Suppose 72 = m x n, where m and n are positive integers such that 1 <

Suppose 72 = m x n, where m and n are positive integers such that 1 < m < n. How many possible values of m are there? [amp_mcq option1="5" option2="6" option3="10" option4="12" correct="option1"]

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UPSC CAPF – 2018
There are 5 possible values for m.
– We are given that 72 = m x n, where m and n are positive integers such that 1 < m < n. - We need to find pairs of factors (m, n) of 72 that satisfy the condition 1 < m < n. - First, list pairs of factors (m, n) of 72 such that m <= n: - 72 = 1 x 72 (m=1, n=72) - 72 = 2 x 36 (m=2, n=36) - 72 = 3 x 24 (m=3, n=24) - 72 = 4 x 18 (m=4, n=18) - 72 = 6 x 12 (m=6, n=12) - 72 = 8 x 9 (m=8, n=9) - Now, apply the condition 1 < m < n: - (1, 72): m=1. Fails 1 < m. - (2, 36): m=2. Satisfies 1 < 2 < 36. m=2 is a possible value. - (3, 24): m=3. Satisfies 1 < 3 < 24. m=3 is a possible value. - (4, 18): m=4. Satisfies 1 < 4 < 18. m=4 is a possible value. - (6, 12): m=6. Satisfies 1 < 6 < 12. m=6 is a possible value. - (8, 9): m=8. Satisfies 1 < 8 < 9. m=8 is a possible value. - The possible values for m are the first elements of the valid pairs: 2, 3, 4, 6, 8. - There are 5 possible values for m.
To find all pairs of factors (m, n) of a number N, list all factors of N. Then pair them up such that their product is N. For the condition m < n, ensure m is the smaller factor in each pair (or check against sqrt(N)). For 72, sqrt(72) is between 8 and 9. Pairs with m <= sqrt(72) are (1,72), (2,36), (3,24), (4,18), (6,12), (8,9).
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