Question

If x and y are positive integers, what is the value of (x + y)?
(1) (x + y - 1)! < 100
(2) y = x^2 - x + 1
 

Comments

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Answer 1

Both x and y are +ve integers i.e. they shall be 1,2,3,4,5 ...... n
Let us pick up statement #1
(x+y-1)! < 100
Let us write the factorials of 1,2,3,4 and see when does it exceed 100.
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
So the max value of x+y-1 can be 4 because when it is 5 it exceeds 100.
So x+y-1 = 1,2,3,4 or
x+y = 2, 3, 4, 5
But does not give us 1 solution. Not Sufficient
Let us Pick up Statement #2
y = x^2 -x + 1
This quadratic has no integer solutions, hence we cannot solve it for answer. Not Sufficient
Umm. Combine them replace y from in statement #1 from #2
x+x^2-x+1 = 2,3,4,5
=> x^2+1 = 2,3,4,5
=> x^2 = 1,2,3,4
this will 2 solutions +ve x = 1,2
corresponding +ve y can be 1,3
The answer should be e, the oa is debatable here. please check again.