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Difference between revisions of "2000 AIME II Problems/Problem 4"

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Since <math>6</math> only has factors from the set <math>1, 2, 3, 6</math>, either <math>a=5</math> and all other variables are <math>0</math>, or <math>a=3</math> and <math>b=2</math>, with again all other variables equal to <math>0</math>.  This gives the two numbers <math>2^2 \cdot 3^5</math> and <math>2^2 \cdot 3^2 \cdot 5</math>.  The latter number is smaller, and is equal to <math>\boxed {180}</math>.
 
Since <math>6</math> only has factors from the set <math>1, 2, 3, 6</math>, either <math>a=5</math> and all other variables are <math>0</math>, or <math>a=3</math> and <math>b=2</math>, with again all other variables equal to <math>0</math>.  This gives the two numbers <math>2^2 \cdot 3^5</math> and <math>2^2 \cdot 3^2 \cdot 5</math>.  The latter number is smaller, and is equal to <math>\boxed {180}</math>.
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==Solution 3==
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We see that the least number with 6 odd factors is <math>3^2*5</math>. Multiplied by <math>2^2</math> (as each factor of 2 doubles the odd factors, as it can be 2n or <math>2^2n</math>. You finally get <math>180</math>
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-dragoon
  
 
== See also ==
 
== See also ==

Revision as of 23:45, 10 May 2022

Problem

What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

Solution 1

We use the fact that the number of divisors of a number $n = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}$ is $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$. If a number has $18 = 2 \cdot 3 \cdot 3$ factors, then it can have at most $3$ distinct primes in its factorization.

Dividing the greatest power of $2$ from $n$, we have an odd integer with six positive divisors, which indicates that it either is ($6 = 2 \cdot 3$) a prime raised to the $5$th power, or two primes, one of which is squared. The smallest example of the former is $3^5 = 243$, while the smallest example of the latter is $3^2 \cdot 5 = 45$.

Suppose we now divide all of the odd factors from $n$; then we require a power of $2$ with $\frac{18}{6} = 3$ factors, namely $2^{3-1} = 4$. Thus, our answer is $2^2 \cdot 3^2 \cdot 5 = \boxed{180}$.

Solution 2

Somewhat similar to the first solution, we see that the number $n$ has two even factors for every odd factor. Thus, if $x$ is an odd factor of $n$, then $2x$ and $4x$ must be the two corresponding even factors. So, the prime factorization of $n$ is $2^2 3^a 5^b 7^c...$ for some set of integers $a, b, c, ...$

Since there are $18$ factors of $n$, we can write:

$(2+1)(a+1)(b+1)(c+1)... = 18$

$(a+1)(b+1)(c+1)... = 6$

Since $6$ only has factors from the set $1, 2, 3, 6$, either $a=5$ and all other variables are $0$, or $a=3$ and $b=2$, with again all other variables equal to $0$. This gives the two numbers $2^2 \cdot 3^5$ and $2^2 \cdot 3^2 \cdot 5$. The latter number is smaller, and is equal to $\boxed {180}$.

Solution 3

We see that the least number with 6 odd factors is $3^2*5$. Multiplied by $2^2$ (as each factor of 2 doubles the odd factors, as it can be 2n or $2^2n$. You finally get $180$

-dragoon

See also

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions

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