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Sub 505 (Easy)|   Multiples and Factors|                              
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Krunaal
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The positive two-digit integers x and y have the same digits, but in 26 Feb 2025, 13:23
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riccardopadrone
Sorry, but with the same reasoning then:

(5a+b)+(a+5b)----> 6a+6b------> 6(a+b) and therefore, 6 must be a factor of x+y, why its not correct?

AkshdeepS
Bunuel
The positive two-digit 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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Since the digit "a" is at tens place, and digit "b" is at units place, two digit integer ab is written as 10a + b

Example: 13, here a = 1, and b = 3
10a + b = 10(1) + 3 = 13

If it was a 3-digit integer abc, it can be written as 100a + 10b +c

Example: 612, here a = 6, b = 1, and c = 2
100a + 10b +c = 100(6) + 10(1) + 2 = 612

You cannot take any random number n and write it as na + b, it won't hold true.
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The positive two-digit integers x and y have the same digits, but in 23 Feb 2026, 13:51
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Bunuel
The positive two-digit 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
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D




Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

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