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If a two-digit positive integer has its digits reversed, the resulting

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If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post Updated on: 07 Feb 2019, 01:04
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If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

Originally posted by salr15 on 14 Apr 2007, 17:54.
Last edited by Bunuel on 07 Feb 2019, 01:04, edited 3 times in total.
Edited the question.
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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 27 Dec 2012, 08:31
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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 14 Apr 2007, 19:33
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A?

Let x be the tenth digit and y be the units digit.

Then, the original number is 10x+y and the reversed number is 10y+x.

The difference between the two is 27. (reversed - original)

Thus, (10y+x)-(10x+y)=27
solve the above eq.
10y+x-10x-y=27
9y-9x=27
9(y-x)=27
y-x=3

Thus the two digits differ by 3.

:)
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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 12 Oct 2012, 09:32
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This is a tricky number properties question. Note that you can use two variables a and b to represent each of the digits.

In terms of expressing them in values - the total value would be (10a + b).

For example, for a number 37, a = 3 and b=7..then the expression 10a+b = 30 + 7 = 37.

Please refer to the video explanation here:
http://www.gmatpill.com/gmat-practice-t ... stion/2397

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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 13 Oct 2013, 14:45
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Bunuel wrote:
Walkabout wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


Given that (10a + b) - (10b + a) = 27 --> 9a - 9b =27 --> a - b = 3.

Answer: A.


Do we use 10 because of the tens digit??
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New post 13 Oct 2013, 14:49
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runningguy wrote:
Bunuel wrote:
Walkabout wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


Given that (10a + b) - (10b + a) = 27 --> 9a - 9b =27 --> a - b = 3.

Answer: A.


Do we use 10 because of the tens digit??


Yes, any two-digit integer ab can be expressed as 10a+b, for example: 45 = 10*4 + 5.

The same for 3, 4, 5, ... digit numbers. For example, 4-digit number 5,432 can be written as 5*1,000 + 4*100 + 3*10 + 2 = 5,432.

Hope it's clear.
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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 12 Jan 2014, 15:28
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Walkabout wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7



The answer has to be a factor of 27, the only option that's a factor of 27 is 3.

Sice \((10x + y) - (10y + x) = 27\), you can simplify this relationship by subtracting with a common factor --> 9x - 9y = 27 ---> 9(x - y) = 27 ---> here, you already notice that the difference has to be a factor of both 9 and 27, but you can simplify further ---> x - y = 3, and thus we have the answer.

But these last steps are superfluous if you already notice that the answer has to be a factor of 27, this way you save time without having to calculate.
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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 18 Feb 2015, 22:12
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Hi All,

Even though this question might seem a little strange, you do NOT need to do any excessive math to get to the correct answer. With just a bit of 'playing around' you can use 'brute force' to get to the answer.

We're told that a 2-digit number has its digits reversed and the difference between those two numbers is 27.

IF we use....
11 and 11, then the difference is 0 - this is NOT a match

12 and 21, then the difference = 9 - this is NOT a match

13 and 31, then the difference = 18 - this is NOT a match (notice the pattern though? The difference keeps increasing by 9!!!!! I wonder what the next one will be???)

14 and 41, then the difference = 27 = this IS a match

The question asks for the difference in the two DIGITS. The difference between 1 and 4 is 3.

Final Answer:

There are actually several ways to get to this answer: 14 and 41, 25 and 52, 36 and 63, 47 and 74, 58 and 85, 69 and 96.

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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 15 Jul 2016, 04:31
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salr15 wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


Let’s first label the original two-digit integer as N. We can then say that N = 10A + B, where A is the tens digit and B is the units digit of N.

If this is hard to see let’s try it with a sample number, say 24. We can say the following:

24 = (2 x 10) + 4

24 = 20 + 4

24 = 24

Getting back to the problem, we are given that if the integer N has its digits reversed the resulting integer differs from the original by 27. First let’s express the reversed number in a similar fashion to the way in which we expressed the original integer.

10B + A = reversed integer

Since we know the resulting integer differs from the original by 27 we can say:

10B + A – (10A + B) = 27

10B + A – 10A – B = 27

9B – 9A = 27

B – A = 3

Since B is the tens digit and A is the units digit, we can say that the digits differ by 3.

The answer is A.
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If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 20 Feb 2019, 10:01
salr15 wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


the difference between reversed integers is always a multiple of 9
dividing that difference by 9 gives the difference between digits
27/9=3
A
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If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 12 May 2019, 22:39
For two-digit reversal questions:

Property-
n and (n+1) will differ by 9
n and (n+2) will differ by 18
n and (n+3) will differ by 27..

n and (n+k) will differ by k*9

Always in multiples of 9.
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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 13 May 2019, 04:06
salr15 wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


(10x+y)-(10y+x)=27
=>9(x-y)=27
=>x-y=3

Answer is A

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Re: If a two-digit positive integer has its digits reversed, the resulting  [#permalink]

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New post 16 Sep 2019, 07:52
salr15 wrote:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?

Let the 2-digit number be xy

|(10x+y) - (10y-x)| = 27
|9x-9y| =27
|x-y| = 3

IMO A
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Re: If a two-digit positive integer has its digits reversed, the resulting   [#permalink] 16 Sep 2019, 07:52
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