Solved Examples

Example 1: Find the number of factors of 360. - 360 = 2³ × 3² × 5¹ - Number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24

Example 2: Find the sum of all factors of 360. - Sum = [(2⁴−1)/(2−1)] × [(3³−1)/(3−1)] × [(5²−1)/(5−1)] - = [15/1] × [26/2] × [24/4] = 15 × 13 × 6 = 1170

Example 3: Find the number of even factors of 1800. - 1800 = 2³ × 3² × 5² - Total factors = (3+1)(2+1)(2+1) = 4×3×3 = 36 - Odd factors (i.e., the power of 2 must be 0) = (1)(3)(3) = 9 - Even factors = Total − Odd = 36 − 9 = 27

Example 4: Find the number of factors of 100 that are perfect squares. - 100 = 2² × 5². A factor is a perfect square only if both exponents used are even: exponent of 2 can be 0 or 2 (2 choices), exponent of 5 can be 0 or 2 (2 choices). - Perfect square factors = 2 × 2 = 4 (these are 1, 4, 25, 100)

Example 5 (Product of all factors): Find the product of all factors of 36. - Number of factors of 36 (=2²×3²) = (2+1)(2+1) = 9 - Formula: Product of all factors of N = N^(number of factors / 2) = 36^(9/2) = 36^4.5 - This equals √(36⁹) — for exam purposes it is usually left in exponential form unless the number of factors is even, in which case it simplifies to a whole power. Since 36 is itself a perfect square (6²), 36^4.5 = 6⁹, a clean integer power — this is the standard form SSC expects as the final answer: 6⁹

Why this formula works: Factors pair up as (d, N/d) — each pair multiplies to N. If there are k factors, there are k/2 such pairs (or, if N is a perfect square, one middle factor √N is left unpaired), so the product of all factors is N raised to the power (number of factors)/2.

Example 6 (Reverse problem — smallest number with a given number of factors): Find the smallest number that has exactly 15 factors. - We need (p+1)(q+1)… = 15. Since 15 = 5 × 3 = 15 × 1, the exponent pattern could be a single prime to the power 14 (giving 2^14, which is huge), or two primes with exponents 4 and 2 (since (4+1)(2+1) = 15). - To minimise the value, assign the larger exponent to the smaller prime: 2⁴ × 3² = 16 × 9 = 144. (The reverse assignment, 2² × 3⁴ = 4 × 81 = 324, is larger.) Answer: 144 (144 = 2⁴×3², number of factors = (4+1)(2+1) = 15 ✓)

Example 7 (Factors divisible by a given number): N = 2³ × 3² × 5¹ × 7². Find how many factors of N are divisible by 10. - A factor divisible by 10 = 2×5 must include at least one 2 and at least one 5. - Choices for exponent of 2 (must be ≥1, up to 3): 1, 2, or 3 → 3 choices - Choices for exponent of 5 (must be ≥1, up to 1): only 1 → 1 choice - Choices for exponent of 3 (0 to 2, unrestricted): 3 choices - Choices for exponent of 7 (0 to 2, unrestricted): 3 choices - Number of factors divisible by 10 = 3 × 1 × 3 × 3 = 27 Answer: 27 (Sanity check: total factors of N = (3+1)(2+1)(1+1)(2+1) = 4×3×2×3 = 72, and 27 < 72 as expected.)