Squares-Cubes-Factors
Let be the least number such that is a perfect square and is a perfect cube.
Then the number of positive divisors of is?
The answer is 72.
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1 solution
10A = 5x2xA and 35A = 5x7xA Now, the question tells us that 10A is a perfect square ......(i) which means, A must have atleast 5x2 (as factors) to make the number a perfect square. Also, 35A is a perfect cube ....(ii) hence: A must have atleast 5x5x7x7 factors.
So, summing up: A = 5x5x7x7x2
But now. We have extra 5 and 2 which does not let us satisfy the given question. So adding three more 5s and two more 2s ( i used hit and trial) .. We get A = 5x5x5x5x5x7x7x2x2x2 ......(iii) which satisfies both (i) and (ii)
→ A = 5x5x5x5x5x7x7x2x2x2x1 (as 1 is always a factor of every number) Therefore, different divisors of A can be given as: 1, 5, 5x5, 5x5x5, 5x5x5x5, 5x5x5x5x5, 7, 7x7, 2, 2x2, 2x2x2, 5x2, 5x7, 5x2x2, 5x2x2x2, 5x7x7........5x5x5x5x5x7x7x2x2x2 Counting above factors, we get no of Divisors: 72 ( final answer)
Shortcut method: Eqn (iii) can be written as, A = 5^5 + 7^2 + 2^3 No of divisors = (5+1)(2+1)(3+1) = 6x3x4 = 72!