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Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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13 Nov 2015, 09:35
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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.



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Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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13 Nov 2015, 12:40
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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.



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Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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13 Nov 2015, 19:16
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.



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Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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13 Nov 2015, 22:53
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
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Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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14 Nov 2015, 08:28
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.
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Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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31 Mar 2016, 07:19
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
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Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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17 Dec 2016, 08:50
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.



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Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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17 Dec 2016, 18:49
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



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Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers [#permalink]
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06 Feb 2018, 09:54
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