2016 AMC 10A Problems/Problem 20
For some particular value of , when
is expanded and like terms are combined, the resulting expression contains exactly
terms that include all four variables
and
, each to some positive power. What is
?
Solution
All the desired terms are in the form , where
(the
part is necessary to make stars and bars work better.)
Since
,
,
, and
must be at least
(
can be
), let
,
,
, and
, so
. Now, we use stars and bars to see that there are
or
solutions to this equation. We notice that
, which leads us to guess that
is around these numbers. This suspicion proves to be correct, as we see that
, giving us our answer of
.
- An alternative is to instead make the transformation
, so
, and all variables are positive integers. The solution to this, by Stars and Bars is
and we can proceed as above.
Solution 2
By Hockey Stick Identity, the number of terms that have all raised to a positive power is
. We now want to find some
such that
. As mentioned above, after noticing that
, and some trial and error, we find that
, giving us our answer of
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 | ||
All AMC 10 Problems and Solutions |
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