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2002 AIME II Problems/Problem 5

Revision as of 02:30, 6 December 2019 by Greenpizza96 (talk | contribs)

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

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 $6^{2^n3^m}$

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 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 $\framebox{42}$.

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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