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# Applied Algebra

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Intern
Joined: 23 May 2008
Posts: 48

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22 Sep 2008, 11:16
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)
Director
Joined: 12 Jul 2008
Posts: 514
Schools: Wharton

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22 Sep 2008, 11:46
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
Intern
Joined: 23 May 2008
Posts: 48

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22 Sep 2008, 13:06
Can you explain (10x + y) - (10y + x) = 27 please?
Director
Joined: 12 Jul 2008
Posts: 514
Schools: Wharton

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22 Sep 2008, 13:20
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.
SVP
Joined: 07 Nov 2007
Posts: 1789
Location: New York

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22 Sep 2008, 13:26
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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Intern
Joined: 23 May 2008
Posts: 48

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22 Sep 2008, 13:58
Gotcha, thank you both.
Re: Applied Algebra   [#permalink] 22 Sep 2008, 13:58
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# Applied Algebra

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