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If a and b are different positive integers and a+b=a(a+b)
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14 Sep 2010, 18:46
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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
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Re: DS  Number theory
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14 Sep 2010, 19:19
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! \(a + b = a(a + b)\) > \(a(a+b)(a+b)=0\) > \((a+b)(a1)=0\) > as \(a\) and \(b\) are positive the \(a+b\neq{0}\), so \(a1=0\) > \(a=1\). 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. Answer: E (I and II only). Hope it's clear.
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Re: DS  Number theory
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14 Sep 2010, 19:31
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.



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Re: DS  Number theory
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14 Sep 2010, 19:33



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Re: DS  Number theory
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15 Sep 2010, 11:02
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



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Re: DS  Number theory
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15 Sep 2010, 11:22
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. Hope it helps.
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Re: DS  Number theory
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20 Sep 2010, 13:21
tricky question, i went with A only to realize that both a and b are positive integers, so b must be greater than a



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Re: DS  Number theory
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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
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Re: DS  Number theory
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01 Oct 2010, 23:46
a+b=a(a+b) 1(a+b)a(a+b)=0 (1a)*(a+b)=0 from this either a1=0 or a+b=0, but since a and b are positive their could not be equal to 0. Thus, 1a=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. hence E.
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Re: DS  Number theory
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03 Oct 2010, 03:18
1 is the smallest positive integer. 0 is a signless integer. These two are required.
from the equation we get a=1 as (a+b) cannot be zero. so I. a = 1 always true II. b = 1 not always true III. a < b always true
so the ans is E



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If a and b are different positive integers and a + b = a(a + b),
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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 TRUEAnswer: E
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Re: If a and b are different positive integers and a+b=a(a+b)
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19 Jun 2013, 03:53



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Re: If a and b are different positive integers and a+b=a(a+b)
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12 Mar 2016, 18:38
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 we can rewrite the equation as: (a+b)/a = a+b a/a=1 a+b/a = a+b we can subtract a from both sides b/a=b this is true only when A is 1. since we are told that the numbers are different, and both positive, we know for sure that b MUST be >1, or b>a. E



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Re: If a and b are different positive integers and a+b=a(a+b)
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16 Sep 2017, 01:32
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 a + b = a(a + b) => (a+b)(a1) = 0 => a=b , a =1 a and b are both positive. So, a = b is not possible. a=1, Also a and b are both different positive integers. So, b >1. and hence a<b So, I & II both are correct. Answer E
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Re: If a and b are different positive integers and a+b=a(a+b)
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02 Nov 2018, 16:06
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 Since a and b are positive integers, a + b ≠ 0. So we can divide the given equation by a + b and obtain: 1 = a Therefore, Roman numeral I is true. Furthermore, since a and b are different positive integers, b > a. So Roman numeral III is also true. Answer: E
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