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15 Jun 2021, 05:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
55% (02:25) correct
45%
(02:08)
wrong
based on 301
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If a and b are positive integers such that \(a^2 + 2b = b^2 + 2a + 5\), find the value of b.
(A) 0
(B) 1
(C) 2
(D) 3
(E) Cannot be determined
(A) 0
(B) 1
(C) 2
(D) 3
(E) Cannot be determined
15 Jun 2021, 09:22
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My approach :
Please find attached image, it was too much to type.
Posted from my mobile device
Please find attached image, it was too much to type.
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15 Jun 2021, 10:25
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Bunuel
\(a^2\) and \(b^2\) in the equation, look if you can use \(a^2-b^2\). Most of the time, the answer will be yes.
\(a^2 + 2b = b^2 + 2a + 5\)
\(a^2 -b^2 = 2a -2b+ 5\)
\((a-b)(a+b)=2(a-b)+5\)
\((a-b)(a+b)-2(a-b)=5\)
\((a-b)(a+b-2)=5=1*5\)
As a and b are positive integers, \(a+b\geq 2\), so \(a+b-2\geq{0}\). This means both (a-b) and (a+b-2) must be some positive integers too. and only possibility that their product is 5 is 1 and 5.
1) a-b=1 and a+b-2=5
Add the two => 2a=5+2+1=8...a=4, and b=3.
2) a-b=5 and a+b-2=1
Add the two => 2a=5+2+1=8...a=4, and b=a-5=-1, but b is positive...Not possible.
Thus b=3.
D
