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12 Apr 2018, 08:25
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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:
95%
(hard)
Question Stats:
25% (01:49) correct
75%
(01:54)
wrong
based on 185
sessions
History
Date
Time
Result
Not Attempted Yet
How many unique values of set (a,b) are possible if a and b are integers and \(a^2-b^2=29\)?
(A) 1
(B) 2
(C) 3
(D) 4
(E) indefinite
(A) 1
(B) 2
(C) 3
(D) 4
(E) indefinite
12 Apr 2018, 10:57
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chetan2u
\(a^2-b^2=29\)
\((a-b)*(a+b)=29\)
since 29 is a prime number... so 29 = 1*29 are the only two prime factors.
Only two possibilities:
\(a+b = 29\)
\(a-b = 1\)
&
\(a+b = 1\)
\(a-b = 29\)
for first on solving a = 15 & b = 14
for second on solving a = 15 & b = -14
Hence we have two unique values for the pair (a,b) : (15,14) & (15,-14)
Hence Option (B) is our answer.
Best,
Gladi
12 Apr 2018, 21:11
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chetan2u
a^2−b^2=29
(a−b)∗(a+b)=29*1
for a-b = 1 and a+b=29, a=15 and b=14
since we are dealing with even power of a and b, both +ve and -ve values are to be considered
so value of (a,b) = (15,14), (15,-14), (-15,14), (-15,-14)
hence 4 values Option D
