Difference between revisions of "2018 AIME I Problems/Problem 1"
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Thus, the answer is: <cmath>\boxed{600}.</cmath> | Thus, the answer is: <cmath>\boxed{600}.</cmath> | ||
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+ | |||
+ | 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 number 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== | ==See Also== | ||
{{AIME box|year=2018|n=I|before=First Problem|num-a=2}} | {{AIME box|year=2018|n=I|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:27, 18 April 2018
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 number 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.