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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
(1) a is a prime.
(2) b^2 is a multiple of 4.
Source: My own
blueseas
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mau5
If a,b,c are different positive integers,and \(a^2+b^2 = c^2\), then what is the value of \((c-b)^2\)?
I.a is a prime.
II.\(b^2\) is a multiple of 4.
I.a is a prime.
II.\(b^2\) is a multiple of 4.
\(a^2+b^2 = c^2\)
\(a^2= c^2-b^2 = (c-b)(c+b)\)
statement1:
\(a\) is prime
therefore a = 1*a = (a cant have any other factor)
\(a^2= a*a = (c-b)(c+b)\) ==>since a is prime \(c-b\) should be \(1\) and \(c+b\) should be some perfect square.(as prime number factors are 1 and prime no.itself)
hence clearly \(c-b=1\)...hence sufficient
statement 2:
\(b^2\) is a multiple of 4
let \(b=4 c=5 a=3\)....therefore \(c-b = 1\)
another case : let\(b=8 c=10 a=6\)...therefore \(c-b=2\)
hence insufficient
hence it is A
03 Aug 2013, 08:16
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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 \((c-b)^2\)?
\(a^2=c^2-b^2=(c-b)(c+b)\). Because \(c\neq{b},c-b\neq{0}\) we get \(c+b=\frac{a^2}{c-b}\).
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 \(c+b=integer=\frac{a^2}{c-b}\). That is true if c-b is a factor of \(a^2\):1, a or \(a^2\). If \(c-b=a^2\) then c+b=1 but this cannot be true as c and b are positive integers.
So we are left with \(c-b=a\) or \(c-b=1\). In the first case \(c-b=a\), substituting in the above equation we find that \((c-b)^2+b^2=c^2\) or \(2b(b-c)=0\), so b=0 (because \(c\neq{b}\)) but this contradicts the statement (b is a positive integer)
We are left with \(c-b=1\), so \((c-b)^2=1\).
Sufficient
II.\(b^2\) is a multiple of 4.
\(a^2+b^2 = c^2,3^2+4^2=5^2\) and \((c-b)^2=1\)
\(a^2+b^2 = c^2,(3*2)^2+(4*2)^2=(5*2)^2,=6^2+8^2=10^2\) and \((c-b)^2=4\). Not sufficient.
\(a^2=c^2-b^2=(c-b)(c+b)\). Because \(c\neq{b},c-b\neq{0}\) we get \(c+b=\frac{a^2}{c-b}\).
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 \(c+b=integer=\frac{a^2}{c-b}\). That is true if c-b is a factor of \(a^2\):1, a or \(a^2\). If \(c-b=a^2\) then c+b=1 but this cannot be true as c and b are positive integers.
So we are left with \(c-b=a\) or \(c-b=1\). In the first case \(c-b=a\), substituting in the above equation we find that \((c-b)^2+b^2=c^2\) or \(2b(b-c)=0\), so b=0 (because \(c\neq{b}\)) but this contradicts the statement (b is a positive integer)
We are left with \(c-b=1\), so \((c-b)^2=1\).
Sufficient
II.\(b^2\) is a multiple of 4.
\(a^2+b^2 = c^2,3^2+4^2=5^2\) and \((c-b)^2=1\)
\(a^2+b^2 = c^2,(3*2)^2+(4*2)^2=(5*2)^2,=6^2+8^2=10^2\) and \((c-b)^2=4\). Not sufficient.
