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Difference between revisions of "2002 AIME II Problems/Problem 5"
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== Solution ==
 
== Solution ==
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Easy Solution:
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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}.
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 +
Harder Solution:
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Substitute <math>a=2^n3^m</math> into <math>a^6</math> and <math>6^a</math>, and find all pairs of non-negative integers (n,m) for which <math>(2^n3^m)^{6}</math> is not a divisor of <math>6^{2^n3^m</math>
 
Substitute <math>a=2^n3^m</math> into <math>a^6</math> and <math>6^a</math>, and find all pairs of non-negative integers (n,m) for which <math>(2^n3^m)^{6}</math> is not a divisor of <math>6^{2^n3^m</math>
  
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Thus there ae six numbers corresponding to (1,0), (2,0), (3,0), (4,0), (0,1), and (0,2).   
 
Thus there ae six numbers corresponding to (1,0), (2,0), (3,0), (4,0), (0,1), and (0,2).   
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Plugging them back into the original expression, these numbers are 2, 4, 8, 16, 3, and 9, respectively.  Their sum is <math>042</math>.
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Plugging them back into the original expression, these numbers are 2, 4, 8, 16, 3, and 9, respectively.  Their sum is <math>42</math>.
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--[[User:Gamjawon|Gamjawon]] 01:51, 2 July 2014 (EDT)gamjawon
  
 
== See also ==
 
== See also ==

Revision as of 00:51, 2 July 2014

Problem

Find the sum of all positive integers $a=2^n3^m$ where $n$ and $m$ are non-negative integers, for which $a^6$ is not a divisor of $6^a$

Solution

Easy Solution:

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}.

Harder Solution:

Substitute $a=2^n3^m$ into $a^6$ and $6^a$, and find all pairs of non-negative integers (n,m) for which $(2^n3^m)^{6}$ is not a divisor of $6^{2^n3^m$ (Error compiling LaTeX. Unknown error_msg)

Simplifying both expressions:

$2^{6n} \cdot 3^{6m}$ is not a divisor of $2^{2^n3^m} \cdot 3^{2^n3^m}$

Comparing both exponents (noting that there must be either extra powers of 2 or extra powers of 3 in the left expression):

$6n > 2^n3^m$ OR $6m > 2^n3^m$


Using the first inequality $6n > 2^n3^m$ and going case by case starting with n $\in$ {0, 1, 2, 3...}:

n=0: $0>1 \cdot 3^m$ which has no solution for non-negative integers m

n=1: $6 > 2 \cdot 3^m$ which is true for m=0 but fails for higher integers $\Rightarrow (1,0)$

n=2: $12 > 4 \cdot 3^m$ which is true for m=0 but fails for higher integers $\Rightarrow (2,0)$

n=3: $18 > 8 \cdot 3^m$ which is true for m=0 but fails for higher integers $\Rightarrow (3,0)$

n=4: $24 > 16 \cdot 3^m$ which is true for m=0 but fails for higher integers $\Rightarrow (4,0)$

n=5: $30 > 32 \cdot 3^m$ which has no solution for non-negative integers m

There are no more solutions for higher $n$, as polynomials like $6n$ grow slower than exponentials like $2^n$.


Using the second inequality $6m > 2^n3^m$ and going case by case starting with m $\in$ {0, 1, 2, 3...}:

m=0: $0>2^n \cdot 1$ which has no solution for non-negative integers n

m=1: $6>2^n \cdot 3$ which is true for n=0 but fails for higher integers $\Rightarrow (0,1)$

m=2: $12>2^n \cdot 9$ which is true for n=0 but fails for higher integers $\Rightarrow (0,2)$

m=3: $18>2^n \cdot 27$ which has no solution for non-negative integers n

There are no more solutions for higher $m$, as polynomials like $6m$ grow slower than exponentials like $3^m$.


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 $42$.

--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

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