IPMAT Indore 2024 · Quantitative Ability · Question 1

Step 1 — factorise.

1800={2}^{3}\times{3}^{2}\times{5}^{2}

Step 2 — force the factor to be a multiple of 6. A divisor d={2}^{a}{3}^{b}{5}^{c} is a multiple of 6=2\times 3 exactly when it carries at least one 2 and at least one 3:

1\le a\le 3, 1\le b\le 2, 0\le c\le 2

Step 3 — count the choices.

3\times 2\times 3=18

The trick worth remembering: instead of counting, just divide out the 6. Every such factor is 6k where k divides 1800/6=300={2}^{2}\times 3\times{5}^{2}, and 300 has \left( 2+1 \right)\left( 1+1 \right)\left( 2+1 \right)=18 divisors. Same answer, one line.

Answer: 18

Shortcut. Divide first, count second. Factors of N that are multiples of m = number of factors of N/m — provided m∣N. Here 1800/6=300={2}^{2}{3}^{1}{5}^{2}, giving 3\times 2\times 3=18. One division and one product; no case work.