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

Question

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

e-GMAT     Question of the Week #34

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chetan2u · Accepted Answer

The single digit that are perfect square are 1, 4 and 9, so the units digit of  A^2 ,  B^2  and  C^2  are 1, 4 and 9.
Which all numbers have their square ending with 1-- 1 and 9
Which all numbers have their square ending with 4-- 2 and 8
Which all numbers have their square ending with 9-- 3 and 7.

(I) Thus A, B and C can be any of 6 but with restriction. 
Without restriction we can fill up A,B and C in 6*5*4 = 120 ways.
Now (1 and 9), (2 and 4) and (3 and 7) cannot be together. Let us take them together.
If 1 and 9 are together, the third can be any of the four, so 4*3!=24. Similarly if 3 and 7 are together or 2 and 8 are together, the ways are 4*3!=24, thus 24+24+24=72.
answer 120-72= 48..

(II) Other way  
(1,9), (2,8), (3,7)..
ways (1,2,3); (1,2,7); (9,2,3); (9,2,7) and another 4 with 8 instead of 2, that is (1,8,3); (1,8,7); (9,8,3); (9,8,7)
All EIGHT can be arranged in 3! ways so answer is 8*3!=8*6=48..

D

999neil · Answer

It should be 8, as distinct square ending number will be 1, 4, 9.

Possible distinct set can be (312, 318, 392, 398, 712,718, 792, 798).

If repetition is allowed than answer can be different this is for distinct only.

Ana4001 · Answer

chetan,

pls elaborate on 
If 1 and 9 are together, the third can be any of the four, so 4*3!=24. Similarly if 3 and 7 are together or 2 and 8 are together, the ways are 4*3!=24, thus 24+24+24=72.
answer 120-72= 48..

Archit3110 · Answer

A,B,C can be 1,4,9 as they are the only single digit perfect square

so A= 1; 1,9
B=4; 2,8
C=9; 3,7

(1,9), ( 2,8), ( 3,7) 
total pairs of these digits can be
(1,2,3), ( 1,8,7), ( 1,2,7), ( 1,8,3) ....

2c1*2c1*2c1 * 6 = 8*6 = 48
IMO D

ankitsood5 · Answer

if 1 and 9 are together, the third number can be selected in 4 ways..( 1 out of 2,8,3,7), and these 3 numbers i.e. 1,9, and the selected number can be arranged in 3! ways. thus total ways = 4*3!.
Similarly for selection when 2 & 8 and 3 & 7 are taken together.

GMATMBA5 · Answer

IMO D

A, B and C are three distinct single-digit positive numbers, so they can hold values from 1-9. 
Also the unit digit of their squared values is a perfect square. This is true only for the square of the numbers 1, 2, 3, 7, 8 and 9.

Therefore, after grouping we have -
1, 9 --> 1
2, 8 --> 4
3, 7 --> 9

LHS are values of A, B and C. And RHS is the unit digit of squared values of A, B and C.
A, B and C can take only one value from each of the LHS. Therefore, A, B and C each can take  C_1^2*C_1^2*C_1^2=8 .
And they can be rearranged into  8*3!=48

-----------------------------------------------------------------------------------
 Kudos if helpful!

gvij2017 · Answer

Hi Friend,

I think question is asking for pairs, not for arrangements. Thus, multiplication with 3! is not required here. 
It is combinations problem,not a permutation problem.

Answer should be 8. 
B.

GMATMBA5 · Answer

Hello,

Combination will give you only 8 possible selections of A, B and Cs to form the pair (A, B, C). For example, (1,2,3), (1,2,7), (1,8,3), (1,8,7).....
But these 8 possible selections can be rearranged among A, B and Cs --> (2,1,3), (2,1,7), (8,1,3), (7,1,8)..... This rearrangement will give 3! times the original 8 selections.

EgmatQuantExpert · Answer

Solution 
  Given:  
 • A, B, and C are three single-digit positive numbers
• The units digit of  A^2 ,  B^2 , and  C^2  are distinct perfect squares

To find:  
 • The number of possible values of (A, B, C)

Approach and Working:   
 • The list of all single digit positive numbers are, {1, 2, 3, 4, 5, 6, 7, 8, 9}
• The units digit of squares of these numbers are, {1, 4, 9, 6, 5, 6, 9, 4, 1}
• Among these the perfect squares are {1, 4, 9}
 o The numbers whose perfect squares end with 1 are {1, 9}
o The numbers whose perfect squares end with 4 are {2, 8}
o The numbers whose perfect squares end with 9 are {3, 7}

• Thus, the total number of ways of selecting one number from each pair = 2 * 2 * 2 = 8 ways.
• These three numbers can be arranged in 3! = 6 ways

Therefore, total number of values of (A, B, C) = 8 * 3! = 48

Hence the correct answer is Option D.

Answer: D

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SudsMeister · Answer

How I went about it:

A, B, C are 1 digit positive numbers.
We need to look at unit digits of  A^2, B^2, C^2

1 digit positive numbers = 1 to 9
Their squares = 1, 4, 9, 16, 25, 36, 49, 64, 81
Out of these numbers, we need to focus on the ones whose unit digit is a perfect square. That can only happen if the unit digit is 1, 4, or 9.
So the possible values of  A^2, B^2, C^2  = 1, 4, 9, 49, 64, 81
Possible values of A, B, C = 1, 2, 3, 7, 8, 9

But we are given that  A^2, B^2, C^2  have distinct unit digits, so if 1 is selected 81 cannot be select (and vice versa), if 4 is selected 64 cannot be selected (and vice versa), and if 9 is selected 81 cannot be selected (and vice versa).
This translates into: if 1 is selected 9 cannot be selected (and vice versa), if 2 is selected 8 cannot be selected (and vice versa), and if 3 is selected 7 cannot be selected (and vice versa).

So for A = there are 6 possible options (1, 2, 3, 7, 8, 9)
For B = there are 4 possible options (for example, if A = 1, then 1 and 9 both are no longer available for any further selection, so 4 other numbers remain).
For C = there are 2 possible options (2 numbers exhausted by A, another 2 by B, so only 2 remain).

Total possible outcomes for A, B, and C = 6 x 4 x 2 = 48.

Answer D.

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

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

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