February 21, 2019 February 21, 2019 10:00 PM PST 11:00 PM PST Kick off your 2019 GMAT prep with a free 7day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th. February 23, 2019 February 23, 2019 07:00 AM PST 09:00 AM PST Learn reading strategies that can help even nonvoracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
Author 
Message 
TAGS:

Hide Tags

eGMAT Representative
Joined: 04 Jan 2015
Posts: 2589

Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
Updated on: 06 Feb 2019, 19:08
Question Stats:
21% (02:46) correct 79% (02:13) wrong based on 77 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Aug 2009
Posts: 7334

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
01 Feb 2019, 09:10
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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html 4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentageincreasedecreasewhatshouldbethedenominator287528.html
GMAT Expert



Intern
Joined: 26 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, 09:22
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: 33
Concentration: International Business, Marketing

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



SVP
Joined: 18 Aug 2017
Posts: 1887
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, 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 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: 8

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
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 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: 59
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, 07:51
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: 12

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
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 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: 08 Aug 2017
Posts: 325

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
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 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: 59
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, 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 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.



Manager
Joined: 12 Sep 2017
Posts: 139

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
05 Feb 2019, 13:49
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: 2589

Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
Show Tags
06 Feb 2019, 19:18
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
_________________
Register for free sessions Number Properties: Get 5 free video lessons, 50 practice questions  Algebra:Get 4 free video lessons, 40 practice questions Quant Workshop: Get 100 practice questions  Free Strategy Weession: Key strategy to score 760
Success Stories Q38 to Q50  Q35 to Q50  More Success Stories
Ace GMAT Articles and Question to reach Q51  Question of the week  Tips From V40+ Scoreres  V27 to V42: GMAT 770  Guide to Get into ISBMBA
Must Read Articles Number Properties – Even Odd  LCM GCD  Statistics1  Statistics2  Remainders1  Remainders2 Word Problems – Percentage 1  Percentage 2  Time and Work 1  Time and Work 2  Time, Speed and Distance 1  Time, Speed and Distance 2 Advanced Topics Permutation and Combination 1  Permutation and Combination 2  Permutation and Combination 3  Probability Geometry Triangles 1  Triangles 2  Triangles 3  Common Mistakes in Geometry Algebra Wavy line  Inequalities Practice Questions Number Properties 1  Number Properties 2  Algebra 1  Geometry  Prime Numbers  Absolute value equations  Sets
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com




Re: Question of the week  34 (A, B and C are three distinct ............)
[#permalink]
06 Feb 2019, 19:18






