Difference between revisions of "2018 AIME I Problems/Problem 9"
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~Formatted by ojaswupadhyay and phoenixfire | ~Formatted by ojaswupadhyay and phoenixfire | ||
− | == Solution | + | == Solution Python == |
This code works: | This code works: |
Revision as of 14:44, 9 November 2020
Problem
Find the number of four-element subsets of with the property that two distinct elements of a subset have a sum of
, and two distinct elements of a subset have a sum of
. For example,
and
are two such subsets.
Solutions
Solution 1
This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set .
Note that there are only two cases: 1 where and
or 2 where
and
. Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you
, which cannot be true.
Case 1.
This is probably the simplest: just make a list of possible combinations for and
. We get
for the first and
for the second. That appears to give us
solutions, right? NO. Because elements can't repeat, take out the supposed sets
That's ten cases gone. So
for Case 1.
Case 2.
We can look for solutions by listing possible values and filling in the blanks. Start with
, as that is the minimum. We find
, and likewise up to
. But we can't have
or
because
or
, respectively! Now, it would seem like there are
values for
and
unique values for each
, giving a total of
, but that is once again not true because there are some repeated values! We can subtract 1 from all pairs of sets that have two elements in common, because those can give us identical sets. Write out all the sets and we get
. That's
for Case 2.
Total gives .
-expiLnCalc
Solution 2
Let's say our four elements in our subset are . We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
and
.
List out possibilities for
but don't list
because those are the same elements and that is restricted.
Then list out the possibilities for but don't list
because they are the same elements.
This will give you elements, which is
. However, as stated above, we have overlap. Just count starting from
all overlap once, which is
, thus
cases in this case. Note that
wasn't included because again, if
,
and
cannot be
.
and
.
Here, is included in both equations. We can easily see that
will never equal each other.
Furthermore, there are 17 choices for (
included elements) for each
. Listing out the possible
s, we go from
. Do not include
or
because if they are included, then
will be the same as
, which is restricted.
There are options there, and thus
. But, if
and
, notice that
. That means that if
is also
, then we have a double-counted set. Starting with
, we have
(where
is
. That means there are
double-counted cases. Thus
cases in this case.
Adding these up, we get
~IronicNinja
~ by AlcBoy1729
~Formatted by ojaswupadhyay and phoenixfire
Solution Python
This code works:
int num = 0; for(int i = 1; i <= 20; i++){ for(int j = i+1; j <= 20; j++){ for(int k = j+1; k <= 20; k++){ for(int m = k+1; m <= 20; m++){ if(i+j==16 || i + k == 16 || i + m == 16 || j + k == 16 || j + m == 16 || k + m == 16){ if(i+j==24 || i+k==24 || i+m==24 || j+k==24 || j+m == 24 || k+m==24){ num++; } } } } } } cout << num << endl;
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.