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Applied Algebra

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Intern
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Joined: 23 May 2008
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Applied Algebra [#permalink]

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New post 22 Sep 2008, 12:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I got this answer correct by guess and check but I would like to understand the algebra behind it.

Q. 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? (OG 183)

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Director
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Re: Applied Algebra [#permalink]

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New post 22 Sep 2008, 12:46
snowbirdskier wrote:
I got this answer correct by guess and check but I would like to understand the algebra behind it.

Q. 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? (OG 183)


let x and y be the digits.

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

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Re: Applied Algebra [#permalink]

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New post 22 Sep 2008, 14:06
Can you explain (10x + y) - (10y + x) = 27 please?

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Re: Applied Algebra [#permalink]

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New post 22 Sep 2008, 14:20
snowbirdskier wrote:
Can you explain (10x + y) - (10y + x) = 27 please?


if x is the 10s digit and y is the units digit, then any two digit number is 10x + y.

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Re: Applied Algebra [#permalink]

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New post 22 Sep 2008, 14:26
snowbirdskier wrote:
Can you explain (10x + y) - (10y + x) = 27 please?


say two digit number XY (e.g 74)
XY = 10*X+1*Y (X- tenth digit number, Y- unit digit number)
reverse number =YX
YX=(10y + x)

XY-YX=(10x + y) - (10y + x) =27
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Kudos [?]: 1032 [0], given: 5

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Kudos [?]: 93 [0], given: 0

Re: Applied Algebra [#permalink]

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New post 22 Sep 2008, 14:58
Gotcha, thank you both.

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Re: Applied Algebra   [#permalink] 22 Sep 2008, 14:58
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