2008 iTest Problems/Problem 84
Contents
[hide]Problem
Let be the sum of all integers for which the polynomial can be factored over the integers. Compute .
Solutions
Solution 1 (credit to official solution)
Let the roots of the quadratic be and . By Vieta's Formulas, and = .
We know that one of the possible values of is 0 because has integer roots. However, adding or removing 0 does not affect the value of , so we can divide both sides by . Doing so results in
WLOG, let be a factor of , so and . Thus,
Since can be positive or negative, the positive values cancel with the negative values. The prime factorization of is , so there are positive factors that are less than . Thus, there are a total of values of , so the absolute value of the sum of all values of equals .
Solution 2
The discriminant of the function is . Since all roots are integers and leading term is 1, the discriminant must equal , where is an integer.
Thus, we know that
Let be a factor of , so
Note that if is odd, then is even, so can not be an integer. Thus, must be even. Let , so .
If , then . Also, since can be positive or negative, the positive values cancel with the negative values. So WLOG, let .
The prime factorization of is , so there are positive factors that are less than . Thus, there are a total of values of , so the absolute value of the sum of all values of equals .
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 83 |
Followed by: Problem 85 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100 |