Difference between revisions of "2002 AIME II Problems/Problem 5"
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== Problem == | == Problem == | ||
− | Find the sum of all positive integers <math>a=2^n3^m</math> where <math>n</math> and <math>m</math> are non-negative integers, for which <math>a^6</math> is not a divisor of <math>6^a</math> | + | Find the sum of all positive integers <math>a=2^n3^m</math> where <math>n</math> and <math>m</math> are non-negative integers, for which <math>a^6</math> is not a divisor of <math>6^a</math>. |
== Solution == | == Solution == | ||
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Thus there are six numbers corresponding to (1,0), (2,0), (3,0), (4,0), (0,1), and (0,2). | 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 <math>\framebox{ | + | Plugging them back into the original expression, these numbers are 2, 4, 8, 16, 3, and 9, respectively. Their sum is <math>\framebox{042}</math>. |
− | == Solution 2 == | + | == Solution 2 (faster and more concise)== |
Notice that the condition is equivalent to saying | Notice that the condition is equivalent to saying | ||
− | <cmath>v_2 | + | <cmath>v_2(a^6) \geq v_2(6^a) \implies 6n \geq a</cmath> |
− | <cmath>v_3 | + | <cmath>v_3(a^6) \geq v_3(6^a) \implies 6m \geq a.</cmath> |
− | Notice that we cannot have both expressions to be equality state, as that would result in <math>a^6 = 6^a.</math> Testing, we see the possible pairs <math>(n, m)</math> are <math>(1, 0), (2, 0), (3, 0), (4, 0), (0, 1), (0, 2).</math> Since the left-hand side of the | + | Notice that we cannot have both expressions to be equality state, as that would result in <math>a^6 = 6^a.</math> Testing, we see the possible pairs <math>(n, m)</math> are <math>(1, 0), (2, 0), (3, 0), (4, 0), (0, 1), (0, 2).</math> Since the <math>a</math> grows much faster than the left-hand side of the above inequalities, these are all possible solutions. Adding, we get <math>\framebox{042}</math>. |
~Solution by Williamgolly | ~Solution by Williamgolly |
Latest revision as of 14:46, 14 February 2021
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 .
Solution 2 (faster and more concise)
Notice that the condition is equivalent to saying
Notice that we cannot have both expressions to be equality state, as that would result in Testing, we see the possible pairs are Since the grows much faster than the left-hand side of the above inequalities, these are all possible solutions. Adding, we get .
~Solution by Williamgolly
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.