Vincen wrote: ↑

Wed Oct 28, 2020 8:52 am

\(A, B\) and \(C\) are three distinct single-digit positive numbers. If the units digit of \(A^2, B^2\) and \(C^2\) are distinct perfect squares, then what is the number of possible values of \((A, B, C)?\)

A. 4
B. 8
C. 20
D. 48
E. 120

Answer: D

Solution:

Let’s look at all the squares of single-digit positive numbers first:

1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2 = 36, 7^2 = 49, 8^2 = 64, and 9^2 = 81

We see that 6 of the 9 numbers have a units digit that is a perfect square. However, since they have to be distinct perfect squares, we can’t, for example, have (A, B, C) = (1, 2, 9) since A and C have the same units digit perfect square. Therefore, if 1 is picked as one of 3 values, 9 can’t be picked as one of the other 2 values. Similarly, if 2 is picked, 8 can’t be picked, and if 3 is picked, 7 can’t be picked.

However, let’s assume there are no restrictions of which 3 numbers we can pick from the 6 available numbers (1, 2, 3, 7, 8, and 9) as long as they are distinct, then we can have 6P3 = 6 x 5 x 4 = 120 ordered triples (A, B, C). Now, all we need to do is to remove the number of ordered triples with 1 and 9, those with 2 and 8, and those with 3 and 7.

If an ordered triple has the numbers 1 and 9, then the third number can be any one of the 4 remaining numbers. However, since there are 3! = 6 ways to permute the three numbers (after all, ordered triple means order matters), there are 4 x 6 = 24 ordered triples with 1 and 9. Using the same argument, there are 24 ordered triples with 2 and 8 and another 24 ordered triples with 3 and 7.

Therefore, there are 120 - 24 - 24 - 24 = 48 ordered triples (A, B, C) such that A, B, C are distinct single-digit positive numbers and A^2, B^2, and C^2 have distinct perfect-square units digits.

Answer: D