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The positive two-digit integers x and y have the same digits, but in

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Bunuel
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The positive two-digit integers x and y have the same digits, but in [#permalink] New post   Updated on: 11 Jul 2021, 05:05
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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

Originally posted by Bunuel on 15 Oct 2015, 21:55.
Last edited by Bunuel on 11 Jul 2021, 05:05, edited 1 time in total.
Edited the question.
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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   16 Oct 2015, 00:00
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Bunuel wrote:
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.

10a+b+10b+a=11(a+b) Answer (D)
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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   17 Oct 2015, 11:13
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Hi All,

This question can be solved by TESTing VALUES:

We're told that X and Y are both 2-digit 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:
Show Spoiler
D


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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   16 Oct 2015, 00:26
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Bunuel wrote:
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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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   16 Oct 2015, 01:00
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10a+b+10b+a
11a+11b
11(a+b)....Ans: 11
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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   18 Oct 2015, 11:05
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WHy not 3,6. 36 and 63 are both divisible by 9.
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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   18 Oct 2015, 12:08
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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.

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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   28 Oct 2015, 13:52
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Bunuel wrote:
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.


Remember: When you take the difference between the two, it will always be 9. e.g 23-32=9, 89-98=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 two-digit integers x and y have the same digits, but in [#permalink] New post   16 Mar 2016, 02:22
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Bunuel wrote:
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.

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 two-digit integers x and y have the same digits, but in [#permalink] New post   02 May 2016, 19:21
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macbookno wrote:
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



Pl post according to guidelines..
Provide OA, proper topic name and post in correct sub-forum..

as for your Q..
let x = ab, a 2-digit 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 two-digit integers x and y have the same digits, but in [#permalink] New post   02 May 2016, 20:09
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macbookno wrote:
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


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 two-digit integers x and y have the same digits, but in [#permalink] New post   03 May 2016, 05:16
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Bunuel wrote:
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.


We can solve this question using the natural relationships that all two-digit 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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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   07 May 2021, 21:46
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Bunuel wrote:
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.


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Answer: Option D

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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   10 Jul 2021, 22:25
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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


This is an interesting question. I think getting that "weird" English in the beginning can be helpful to get a firm idea what's going on.

x is a two-digit positive integer

y is a two-digit positive interger

They are mirrors of each other.

Noting this means that neither can have a zero in it may be helpful.

A few random examples

23 + 32 = 55 ---> Divisible by 11

29 + 92 = 121 ---> Divisible by 11

Click (D) and move on.

P.S. There is an algebra way to solve this that may suit a person more, though.

The 10 multiplied by M thing you may come across is basically to "algebraically" make that digit a ten's digit.

53 is basically ten 5s and one 3, for example.
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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   01 Sep 2021, 05:55
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Bunuel wrote:
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


Solution -

General representation of a two-digit number = 10a + b = X
Number obtained by reversing digits of a number X = 10b + a = Y
Let, ab = 32
3×10+2=X
Now, by reversing the digits, ba =23
10×2+3=Y
Sum of X and Y (X+Y) = (10a+b) + (10b+a) = 11(a+b) => multiple of 11
So, option D is the answer.
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The positive two-digit integers x and y have the same digits, but in [#permalink] New post   24 May 2022, 19:45
X+Y
18+81
27+72
36+63
45+54

They all equal to 99 and are divisible by 9 as well. So why are we certain of 11 being the answer ?

What am I missing :?
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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   25 May 2022, 14:20
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nishithpatnaik wrote:
X+Y
18+81
27+72
36+63
45+54

They all equal to 99 and are divisible by 9 as well. So why are we certain of 11 being the answer ?

What am I missing :?


Hi nishithpatnaik,

The numbers that you've listed as examples all fit a particular pattern: each individual number is divisible by 9. However, the prompt does NOT list that as a requirement when selecting the values of X and Y. If you were to start with the easiest possible values for X and Y though, you'd use 12 and 21 - and with a total of 33, you'd then be able to find the correct answer without needing any other examples.

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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   17 Aug 2022, 09:37
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: The positive two-digit integers x and y have the same digits, but in [#permalink] New post   21 Jan 2024, 15:48
Nice property to note , when 2 digit numbers are reversed i.e. ab and ba

Their sum is always div by 11 (x+y)
their diff is always div by 9 (x-y)
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Re: The positive two-digit integers x and y have the same digits, but in [#permalink]
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