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# Algebra Anyone?

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Eternal Intern
Joined: 07 Jun 2003
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Location: Lone Star State

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10 Jun 2003, 07:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

uestion
I think algebra is the easiest way to solve these: don't you think, I don't know I'm not a tutor.

a) If a 2-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. What is the original number?

b) R is the set of positive odd integers less than 50.

How many are in R ?

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Manager
Joined: 25 Apr 2003
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10 Jun 2003, 20:32
a) I can't solve for a unique solution
b) this is 25

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Manager
Joined: 03 Jun 2003
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Location: Uruguay

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11 Jun 2003, 11:27
I agree with anupag

a) can be 41, 52, 63, 74, 85, 96
b) 25

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Intern
Joined: 04 Feb 2008
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21 Feb 2008, 12:23
(1) 10p+q - (10q+p) = 27. Simplify to get p-q = 3. Any number of solutions possible.
(2) {1, 3, ..., 49}. This can be written as x = 2n-1. For n = 1, x = 1; n=2, x = 3;... n =25, x = 49.

So, that's that...

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Intern
Joined: 11 Aug 2004
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Location: Silicon Valley now Washington D.C.

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21 Feb 2008, 13:26
I agree with all the responses posted.
_________________

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CEO
Joined: 29 Mar 2007
Posts: 2553

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21 Feb 2008, 19:25
Curly05 wrote:
uestion
I think algebra is the easiest way to solve these: don't you think, I don't know I'm not a tutor.

a) If a 2-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. What is the original number?

b) R is the set of positive odd integers less than 50.

How many are in R ?

Please repost A. I get 10x+y - (10y+x) --> x-y=3

B: 50-1=49+1 = 50 thus 50/2 =25 thats how many odds there are.

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SVP
Joined: 28 Dec 2005
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21 Feb 2008, 21:06
agree with the responses.

for a, you cant get a unique solution based on teh information given. the most we can determine is that the digits differ by 3, so you can have 41, 52,30, etc.

for b, half of the integers will be even and half will be odd...

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Re: Algebra Anyone?   [#permalink] 21 Feb 2008, 21:06
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# Algebra Anyone?

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