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Bunuel
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Bunuel
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20 can be written as -

(20 * 1) OR (2 * 10) OR (4 * 5)

As we want the smallest integers, we have to choose small prime numbers

Case 1 :

Say, N = \(2^{19}\) | The number of factors of N = 20

N is greater than 1000

Case 2 :

Say, N = \(3^1 * 2^{10}\) | The number of factors of N = 20

N is greater than 1000

Case 3 :

Say, N = \(3^4 * 2^{5}\) | The number of factors of N = 20

N is greater than 1000

IMO - E
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its 240
(1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, and 240)
lies in range 1 to 250
Option A
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\(­G = a^x b^yc^z ...\)

total number of factors = 20 = (x+1)(y+1)(z+1)...

G is smallest when x, y, z, ... = 1 (if possible)

20 = 2 * 2 * 5 = (1+1) * (1+1) * (4+1)

Let 2 (the smallest prime number) takes the highest exponent (4)
Then the next two, 3 and 5 take the exponent 1

=> G = 2^4 * 3 * 5 = 240

=> Answer is A­
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