Difference between revisions of "2002 AIME II Problems/Problem 5"
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Revision as of 03:30, 6 December 2019
Problem
Find the sum of all positive integers where and are non-negative integers, for which is not a divisor of
Solution
Substitute into and , and find all pairs of non-negative integers (n,m) for which is not a divisor of
Simplifying both expressions:
is not a divisor of
Comparing both exponents (noting that there must be either extra powers of 2 or extra powers of 3 in the left expression):
OR
Using the first inequality and going case by case starting with n {0, 1, 2, 3...}:
n=0: which has no solution for non-negative integers m
n=1: which is true for m=0 but fails for higher integers
n=2: which is true for m=0 but fails for higher integers
n=3: which is true for m=0 but fails for higher integers
n=4: which is true for m=0 but fails for higher integers
n=5: which has no solution for non-negative integers m
There are no more solutions for higher , as polynomials like grow slower than exponentials like .
Using the second inequality and going case by case starting with m {0, 1, 2, 3...}:
m=0: which has no solution for non-negative integers n
m=1: which is true for n=0 but fails for higher integers
m=2: which is true for n=0 but fails for higher integers
m=3: which has no solution for non-negative integers n
There are no more solutions for higher , as polynomials like grow slower than exponentials like .
Thus there are six numbers corresponding to (1,0), (2,0), (3,0), (4,0), (0,1), and (0,2).
Plugging them back into the original expression, these numbers are 2, 4, 8, 16, 3, and 9, respectively. Their sum is .
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.