2018 AIME I Problems/Problem 1
Problem 1
Let be the number of ordered pairs of integers with and such that the polynomial can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when is divided by .
Solution
You let the linear factors be as .
Then, obviously and .
We know that and , so and both have to be positive.
However, cannot be , so at least one of and must be greater than , ie positive.
Also, cannot be greater than , so must be less than or equal to .
Essentially, if we plot the solutions, we get a triangle on the coordinate plane with vertices and . Remember that does not work, so there is a square with top right corner .
Note that and are interchangeable, since they end up as and in the end anyways. Thus, we simply draw a line from to , designating one of the halves as our solution (since the other side is simply the coordinates flipped).
We note that the pattern from to is solutions and from to is solutions, since we can decrease the -value by until for each coordinate.
Adding up gives This gives us , and
Thus, the answer is:
Solution 2
a-value b-value(s) amount of b-value(s)
a=1 b=0 1
a=2 b=1, 0 2 a=3 b=2, 0 2 a=4 b=4, 3, 0 3 a=5 b=6, 4, 0 3 a=6 b=9, 8, 5, 0 4 a=7 b=12, 10, 6, 0 4 a=8 b=16, 15, 12, 7, 0 5 a=9 b=20, 18, 14, 8, 0 5 a=10 b=25, 24, 21, 16, 9, 0 6 a=11 b=30, 28, 24, 18, 10, 0 6 Excluding a=1, we can see there is always a pair of 2 a-values for a certain amount of b-values.
For instance, a=2 and a=3 both have 2 b-values. a=4 and a=5 both have 3 b-values.
We notice the pattern of the amount of b-values in relation to the a values: 1, 2, 2, 3, 3, 4, 4…
The pattern continues until a=100, and in total, there are 49 pairs of a-value with the same amount of b-values. The two lone a-values without a pair are, the (a=1, amount of b-values=1) in the beginning, and (a=100, amount of b-values=51) in the end.
Then, we add numbers from the opposite ends of the spectrum, and quickly notice that there are 50 pairs each with a sum of 52. 52x50 gives 2600 ordered pairs:
1+51, 2+50, 2+50, 3+49, 3+49, 4+48, 4+48…
When divided by 1000, it gives the remainder 600, the answer.
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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