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

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08 Jan 2014, 21:10

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

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15 Feb 2015, 10:42

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The "math" behind this question is such that there are several different numbers that "fit" the given "restrictions"; in that way, we can use a bit of "brute force" to quickly come up with the answer.

We're told that a two-digit positive integer will differ from its "reverse" by 27.

I'm going to 'play' with this math a bit.....

IF we use.... 12 and 21 then the difference is 21-12 = 9 This is NOT a match 13 and 31 then the difference is 31-13 = 18 This is NOT a match. Notice how the difference increases by 9 though!!!!! 14 and 41 then the difference is 41-14 = 27 This IS a match.

The question asks for the difference in the two digits involved. Using these values, the answer is 4-1 = 3

I mentioned at the beginning that there were several numbers that "fit" what this question was asking for. They are 14 and 41, 25 and 52, 36 and 63, 47 and 74, 58 and 85, 69 and 96.

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

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09 Jun 2016, 13:19

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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

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

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18 May 2017, 08:09

I find this problem simple and a little confusing.

So we have here ab – ba = 27 Let’s say ab = N, We can formulate it as N = 10a + b, a – is tenths digit, and b – unit digit. For example: N = 4*10 +5 = 45 Now reversed number will be as follows:

10b+a

we know: ab – ba = 27 => 10a + b – (10b +a) = 27 10a + b – 10b – a = 27 9a – 9b = 27 a – b = 3. => two digits differ by 3.

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

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04 Sep 2017, 04:02

ScottTargetTestPrep wrote:

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.