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factorization

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Source: — Problem Solving |

by adthedaddy » Fri Aug 10, 2012 3:49 am
Can you please post the OA ?
This will help.
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by manihar.sidharth » Fri Aug 10, 2012 6:11 am
Here's something I got after 20 mins of wasted effort and after that copying it from google
Equate x coefficients
b + c = 1991+a
bc = 1991a+1
Eliminate a
1991(b + c) - bc = 3964080
Let c = b + k, then
b² + (k-3982)b - 1991k + 3964080 = 0
with solutions
b = -k/2 + 1991+ (1/2)√(k²+4),
b = -k/2 +1991 - (1/2)√(k²+4)
Since √(k²+4) is an integer only if k = 0, the only two solutions are
b = c = 1992, a = 1993
b = c = 1990, a = 1989
Answer: 2
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