Two distinct numbers are randomly selected from among the first 64 positive integral perfect squares. Find the probability that these two numbers are both cubes of integers.

bigmac Dec 11, 2019

#1**+1 **

These are the first 64 "perfect squares":

(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096).

Of the above 64, there are 4 cubes:1, 64, 729, 4096.

The probability that the first number is a cube is:

4/64 =1 / 16

The probability that the 2nd is also a cube is:

3/63 =1 / 21

Then the probability that both are cubes is:

1/16 x 1/21 =1 / 336

P.S. Somebody should check this, please. Thanks.

Guest Dec 11, 2019