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If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]
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03 Aug 2013, 06:56
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If a, b, c are different positive integers, and a^2 + b^2 = c^2, then what is the value of (c  b)^2? (1) a is a prime. (2) b^2 is a multiple of 4. Source: My own
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Re: If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]
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Updated on: 03 Aug 2013, 08:21
mau5 wrote: If a,b,c are different positive integers,and \(a^2+b^2 = c^2\), then what is the value of \((cb)^2\)?
I.a is a prime.
II.\(b^2\) is a multiple of 4. \(a^2+b^2 = c^2\) \(a^2= c^2b^2 = (cb)(c+b)\) statement1: \(a\) is prime therefore a = 1*a = (a cant have any other factor) \(a^2= a*a = (cb)(c+b)\) ==>since a is prime \(cb\) should be \(1\) and \(c+b\) should be some perfect square.(as prime number factors are 1 and prime no.itself) hence clearly \(cb=1\)...hence sufficient statement 2: \(b^2\) is a multiple of 4 let \(b=4 c=5 a=3\)....therefore \(cb = 1\) another case : let\(b=8 c=10 a=6\)...therefore \(cb=2\) hence insufficient hence it is A
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Originally posted by blueseas on 03 Aug 2013, 08:15.
Last edited by blueseas on 03 Aug 2013, 08:21, edited 1 time in total.



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Re: If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]
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03 Aug 2013, 08:16
If a,b,c are different positive integers,and \(a^2+b^2 = c^2\), then what is the value of \((cb)^2\)? \(a^2=c^2b^2=(cb)(c+b)\). Because \(c\neq{b},cb\neq{0}\) we get \(c+b=\frac{a^2}{cb}\). I.a is a prime.If a is prime, then \(a^2\) has three factors: 1,a and \(a^2\). Because c and b are integers, their sum must be an integer so \(c+b=integer=\frac{a^2}{cb}\). That is true if cb is a factor of \(a^2\):1, a or \(a^2\). If \(cb=a^2\) then c+b=1 but this cannot be true as c and b are positive integers. So we are left with \(cb=a\) or \(cb=1\). In the first case \(cb=a\), substituting in the above equation we find that \((cb)^2+b^2=c^2\) or \(2b(bc)=0\), so b=0 (because \(c\neq{b}\)) but this contradicts the statement (b is a positive integer) We are left with \(cb=1\), so \((cb)^2=1\). Sufficient II.\(b^2\) is a multiple of 4.\(a^2+b^2 = c^2,3^2+4^2=5^2\) and \((cb)^2=1\) \(a^2+b^2 = c^2,(3*2)^2+(4*2)^2=(5*2)^2,=6^2+8^2=10^2\) and \((cb)^2=4\). Not sufficient.
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Re: If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]
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25 Mar 2015, 12:10
Hi All, While the math involved in this question is certainly algebrabased, the patterns are rooted more in Geometry (specifically, the Pythagorean Theorem). You can use that knowledge to your advantage here. We're told that A, B and C are DIFFERENT positive integers and that A^2 + B^2 = C^2. We're asked for the value of (CB)^2. Since the variables are INTEGERS, and if you recognize the Pythagorean Theorem, you might choose to jot down a few examples.... eg 3/4/5 5/12/13 And the multiples of those Triplets.... Fact 1: A is PRIME. Right away, we have a couple of TESTs that we can run.... IF... A = 3, B = 4, C = 5 The answer to the question is .... (CB)^2 = (54)^2 = 1 IF... A = 5, B = 12, C = 13 The answer to the question is .... (CB)^2 = (1312)^2 = 1 Since we only have two TESTs so far, I'd look to also TEST a rarer Pythagorean Triplet.... IF.... A = 7, B = 24, C = 25 The answer to the question is... (CB)^2 = (2524)^2 = 1 At this point, a pattern clearly exists. The answer to the question is ALWAYS 1. Fact 1 is SUFFICIENT Fact 2: B^2 is a multiple of 4 IF... A = 3, B = 4, C = 5 The answer to the question is .... (CB)^2 = (54)^2 = 1 IF... A = 6, B = 8, C = 10 The answer to the question is .... (CB)^2 = (108)^2 = 4 Fact 2 is INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]
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09 Aug 2017, 18:49
a, b & c =+ve distinct integers a²+b²=c² or a²=c²b²=(cb)(c+b) 1) a is prime plugin a=3 a²=9 9=(cb)(c+b) now 9 is the product of (cb)(c+b) so it can be 1X9 or 9x1 or 3x3 cb will be always less than c+b (sum of 2 positive distinct no will always be greater than their difference) hence cb can only be 1 and (cb)²=(1)²sufficient. 2) b² is multiple of 4 multiple values are possible b²=4, 8, 16..... b²=(ca)(c+a)=1x4 or1x8 or 2x4 we can stop hereNot sufficient




Re: If a, b, c are different positive integers, and a^2+b^2 = c^
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09 Aug 2017, 18:49






