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JEE Advanced 2023 Paper 1 · Mathematics · Permutations and Combinations · medium

Let N be the number of ordered pairs (a,b) of positive integers with a<b and lcm(a,b)=360. Find N.

Numerical answer

Let NN be the number of ordered pairs (a,b)(a,b) of positive integers with a<ba<b and lcm⁡(a,b)=360\operatorname{lcm}(a,b)=360. Find NN.
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Answer: 52

Write 360=23⋅32⋅5360=2^3\cdot3^2\cdot5. For lcm⁡(a,b)=360\operatorname{lcm}(a,b)=360, in each prime pep^e the exponent pair (u,v)(u,v) of (a,b)(a,b) must satisfy max⁡(u,v)=e\max(u,v)=e. The number of such pairs is 2e+12e+1. So the number of ordered pairs (a,b)(a,b) is (2⋅3+1)(2⋅2+1)(2⋅1+1)=7⋅5⋅3=105(2\cdot3+1)(2\cdot2+1)(2\cdot1+1)=7\cdot5\cdot3=105. Exactly one of these has a=ba=b, namely a=b=360a=b=360 (since 360360 is the only number whose lcm with itself is 360360). The remaining 104104 pairs split equally into a<ba<b and a>ba>b, so N=105−12=52N=\frac{105-1}{2}=52. A brute-force count over 1≤a<b≤3601\le a<b\le360 confirms 5252.
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