Difference between revisions of "2002 AIME II Problems/Problem 5"
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− | We have that a^6 = 2^{6n} 3^{6m} and 6^a = 2^a 3^a. So we need at least one of 6n > a or 6m > a to hold. Notice that a \geq 2^n and a \geq 3^m. It can easily be shown by induction that 6n \leq 2^n for all n \geq 5 and 6m \leq 3^m for all m \geq 3, so we must have n \leq 4 and m \leq 2. Testing all of these values, we find it only holds when one of m,n is zero: 2^4, 2^3, 2^2, 2^1, 3^2, 3^1 is the complete list of solutions. Their sum is 16 + 8 + 4 + 2 + 9 + 3 = \boxed{42}. | + | We have that <math>a^6 = 2^{6n} 3^{6m}</math> and <math>6^a = 2^a 3^a</math>. So we need at least one of <math>6n > a</math> or <math>6m > a</math> to hold. Notice that <math>a \geq 2^n</math> and <math>a \geq 3^m</math>. It can easily be shown by induction that <math>6n \leq 2^n</math> for all <math>n \geq 5</math> and <math>6m \leq 3^m</math> for all <math>m \geq 3</math>, so we must have <math>n \leq 4 and m \leq 2</math>. Testing all of these values, we find it only holds when one of <math>m,n</math> is zero: <math>2^4, 2^3, 2^2, 2^1, 3^2, 3^1</math> is the complete list of solutions. Their sum is <math>16 + 8 + 4 + 2 + 9 + 3 = \boxed{42}</math>. |
Harder Solution: | Harder Solution: |
Revision as of 01:53, 2 July 2014
Problem
Find the sum of all positive integers where and are non-negative integers, for which is not a divisor of
Solution
Easy Solution:
We have that and . So we need at least one of or to hold. Notice that and . It can easily be shown by induction that for all and for all , so we must have . Testing all of these values, we find it only holds when one of is zero: is the complete list of solutions. Their sum is .
Harder Solution:
Substitute into and , and find all pairs of non-negative integers (n,m) for which is not a divisor of $6^{2^n3^m$ (Error compiling LaTeX. ! Missing } inserted.)
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 ae 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 .
--Gamjawon 01:51, 2 July 2014 (EDT)gamjawon
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.