Simple yet complicated
a , b , c are integers such that they satisfy the constraints: 0 < a < b , and the polynomial x ( x − a ) ( x − b ) − 1 7 is divisible by ( x − c ) .
What is a + b + c ?
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1 solution
Discussions for this problem are now closed
Wouldn't a=8, b=10, c=7, x=11 also work?
x here is a considered as a variable. So taking only one value of x won't work. You have to show that it works for all values of x .
Thanks Siddhartha Srivastava.
I couldn't follow how did you state c>0 .Please explain.
Because if c < 0 , then
0 < a < b
⟹ 0 > − a > − b
⟹ c > c − a > c − b
⟹ 0 > c > c − a > c − b .
Therefore c − a and c − b are < 0
Multiplying these together, we have c ( c − a ) ( c − b ) < 0 , which is a contradiciton
I got it .Nicely explained. Sir , can you tell me the link of the same type of problems ?
Since it is divisible by (x − c), we have
c ( c − a ) ( c − b ) = 1 7 .
Since c ( c − a ) ( c − b ) = 1 7 > 0 , it follows that c > 0 and we have the following two cases:
Case 1: 0 < ( c − b ) < ( c − a ) < c
Case 2: ( c − b ) < ( c − a ) < 0 < c .
Since 1 7 is a prime number, case 1 does not occur . In case 2, c = 1 , c − a = − 1 , c − b = − 1 7 .
Hence a = 2 , b = 1 8 , c = 1 .
Thus, a + b + c = 2 1 .