1996 AIME Problems/Problem 3
Problem
Find the smallest positive integer for which the expansion of , after like terms have been collected, has at least 1996 terms.
Solution 1
Using Simon's Favorite Factoring Trick, we rewrite as . Both binomial expansions will contain non-like terms; their product will contain terms, as each term will have an unique power of or and so none of the terms will need to be collected. Hence , the smallest square after is , so our answer is .
Alternatively, when , the exponents of or in can be any integer between and inclusive. Thus, when , there are terms and, when , there are terms. Therefore, we need to find the smallest perfect square that is greater than . From trial and error, we get and . Thus, .
Solution 2 (Generating Functions)
Notice that the coefficients in the problem statement have no effect on how many unique terms there will be in the expansion. Therefore this problem is synonymous with finding the amount of terms in the expansion of (we do this to simplify the problem).
If we expand the exponent the expression becomes .
This is equivalent to starting off with a terms and choosing between options different times:
Adding nothing to either exponent (choosing ).
Adding to the exponent (choosing ).
Adding to the exponent (choosing ).
Adding to the exponent and adding to the exponent (choosing ).
Doing this times, you can see that you end up with a term in the form where is some coefficient (which we don't care about) and and .
Repeating this for all possible combinations of choices yields options for each of and which means there are a total of possible terms in the form . Therefore has terms.
which yields .
~coolishu
See also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.