Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 10 Nov 2015
Posts: 4

Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
13 Nov 2015, 10:35
2
This post received KUDOS
D: I and II only. Easy to realize that P must be 1 because maximum 99+99 = 198 < 200, P cannot be 2 or greater. XY + YX = 10X + Y + 10Y + X = 11(X+Y) = 1Q5 => 1Q5 divisible by 11 => Q = 1+5 = 6, PQ5 = 165 => X+Y = 15 => (X,Y) e {(9,6),(8,7),(7,8),(6,9)}. But notice that X,Y,P,Q is different from each other so we cannot take 6 twice. Conclusion : X e (7,8). Choice D.



Current Student
Joined: 13 Apr 2015
Posts: 6
Location: India
GPA: 3.7

Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
13 Nov 2015, 13:40
1
This post received KUDOS
X Y + Y X = PQ5
This implies that, X + Y should have a units digit as 5. there are 2 options: 1) One of the numbers is 1 and the other is 4. But this condition is ruled out as it is mentioned in the question that X, Y, P, Q are different nonzero digits, hence P cannot be equal to 0 2) The only other option is when one of the numbers is 7 and the other is 8. So when we add X Y + Y X, we get the desired 3 digit number with the units digit as 5.
So the correct answer is D.



Manager
Joined: 20 Sep 2008
Posts: 78

Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
13 Nov 2015, 20:16
Answer is E:
x,y single digits > 0 because only sum of two digit numbers can give a three digit number.
Also, x + y <= 18
Unit digit = 5 Tens digit = Q Hundreds digit = P
I) If x = 7. For the unit digit of (x + y) to end on 5, the value of y = 15  7 = 8. So, the two numbers are 78 and 87. Sum is 165. P=1, Q = 6. Values are all different, which is a solution. II) If x = 8. For the unit digit of (x + y) to end on 5, the value of y = 15  8 = 7. So, the two numbers are 87 and 78. Sum is 165. P=1, Q = 6. Values are all different, which is a solution. III) If x = 9. For the unit digit of (x + y) to end on 5, the value of y = 15  9 = 6. so, the two numbers are 69 and 96. Sum is 165. All values are different, which is a solution.



Manager
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 249
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32 GMAT 2: 680 Q47 V35

Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
13 Nov 2015, 23:53
It should be E. All the 3 values are possible for X. X Y + Y X _____ P Q 5 Possibilities for adding units digits:Let, Y + X = N5 (where N can be any number from 0 to 9) Now, since, Y and X both can only range between 19, Y+X will can only range between 218 So, Y + X can only yield 5 or 15 Possibilities for adding tens digits:Given that X + Y at tens place yield a 2digit number (P cant be 0). If Y + X from units yield only a 5 then X + Y from tens will never yield a twodigit number. Therefore Y + X = 15Possible values of (X,Y): (7,8) or (8,7) or (9,6) or (6,9) Substituting values of and summing XY + YX gives 165. Since, all the numbers X, Y, P, Q have to be different, X cannot be equal to 6 (as Q=6), therefore X can take values 7, 8 or 9. Answer E
_________________
One Kudos for an everlasting piece of knowledge is not a bad deal at all...
 Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover. Mark Twain



Math Expert
Joined: 02 Sep 2009
Posts: 44596

Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
14 Nov 2015, 09:28
Bunuel wrote: Jamboree and GMAT Club Contest Starts QUESTION #1:X Y +Y X ________ The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X? I) 7 II) 8 III) 9 (A) I only (B) II only (C) III only (D) I and II only (E) I , II and III Check conditions below: For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend. To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:  GMAT Online Comprehensive ( If the student wants an online GMAT preparation course)  GMAT Classroom Program ( Only if he/she has a Jamboree center nearby and is willing to join the classroom program) Bookmark this post to come back to this discussion for the question links  there will be 2 on Saturday and 2 on Sunday! There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
Thank you! JAMBOBREE OFFICIAL SOLUTION:Since XY and YX are 2 digit numbers, there sum has to be less than 200. Thus, P = 1. Also sum of the digits X and Y should end with 5. Hence X + Y = 5 or 15 or 25… so that 5 is the units digit and tens digit is carried over. However, X + Y is less than 18 (as they are digits) and greater than 10 (else sum won’t be a 3digit number). Thus, X + Y = 15, which gives X Y + Y X  P Q 5 => 1 6 5 Alternatively: when a two digit number is added to another two digit number formed by reversing the digits of the first two digit number,the result is a multiple of 11. So in this case as PQ5 has to be a multiple of 11. So, using the divisibility rule of 11, P =1 and Q = 6. Now, X+Y = 15,so all possible values of x and Y are Case 1 X = 6 Y = 9 Case 2 X = 7 Y = 8 Case 3 X = 8 Y =7 Case 4 X =9 Y = 6 But as it is given that each letter X, Y, P, and Q represents a different non zero digit AND we know that Q = 6 so case 1 and case 4 are not possible So the only possible values of x are 7 and 8 Hence the correct answer is option D.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 13 Mar 2011
Posts: 21

Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
31 Mar 2016, 08:19
MelodyofMew wrote: D: I and II only. Easy to realize that P must be 1 because maximum 99+99 = 198 < 200, P cannot be 2 or greater. XY + YX = 10X + Y + 10Y + X = 11(X+Y) = 1Q5 => 1Q5 divisible by 11 => Q = 1+5 = 6, PQ5 = 165 => X+Y = 15 => (X,Y) e {(9,6),(8,7),(7,8),(6,9)}. But notice that X,Y,P,Q is different from each other so we cannot take 6 twice. Conclusion : X e (7,8). Choice D. I like the solution above. However, for me, it isn't understandable why Q = 1 + 5 = 6 ??? So, In my humble opinion, much easier will be to solve this problem like below : the system of thoughts will be : Easy to realize that P must be 1 because maximum 99+99 = 198 < 200, P cannot be 2 or greater. (1) and X and Y are integer (1.1) XY + YX = 10X + Y + 10Y + X = 11(X+Y) = 1Q5 (2) and the unit digit is 5 (2.1) because of (1) and (1.1) => 100 < 11(X+Y) < 200 => 9 < X+Y < 19 (2.2) because of (2.1) and (2.2) and (1.1) => X + Y = 15 (3)  in other words 15 will be the single integer from 9 to 19 that gives unit digit of 5 because of (2) and (3) 11(X+Y) = 11*15 = 165 (4) X+Y = 15 => (X,Y) e {(9,6),(8,7),(7,8),(6,9)}. (5) But because of (4) => X e (7,8). Choice D
_________________
I’m not afraid of the man who knows 10,000 kicks and has practiced them once. I am afraid of the man who knows one kick & has practiced it 10,000 times!  Bruce Lee
Please, press the +1 KUDOS button , if you find this post helpful



Intern
Joined: 10 Aug 2015
Posts: 34
Location: India
GPA: 3.5
WE: Consulting (Computer Software)

Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
17 Dec 2016, 09:50
Bunuel wrote: Jamboree and GMAT Club Contest Starts QUESTION #1:X Y +Y X ________ The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X? I) 7 II) 8 III) 9 (A) I only (B) II only (C) III only (D) I and II only (E) I , II and III Thank you![/header3] Every one is missing a small fact that XY and YX are 2 digit numbers one with reversed digit. Now their sum will always be divisible by 11. As both are 2 digit numbers then P=1 so only possibility where 1Q5 is divisible by 11 is when Q=6. So Essentially we have XY +YX _____ 165 Now you can check numbers that give X+Y =15 to get answer.



Director
Joined: 07 Dec 2014
Posts: 962

Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
17 Dec 2016, 19:49
X Y +Y X ________
The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X?
I) 7 II) 8 III) 9
(A) I only (B) II only (C) III only (D) I and II only (E) I , II and III
XY+YX is a multiple of 11 and 5 100<XY+YX<187 XY+YX=11*15=165 XY=78 and YX=87; or XY=87 and YX=78 D I and II only



Intern
Joined: 17 Jan 2018
Posts: 38

Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
Show Tags
22 Mar 2018, 05:00
Given  10X + Y + 10Y + X = 100P + 10Q + 5 11(X+Y) = 5(20P +2Q + 1) Now, common sense says that P needs to be 1. (Max 99 + 99 is still 198 only) So, 11(X + Y) = 5(21 + 2Q) Now, we can see that 21 + 2Q should be a multiple of 11, to satisfy this equation. We also know that Q can be at max 9, (So, 21 + 18 = 39), so a multiple of 11, less than 39 is 33. Hence, 21 + 2Q = 33..... Shows that Q = 6 Hence, X + Y = 5*3 = 15 Also, P and Q are taken, 1 and 6 respectively. The only way to make (X,Y) as 15 are (9,6) and (8,7). As, 6 is taken by Q, we are left with just (8,7) So, X can be either 7 or 8. Ans) D Bunuel wrote: Jamboree and GMAT Club Contest Starts QUESTION #1:X Y +Y X ________ The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X? I) 7 II) 8 III) 9 (A) I only (B) II only (C) III only (D) I and II only (E) I , II and III Check conditions below: For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend. To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:  GMAT Online Comprehensive ( If the student wants an online GMAT preparation course)  GMAT Classroom Program ( Only if he/she has a Jamboree center nearby and is willing to join the classroom program) Bookmark this post to come back to this discussion for the question links  there will be 2 on Saturday and 2 on Sunday! There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
Thank you!




Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers
[#permalink]
22 Mar 2018, 05:00



Go to page
Previous
1 2 3
[ 49 posts ]



