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If a two-digit positive integer has its digits reversed...

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by kminnesota14 » Wed Sep 12, 2012 8:42 am
I need a better explanation than the OG is giving me on how they came up with the answer to the following question:

If a two-digit positive integer has its digits reveresed the resulting integer differs from the original by 27. By how much do the two digits differ? (OG 13th Ed, pg 182)

A 3
B 4
C 5
D 6
E 7
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by GMATGuruNY » Wed Sep 12, 2012 8:56 am
kminnesota14 wrote:I need a better explanation than the OG is giving me on how they came up with the answer to the following question:

If a two-digit positive integer has its digits reveresed the resulting integer differs from the original by 27. By how much do the two digits differ? (OG 13th Ed, pg 182)

A 3
B 4
C 5
D 6
E 7
Let T = the tens digit and U = the units digit.

Original integer = 10T + U.
To illustrate:
53 = 10(5) + 3.

The original integer with the digits reversed = 10U + T.

Since the difference is 27, we get:
(10T + U) - (10U + T) = 27.
9T - 9U = 27.
9(T-U) = 27.
T-U = 3.

The correct answer is A.
Last edited by GMATGuruNY on Wed Sep 12, 2012 12:42 pm, edited 1 time in total.
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by kminnesota14 » Wed Sep 12, 2012 11:29 am
Thank you! I think the 10 was where I was confused but that is related to the place of the number! Got it!
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by jervizeloy » Thu Sep 08, 2016 10:09 am
GMATGuruNY wrote:
kminnesota14 wrote:I need a better explanation than the OG is giving me on how they came up with the answer to the following question:

If a two-digit positive integer has its digits reveresed the resulting integer differs from the original by 27. By how much do the two digits differ? (OG 13th Ed, pg 182)

A 3
B 4
C 5
D 6
E 7
Let T = the tens digit and U = the units digit.

Original integer = 10T + U.
To illustrate:
53 = 10(5) + 3.

The original integer with the digits reversed = 10U + T.

Since the difference is 27, we get:
(10T + U) - (10U + T) = 27.
9T - 9U = 27.
9(T-U) = 27.
T-U = 3.

The correct answer is A.
I know this is an old post but I have a small question.

I fully understand how you solved this exercise, in fact, I did it in the same way the first time I tried it. However, now that I'm doing some review I came up with another way of solving it but it seems to be wrong and I dont get why.

Let AB be the original number. Applying the place digit concept we have that it can be expressed as 10A + B, thus, its reversed version would be 10B + A. Given that their difference is 27, we can state that:

10B + A - (10A + B) = 27

If we group the terms in the tens and the terms in the units we have:

10 (B-A) + (A-B) = 27

Thus,

B - A = 2, and
A - B = 7

At this point I reach a system of equations I don't quite understand. This solution is obviously flawed but I can't see why. Hope you can help me.
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by [email protected] » Thu Sep 08, 2016 3:16 pm
Hi jervizeloy,

The pattern that you're trying to use only works when dealing with individual digits; you're trying to use the DIFFERENCE of two digits.

As proof, when considering the terms (A-B) and (B-A), if A and B are not equal, then one of those terms will be POSITIVE and the other will be NEGATIVE.

For example:
7 - 4 = 3
4 - 7 = -3

However, you cannot have a 'negative digit', so you cannot approach the math in this way.

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by regor60 » Wed Sep 14, 2016 8:58 am
jervizeloy wrote:
GMATGuruNY wrote:
kminnesota14 wrote:I need a better explanation than the OG is giving me on how they came up with the answer to the following question:

If a two-digit positive integer has its digits reveresed the resulting integer differs from the original by 27. By how much do the two digits differ? (OG 13th Ed, pg 182)

A 3
B 4
C 5
D 6
E 7
Let T = the tens digit and U = the units digit.

Original integer = 10T + U.
To illustrate:
53 = 10(5) + 3.

The original integer with the digits reversed = 10U + T.

Since the difference is 27, we get:
(10T + U) - (10U + T) = 27.
9T - 9U = 27.
9(T-U) = 27.
T-U = 3.

The correct answer is A.
I know this is an old post but I have a small question.

I fully understand how you solved this exercise, in fact, I did it in the same way the first time I tried it. However, now that I'm doing some review I came up with another way of solving it but it seems to be wrong and I dont get why.

Let AB be the original number. Applying the place digit concept we have that it can be expressed as 10A + B, thus, its reversed version would be 10B + A. Given that their difference is 27, we can state that:

10B + A - (10A + B) = 27

If we group the terms in the tens and the terms in the units we have:

10 (B-A) + (A-B) = 27

Thus,

B - A = 2, and
A - B = 7

At this point I reach a system of equations I don't quite understand. This solution is obviously flawed but I can't see why. Hope you can help me.
By doing this, you're essentially arbitrarily "assigning" these values instead of solving the equation, so you shouldn't be surprised that it is not correct.

In other words, 20 + 7 "works" in that it equals the desired 27, but neither have anything to do with A-B or B-A.
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by Scott@TargetTestPrep » Thu Sep 15, 2016 8:10 am
kminnesota14 wrote: If a two-digit positive integer has its digits reveresed the resulting integer differs from the original by 27. By how much do the two digits differ? (OG 13th Ed, pg 182)

A 3
B 4
C 5
D 6
E 7
Let's 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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by MartyMurray » Thu Sep 15, 2016 10:12 am
Scoring high on GMAT quant can be done with via using of the box thinking to get answers any way you can.

So, if you were taking the test and didn't immediately see the algebraic method of getting the answer to this question, an alternative way to get the answer would be to quickly hack your way to one example.

Then, since the difference between the values of the digits in any one example that works has to be the same as the difference in all cases, the difference between the digits in the example that you hacked to is the answer.

To get the hack going, you could just try one number to see what you generate by adding it to 27.

53 + 27 = 80 Not even close. The seven added to the ones digit has to either create a high digit that matches the tens digit, such as 8, or a low digit that matches the tens digit, such as 2.

Try an 8 in the tens column.

81 + 27 = 109. Nope. So the highest possible tens digit of a number that when added to 27 creates a two digit number is 7, but 7 won't work either, because then you would need a 0 in the ones place, and you can't have a 0 in the tens place.

Closing in. There are actually not that many possibilities left.

Try a 6 in the tens place and a 9 in the ones place, because 9 + 7 generates a 6 in the ones place.

69 + 27 = 96 Bingo.

The difference between the digits of 69 is 3.

Yup, the algebraic method is better for getting the answer to this question, but taking a think fast and GET IT DONE attitude works pretty well too.

The correct answer is A.
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by Matt@VeritasPrep » Thu Sep 15, 2016 6:37 pm
Marty Murray wrote:Scoring high on GMAT quant can be done with via using of the box thinking to get answers any way you can.
Sort of ... but you still need to have the interior of the box pretty well mapped out!
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by Matt@VeritasPrep » Thu Sep 15, 2016 6:48 pm
regor60 wrote:By doing this, you're essentially arbitrarily "assigning" these values instead of solving the equation, so you shouldn't be surprised that it is not correct.

In other words, 20 + 7 "works" in that it equals the desired 27, but neither have anything to do with A-B or B-A.
No, he should be very surprised if it's incorrect, since it's conceptually sound. The issue here is the arithmetic, I think.

If BA - AB = 27, then we've got

(10B + A) - (10A + B) = 27, or

9B - 9A = 27, or

B - A = 3

Since B is one digit and A is the other, this tells that we'll always get this result if B is three greater than A. Checking a few numbers to see this for ourselves, we get

A = 1, B = 4, BA - AB = 41 - 14 = 27
A = 2, B = 5, BA - AB = 52 - 25 = 27
A = 3, B = 6, BA - AB = 63 - 36 = 27

etc.
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by Matt@VeritasPrep » Thu Sep 15, 2016 6:51 pm
jervizeloy wrote:10B + A - (10A + B) = 27

If we group the terms in the tens and the terms in the units we have:

10 (B-A) + (A-B) = 27

Thus,

B - A = 2, and
A - B = 7

At this point I reach a system of equations I don't quite understand. This solution is obviously flawed but I can't see why. Hope you can help me.
I like the idea, but there's one annoying detail left! You don't have to have 10*(B-A) = 2 and (A-B) = 7, since (A-B) could be negative. For instance, suppose B = 4 and A = 1. In that case, BA - AB does equal 27, but 10*(B-A) = 30, not 20. This is because (A-B) is negative, so we when we add it to 10(B-A), we end up subtracting:

10*(4 - 1) + (1 - 4) = 27
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by regor60 » Fri Sep 16, 2016 8:05 am
Matt@VeritasPrep wrote:
regor60 wrote:By doing this, you're essentially arbitrarily "assigning" these values instead of solving the equation, so you shouldn't be surprised that it is not correct.

In other words, 20 + 7 "works" in that it equals the desired 27, but neither have anything to do with A-B or B-A.
No, he should be very surprised if it's incorrect, since it's conceptually sound. The issue here is the arithmetic, I think.

If BA - AB = 27, then we've got

(10B + A) - (10A + B) = 27, or

9B - 9A = 27, or

B - A = 3

Since B is one digit and A is the other, this tells that we'll always get this result if B is three greater than A. Checking a few numbers to see this for ourselves, we get

A = 1, B = 4, BA - AB = 41 - 14 = 27
A = 2, B = 5, BA - AB = 52 - 25 = 27
A = 3, B = 6, BA - AB = 63 - 36 = 27

etc.
The approach is sound if she😉 follows through with the arithmetic but it appeared to me she just assigned 20&7 because they appeared to work.

In other words, treating B-A and A-B as if they were independent X and Y where in reality X=-Y
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by Matt@VeritasPrep » Fri Sep 16, 2016 2:23 pm
regor60 wrote: The approach is sound if she😉 follows through with the arithmetic but it appeared to me she just assigned 20&7 because they appeared to work.

In other words, treating B-A and A-B as if they were independent X and Y where in reality X=-Y
Meh, not really. jervizeloy (who doesn't look like a she in that (albeit small and hard to see) avatar photo, and has an Instagram by that name that I just Googled that doesn't look like a woman's, so I'm sticking with 'he', but my apologies if I'm wrong) reached the conclusion

10*(B - A) + 1*(A - B) = 10*2 + 1*7

and not unintuitively concluded that

10*(B - A) = 10*2

and

1*(A - B) = 1*7

That's a subtle enough mistake that's worth elucidating: it's very teachable.
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