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Question of the week - 34 (A, B and C are three distinct ............)
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Updated on: 06 Feb 2019, 19:08
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Question Stats:
21% (02:46) correct 79% (02:13) wrong based on 77 sessions
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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)?
Re: Question of the week - 34 (A, B and C are three distinct ............)
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01 Feb 2019, 09:10
1
EgmatQuantExpert wrote:
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 pairs of (A, B, C)?
A. 4 B. 8 C. 20 D. 48 E. 120
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..
Re: Question of the week - 34 (A, B and C are three distinct ............)
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02 Feb 2019, 02:27
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..
Re: Question of the week - 34 (A, B and C are three distinct ............)
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02 Feb 2019, 03:19
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
EgmatQuantExpert wrote:
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 pairs of (A, B, C)?
A. 4 B. 8 C. 20 D. 48 E. 120
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Re: Question of the week - 34 (A, B and C are three distinct ............)
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02 Feb 2019, 07:01
Ana4001 wrote:
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..
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.
Re: Question of the week - 34 (A, B and C are three distinct ............)
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02 Feb 2019, 07:51
2
EgmatQuantExpert wrote:
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 pairs of (A, B, C)?
A. 4 B. 8 C. 20 D. 48 E. 120
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!
Re: Question of the week - 34 (A, B and C are three distinct ............)
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02 Feb 2019, 08:40
chetan2u wrote:
EgmatQuantExpert wrote:
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..
I am sorry, but I don't get why (1 and 9), (2 and 4) and (3 and 7) cannot be together, could you please help me clarify this doubt?
Re: Question of the week - 34 (A, B and C are three distinct ............)
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02 Feb 2019, 18:20
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 wrote:
EgmatQuantExpert wrote:
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 pairs of (A, B, C)?
A. 4 B. 8 C. 20 D. 48 E. 120
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!
Re: Question of the week - 34 (A, B and C are three distinct ............)
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02 Feb 2019, 20:19
gvij2017 wrote:
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 wrote:
EgmatQuantExpert wrote:
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 pairs of (A, B, C)?
A. 4 B. 8 C. 20 D. 48 E. 120
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!
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.
Re: Question of the week - 34 (A, B and C are three distinct ............)
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06 Feb 2019, 19:18
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
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21% (02:46) correct 
