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Re: DS - Number theory [#permalink]
14 Sep 2010, 19:19
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hemanthp wrote:
If a and b are different positive integers and a + b = a(a + b), then which of the following must be true? I. a = 1 II. b = 1 III. a < b
A. I only B. II only C. III only D. I and II E. I and III
This is slightly tricky. Watch out! Don't get angry if your answer didn't match the OA. I want to have a discussion around this. ---- Kudos if you like this!
\(a + b = a(a + b)\) --> \(a(a+b)-(a+b)=0\) --> \((a+b)(a-1)=0\) --> as \(a\) and \(b\) are positive the \(a+b\neq{0}\), so \(a-1=0\) --> \(a=1\). Also as \(a\) and \(b\) are different positive integers then \(b\) must be more than \(a=1\) --> \(a<b\) (\(b\) can not be equal to \(a\) as they are different and \(b\) can not be less than \(a\) as \(b\) is positive integerand thus can not be less than 1).
So we have that: \(a=1\) and \(a<b\).
I. a = 1 --> true; II. b = 1 --> not true; III. a < b --> true.
Re: DS - Number theory [#permalink]
14 Sep 2010, 19:31
2
This post received KUDOS
Awesome! This was the exact point I wanted to bring up. And I was hoping for Bunuel to reply to the post. Thanks dude.
A and B are distinct positive integers. Why can't B be 0? Isn't 0 a positive integer? It sure as hell isn't negative. And indeed WIKI says "An integer is positive if it is greater than zero and negative if it is less than zero. Zero is defined as neither negative nor positive."
So per GMAT ..is zero positive or neither positive or negative?
Re: DS - Number theory [#permalink]
14 Sep 2010, 19:33
Expert's post
hemanthp wrote:
Awesome! This was the exact point I wanted to bring up. And I was hoping for Bunuel to reply to the post. Thanks dude.
A and B are distinct positive integers. Why can't B be 0? Isn't 0 a positive integer? It sure as hell isn't negative. And indeed WIKI says "An integer is positive if it is greater than zero and negative if it is less than zero. Zero is defined as neither negative nor positive."
So per GMAT ..is zero positive or neither positive or negative?
Thank you again ..Bunuel.
Zero is neither positive not negative integer. _________________
Re: DS - Number theory [#permalink]
15 Sep 2010, 11:22
Expert's post
TomB wrote:
I pick numbers for this problem
a+b = a(a+b). to make the statement true a=1,b=2 1+2=1(1+2) 3 =3.
from this we know ,a=1 & a<b. answer should be E
Number picking might not be the best way to solve MUST BE TRUE questions.
The question asks which of the following MUST be true, or which of the following is ALWAYS true no matter what set of numbers you choose. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.
So the set you chose just proves that II is not always true and hence it's not a part of a correct choice. As for I and III: they might be true for this particular set of numbers but not true for another set, so you can not say that I and III are always true just based on one set of numbers (it just happens to be that I and III are always true).
As for "COULD BE TRUE" questions: The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.
Re: DS - Number theory [#permalink]
20 Sep 2010, 17:48
Tricky one indeed. I marked E but was wrong in solutioning. I thought a=-b means a is smaller than b. Didn't realise that this equation is not possible since both a and b are positive _________________
If you like my post, consider giving me some KUDOS !!!!! Like you I need them
Re: DS - Number theory [#permalink]
01 Oct 2010, 23:46
a+b=a(a+b) 1(a+b)-a(a+b)=0 (1-a)*(a+b)=0 from this either a-1=0 or a+b=0, but since a and b are positive their could not be equal to 0. Thus, 1-a=0 or a=1. 1. holds true 2.not true because a and b are different positive integers and we know that a=1 3. true- As a, b are "different positive integers" when 1 is the smallest , thus a<b.
If a and b are different positive integers and a + b = a(a + b), [#permalink]
12 Dec 2012, 18:32
\(a+b = a (a+b)\) \(\frac{a+b}{a+b}=a\) \(a=1\)
I. a = 1 always TRUE II. b = 1 b must not be the same as a. Thus, b is not equal to 1. FALSE III. a < b b is a positive integer but not equal to 1 then it must be 2 onwards. always TRUE
Re: If a and b are different positive integers and a+b=a(a+b) [#permalink]
11 Feb 2015, 07:38
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