and 7th factors. Factors pair up with product 60 and the middle pair is the pair closest to √60 ≈ 7.7, which is (6, 10). Median = (6 + 10)/2 = 8. Listing confirms: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
Answer: 8
Pattern 5: Prime identification and counting prime factors
[PYQ: NTPC CBT-1 15-Mar-2021 Shift-1]
EXAM LEVEL
Q. Which of the following is a prime number? (a) 221 (b) 323 (c) 391 (d) 401
221 = 13 × 17, 323 = 17 × 19, 391 = 17 × 23. For 401, √401 ≈ 20 and none of 2, 3, 5, 7, 11, 13, 17, 19 divides it, so 401 is prime.
Answer: (d) 401
EXAMATLAS LEVEL
Q. Find the total number of prime factors in 6⁵ × 15³ × 35².
Break each base into primes: 6⁵ = 2⁵ × 3⁵, 15³ = 3³ × 5³, 35² = 5² × 7². Combine: 2⁵ × 3⁸ × 5⁵ × 7². Total prime factors (counted with repetition) = 5 + 8 + 5 + 2 = 20. The distinct primes are only 4, which is the trap option.
Answer: 20
Pattern 6: Trailing zeros and highest power in a factorial
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. How many zeros are there at the end of 250!?
⌊250/5⌋ + ⌊250/25⌋ + ⌊250/125⌋ = 50 + 10 + 2 = 62.
Answer: 62
EXAMATLAS LEVEL
Q. How many zeros are there at the end of 5 × 10 × 15 × 20 × ... × 500?
There are 100 terms and each is 5 × k for k = 1 to 100, so the product is 5¹⁰⁰ × 100!. Count 5s: 100 + (20 + 4) = 124. Count 2s: they come only from 100!, which has 50 + 25 + 12 + 6 + 3 + 1 = 97. Zeros = min(97, 124) = 97. Counting only the 5s gives 124, the trap.
Answer: 97
60-Second Revision
- Factor count = product of (exponent + 1); odd factors ignore 2; even = total − odd.
- Sum of factors = product of (1 + p + p² + ...) for each prime.
- Perfect-square factors: count even exponents only; multiplier/divisor fixes odd exponents.
- Factor pairs multiply to N; odd factor count only for perfect squares; median of factors = average of the pair nearest √N.
- Trailing zeros = min(2s, 5s); in n! it is ⌊n/5⌋ + ⌊n/25⌋ + ...
- Test primality only up to √N.