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The positive twodigit integers x and y have the same digits, but in [#permalink]
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15 Oct 2015, 21:55
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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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16 Oct 2015, 00:00
Bunuel wrote: The positive twodigit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?
(A) 6 (B) 9 (C) 10 (D) 11 (E) 14
Kudos for a correct solution. 10a+b+10b+a=11(a+b) Answer (D)
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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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16 Oct 2015, 00:26
Bunuel wrote: The positive twodigit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?
(A) 6 (B) 9 (C) 10 (D) 11 (E) 14
Kudos for a correct solution. My Solution:
Lets say integer x has two digits "ab" then as per question y has two digits in reverse i.e, "ba" ,
Then x+y will be,
(10a+b)+(10b+a)> 11a+11b>11(a+b)
Therefore 11 must be a factor of x+y
Answer D
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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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16 Oct 2015, 01:00
10a+b+10b+a 11a+11b 11(a+b)....Ans: 11



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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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17 Oct 2015, 11:13
Hi All, This question can be solved by TESTing VALUES: We're told that X and Y are both 2digit positive integers with the digits reversed. We're asked for what MUST be a factor of X+Y IF.... X=12 Y=21 X+Y = 12+21 = 33 Only one of the answers is a factor of 33... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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18 Oct 2015, 11:05
WHy not 3,6. 36 and 63 are both divisible by 9.



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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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18 Oct 2015, 12:08
Hi alice7, The prompt asks us for what MUST be a factor of X+Y... Using your example (36 and 63), we would have a total of 99. In this case, TWO of the answers 'fit'  both 9 and 11 are factors of 99. So one of these MUST be the solution, but we won't know which one until we find another example that is NOT divisible by one of the two options. GMAT assassins aren't born, they're made, Rich
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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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28 Oct 2015, 13:52
Bunuel wrote: The positive twodigit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?
(A) 6 (B) 9 (C) 10 (D) 11 (E) 14
Kudos for a correct solution. Remember: When you take the difference between the two, it will always be 9. e.g 2332=9, 8998=9 and when you add both integers, the sum will always be a multiple of 11 e.g 23+32=55, 89+98= 187 so the answer is 11



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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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16 Mar 2016, 02:22
Bunuel wrote: The positive twodigit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?
(A) 6 (B) 9 (C) 10 (D) 11 (E) 14
Kudos for a correct solution. Supposing that the numbers are ab and ba, then ab can be written as 10a+b and ba can be written as 10b+a Adding we get 11a+11b=11*(a+b) Therefore, the sum is divisible by 11 as well as by sum of the digits.



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Re: The positive twodigit integers x and y [#permalink]
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02 May 2016, 19:21
macbookno wrote: The positive twodigit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?
A. 6 B. 9 C. 10 D. 11 E. 14 Pl post according to guidelines.. Provide OA, proper topic name and post in correct subforum.. as for your Q.. let x = ab, a 2digit integer so y= ba.. x= ab = 10a+b and y=ba=10b+a..so \(x+y = 10a+b+10b+a = 11(a+b)\).. therefore x+y will always have a factor 11.. D
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Re: The positive twodigit integers x and y [#permalink]
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02 May 2016, 20:09
macbookno wrote: The positive twodigit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?
A. 6 B. 9 C. 10 D. 11 E. 14 Take a simple example. Two integers could be 12 and 21. x + y = 33. Only 11 is a factor. So 11 must be a factor of (x+y) Answer (D)
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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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03 May 2016, 05:16
Bunuel wrote: The positive twodigit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?
(A) 6 (B) 9 (C) 10 (D) 11 (E) 14
Kudos for a correct solution. We can solve this question using the natural relationships that all twodigit numbers have. As an example, we can express 37 as (10 x 3) + 7. We multiply the digit in the tens position by 10 and then add the digit in the ones position. If we let a = the tens digit of x and b = the ones digit of x, we know: x = 10a + b Since the digits of y are the reverse of those of x, we can express y as: y = 10b + a When we sum x and y we obtain: x + y = 10a + b + 10b + a = 11a + 11b x + y = 11(a + b) The final expression 11(a + b) is a multiple of 11, and therefore 11 divides evenly into it. We see, therefore, that 11 must be a factor of x + y. Answer: D
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The positive twodigit integers x and y have the same digits, but in [#permalink]
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23 Jun 2016, 10:25
EMPOWERgmatRichC wrote: Hi All, This question can be solved by TESTing VALUES: We're told that X and Y are both 2digit positive integers with the digits reversed. We're asked for what MUST be a factor of X+Y IF.... X=12 Y=21 X+Y = 12+21 = 33 Only one of the answers is a factor of 33... Final Answer: GMAT assassins aren't born, they're made, Rich I thought same digits were 11, 22, 33, 44, 55, 66..... 99. I am confused now.



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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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23 Jun 2016, 11:09
bimalr9 wrote: I thought same digits were 11, 22, 33, 44, 55, 66..... 99. I am confused now. Ok Let me try once 
X = 14 , Y = 41 X + Y = 14 + 41 =>55 55 = 11 * 5 Check again 
X = 12 , Y = 21 X + Y = 12 + 21 =>33 33 = 11 * 3 Check again 
X = 16 , Y = 61 X + Y = 16 + 61 =>77 77 = 11 * 7 Check in each case the common factor is 11 , hence the answer must be 11......Feel free to revert in case of the slightest doubt, I will love to explain it again... PS: IMHO the best method for this problem will be (10a + b ) + (10b + a ) => 11(a+b) as posted earlier...
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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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04 Dec 2016, 01:16
Here is my solution for this one => x=MN => 10M+N y=NM=>10N+M x+y=> Must be a multiple of 11 OR in other words => 11 must be a factor of x+y
Alternatively Let x=23 y=32 x+y=55 Only option that is a factor of 55 is 11 i.e Option D
Hence D
Additionally xy will always be multiple of 9
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The positive twodigit integers x and y have the same digits, but in [#permalink]
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28 Sep 2017, 23:29
VeritasPrepKarishma Bunuel Engr2012 Quote: Take a simple example. Two integers could be 12 and 21. Is not the question stem ambiguous, leaving scope for discrepancy ? Ideal Q should be : no of digits are the same if no if digits of x and y are same, x = y = 10x + y Is q says reversing digits makes x and y same, it should be inferred as 10x + y = 10 y + x Let me know your understanding.



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Re: The positive twodigit integers x and y have the same digits, but in [#permalink]
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08 Oct 2017, 02:58
take x as 23 and y as 32 x+y= 55 prime factors of 55 is 5 and 11. thus 11 is the answer!




Re: The positive twodigit integers x and y have the same digits, but in
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