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 Post subject: Advanced World Problems & Quant Review p. 32 #25Posted: Tue Aug 03, 2010 4:20 pm
How many integers n greater than 10 and less than 100 are there 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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 Post subject: Re: Advanced World Problems & Quant Review p. 32 #25Posted: Tue Aug 03, 2010 4:56 pm

Joined: Thu Jul 03, 2008 2:13 pm
Posts: 117
Great question - I love this one.

First, let's get an idea of what kind of number would, when you reverse the digits, increase by 9.

One fairly easy way to do that is to just start with numbers above 10 until you find one that catches:

10 --> 01
11 -->11
12 --> 21 we have a match!

You could keep doing that until you find a pattern, but it will start to become clear that it won't work for many numbers:

13 --> 31 ( a difference of 18)
14 --> 41 ( a difference of 27)
15 --> 51 (a difference of 36)

and you may notice a few things - the differences keep growing, so the farther apart the numbers the less likely it is that we'll only have a difference of 9. And the differences are multiples of 9 each time, which is in line with what we're looking for.

It worked when the digits were adjacent to each other, so let's start in the 20s:

21 ---> 12 (difference of 9, but it's -9, not +9)
22 will just stay the same
23 ---> 32 (+9)

Do you see a pattern? If the digits are consecutive, increasing integers, we have a difference of 9, so let's try to prove it moving higher:

34 --> 43 (+9)
45 --> 54 (+9)
56 ---> 65 (+9)
67 ---> 76 (+9)
78 ---> 87 (+9)
89 ---> 98 (+9)

So this pattern works for: 12, 23, 34, 45, 56, 67, 78, and 89. The correct answer, then, is 8, answer choice D.

This is a great example of the "find the pattern" style of number properties problem.

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