n reversed

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n reversed

by krishnasty » Mon Aug 01, 2011 12:58 am
How many integers n greater than 10 and less than 100 are such that, if the digits of n are reversed, the resulting integer is n+9 ?

(a) 5
(b) 6
(c) 7
(d) 8
(e) 9
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by GMATGuruNY » Mon Aug 01, 2011 2:21 am
krishnasty wrote:How many integers n greater than 10 and less than 100 are such that, if the digits of n are reversed, the resulting integer is n+9 ?

(a) 5
(b) 6
(c) 7
(d) 8
(e) 9
Let T = tens digit and U = units digit.
N = 10T + U.
When the digits are reversed, New N = 10U + T.

Since the difference between new N and old N is 9, we get:
(10U + T) - (10T + U) = 9.
9U - 9T = 9
9(U-T) = 9
U-T = 1
U = T+1.

Thus, the units digit of N is 1 more than the tens digit, yielding the following options:
12, 23, 34, 45, 56, 67, 78, 89.
8 integers.

The correct answer is D.
Last edited by GMATGuruNY on Mon Aug 01, 2011 11:17 am, edited 2 times in total.
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by MBA.Aspirant » Mon Aug 01, 2011 7:00 am
let t be tens digit and u unit digit

10t + u = 10u + t +9

9t = 9u +9

t = u+1,

21 = 12 +9

32 = 23 +9

43

54

65

76

87

98 = 89 +9

total 8

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by cuteprince1989 » Mon Nov 07, 2016 1:32 am
he is asking for the integer not the 2 digit number so why did not u include '01'so that makes the answer 9,
01 also satisfies the condition given inthe question.

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by Brent@GMATPrepNow » Mon Nov 07, 2016 4:43 am
cuteprince1989 wrote:he is asking for the integer not the 2 digit number so why did not u include '01'so that makes the answer 9,
01 also satisfies the condition given inthe question.
The question limits the integers to those greater than 10 ("How many integers n greater than 10....")
So, n = 01 does not count.

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by Scott@TargetTestPrep » Tue Nov 08, 2016 4:19 pm
krishnasty wrote:How many integers n greater than 10 and less than 100 are such that, if the digits of n are reversed, the resulting integer is n+9 ?

(a) 5
(b) 6
(c) 7
(d) 8
(e) 9
Since n is greater than 10 and less than 100, n is a two-digit number. To start, we can represent n as 10A + B, in which A = the tens digit of n and B = the units digit of n.

Since we are given that when the digits of n are reversed the resulting integer is n + 9, we can say:

10A + B + 9 = 10B + A

9A - 9B + 9 = 0

A - B + 1 = 0

A = B - 1

Our possible values of n are such that the tens digit is one less than the units digit. Thus n can be the following values:

12, 23, 34, 45, 56, 67, 78, 89

Answer: D

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by Matt@VeritasPrep » Fri Nov 11, 2016 3:11 pm
I'm a little surprised that nobody has pointed out that anyone asking this question isn't likely to think of the algebraic approach at first.

With that in mind, under test conditions, TRY NUMBERS and look for a pattern.

Start with n = 24. Reversed, that gives 42, and 42 ≠ 24 + 9.

If we try n = 25, things get worse! Reversed we've got 52, and the distance is even further.

That means we should go the other way: n = 23. Aha! Success.

It seems that this works when the numbers are close together, so let's test to be sure. n = 34 works (reverse is 43), n = 45 works (reverse is 54), etc. etc.

So it should always work if the tens digit is one more than the units digit. That gives us 12, 23, ..., 89, and we're set.

This approach always helps when you can't think of the algebra, and since the GMAT is an adaptive test, you'll often get questions hard enough for you that you won't be able to think of the algebra! Have a backup plan, and you can work yourself out of the jam.