A, B and C are three distinct single-digit positive numbers. If the un

A late response, but here goes. Even I followed the approach you outlined initially, but upon closer inspection of the question saw that even the units digits of the squares are
DISTINCT perfect squares.
Now the relevant set of values we have for (A,B,C) is {1,2,3,7,8,9}.
Say our 6C3 results in the subset {1,3,9}. Now the squares of those will be 1,9,81. Notice that the units digit is the same for 1^2 and 9^2, and this does not match the criteria required in the question.
Therefore, the grouping needs to be done on the basis of the squares that can be obtained from the single digit positive integers, namely 1,4 and 9.
1 can be obtained from 1^2 and 9^2
4 can be obtained from 2^2 and 8^2
9 can be obtained from 3^2 and 7^2.
So 2 cases for each -> 2*2*2 = 8 ; multiply by 3! to account for all arrangements and you get 48.

A B and C are three distinct single-digit positive numbers.

=> Possible choice to pick number { 1,2,3,4,5,6,7,8,9}

The units digit of \(A^2\), \(B^2\) and \(C^2\) are distinct perfect squares

=> Single digit perfect squares are : 1,4,9

Possible combinations for 1 as unit digit: \(1^2\) or \(9^2\) = 2 ways
Possible combinations for 4 as unit digit: \(2^2\) or \(8^2\) = 2 ways
Possible combinations for 9 as unit digit: \(3^2\) or \(7^2\) = 2 ways

Total ways : 2*2*2 = 8.

'3' numbers that satisfy the condition can themselves be arranged in 3! ways.

So, overall ways will be : 8 * 3! = 8 * 6 = 48

Answer D

A^2 B^2 & C^2 are perfect sqaure means they must be 1, 4, 9

For a number's sqaure to end at 1= possible single digit integer is 1, 9
For a number's sqaure to end at 4=possible single digit integer is 2, 8
For a number's sqaure to end at 9=possible single digit integer is 3, 7

So there are 2 each integer possible values.
So the number of ways will be 2*2*2*3!(because it can be interchanged between ABC).

Asnwer-D

Each of the Units Digit of A^2nd, B^2nd, C^2nd have to be Distinct Perfect Squares.

From the Single Digits

[1'2 = 1 ---- 9'2 = 81] ----- have Perfect Square 1 in its Units Digit, so either 1 or 9 can be in an Ordered Solution (A , B , C), but NOT Both

[2'2 = 4 --- 8'2 = 64] ---- have Perfect Square 4 in its Units Digit. Same logic as above.

[3'2 = 9 ---- 7'2 = 49] ---- have Perfect Square 9 in its Units Digit. Same logic as above.



To Find the Number of Ordered Solutions, I used the "Options Available" approach and went through the Possible Permutations.


If Variable A were a person and had to choose 1 of the 6 possible numbers above, he would have 6 available options.

A ----- 6 Options Available

assume A picked 1

Now Variable B can choose any of the 4 options OTHER THAN 9, because 9'2 = 81 has the same Perfect Square in its Units Digit.

B ----- 4 Options Available

assume B picked 4.

Lastly, Variable C will only have 2 options left to choose from - either 3 or 7

C ------ 2 Options Available

Multiplying the Options Available will give the No. of Permutations

6 * 4 * 2 = 48 Ordered Solutions in which Each Variable in (A, B, C) have a DISTINCT Perfect Square as its Units Digit.

Answer D

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