Oct 14 08:00 PM PDT  11:00 PM PDT Join a 4day FREE online boot camp to kick off your GMAT preparation and get you into your dream bschool in R2.**Limited for the first 99 registrants. Register today! Oct 15 12:00 PM PDT  01:00 PM PDT Join this live GMAT class with GMAT Ninja to learn to conquer your fears of long, kooky GMAT questions. Oct 16 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

eGMAT Representative
Joined: 04 Jan 2015
Posts: 3074

Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
Updated on: 28 Feb 2019, 00:28
Question Stats:
21% (02:43) correct 79% (02:16) wrong based on 89 sessions
HideShow timer Statistics
eGMAT Question of the Week #34A, B and C are three distinct singledigit 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)?
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Math Expert
Joined: 02 Aug 2009
Posts: 7957

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
01 Feb 2019, 10:10
EgmatQuantExpert wrote: A, B and C are three distinct singledigit 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)? 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 12072= 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
_________________



Intern
Joined: 27 Jan 2019
Posts: 4
GPA: 3.04

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
01 Feb 2019, 10:22
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.



Intern
Joined: 30 Dec 2018
Posts: 34
Concentration: International Business, Marketing

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
02 Feb 2019, 03:27
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 12072= 48..



GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4987
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
02 Feb 2019, 04:19
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 singledigit 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)?
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.



Intern
Joined: 28 Apr 2016
Posts: 16

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
02 Feb 2019, 08:01
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 12072= 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.



Manager
Joined: 07 Aug 2017
Posts: 86
Location: India
GPA: 4
WE: Information Technology (Consulting)

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
02 Feb 2019, 08:51
EgmatQuantExpert wrote: A, B and C are three distinct singledigit 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)? IMO D A, B and C are three distinct singledigit positive numbers, so they can hold values from 19. 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!



Intern
Joined: 22 Oct 2017
Posts: 19

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
02 Feb 2019, 09:40
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 12072= 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?



Senior Manager
Joined: 09 Aug 2017
Posts: 485

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
02 Feb 2019, 19:20
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 singledigit 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)? IMO D A, B and C are three distinct singledigit positive numbers, so they can hold values from 19. 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!



Manager
Joined: 07 Aug 2017
Posts: 86
Location: India
GPA: 4
WE: Information Technology (Consulting)

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
02 Feb 2019, 21:19
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 singledigit 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)? IMO D A, B and C are three distinct singledigit positive numbers, so they can hold values from 19. 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.



Senior Manager
Joined: 12 Sep 2017
Posts: 301

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
05 Feb 2019, 14:49
Helo chetan2u !
Which is the concept behind considering just 1,4 and 9 as perfect squares digits?
Because 1*1 =1, 2*2 = 4 and 3*3 = 9?
So, for ex 4, 4*4 = 16 and because 16 is not a single digit then 4 is not considered as a perfect square single digit?
Kind regards!



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3074

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
06 Feb 2019, 20:18
Solution Given:• A, B, and C are three singledigit 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
_________________




Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
06 Feb 2019, 20:18






