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Difference between revisions of "2000 AIME II Problems/Problem 4"
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What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
 
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
  
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== Solution ==
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== Solution 1==
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If a number has twice as many even divisors as odd divisors, then the number has two factors of 2 in it.(a.k.a. 4 is a factor of the number, but not 8.)
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We use the fact that the number of divisors of a number <math>n = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}</math> is <math>(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)</math>. If a number has <math>18 = 2 \cdot 3 \cdot 3</math> factors, then it can have at most <math>3</math> distinct primes in its factorization.  
  
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So we have <math>2^2*whatever</math>.
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Dividing the greatest power of <math>2</math> from <math>n</math>, we have an odd integer with six positive divisors, which indicates that it either is (<math>6 = 2 \cdot 3</math>) a prime raised to the <math>5</math>th power, or two primes, one of which is squared. The smallest example of the former is <math>3^5 = 243</math>, while the smallest example of the latter is <math>3^2 \cdot 5 = 45</math>.  
  
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Since <math>whatever</math> has 6 factors(as stated in the problem), <math>whatever</math> is either the product of a prime and a perfect cube, or a perfect 5th power. We see which gives out the smallest:
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Suppose we now divide all of the odd factors from <math>n</math>; then we require a power of <math>2</math> with <math>\frac{18}{6} = 3</math> factors, namely <math>2^{3-1} = 4</math>. Thus, our answer is <math>2^2 \cdot 3^2 \cdot 5 = \boxed{180}</math>.
  
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<math>3*5^3=375</math>
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== Solution 2 ==
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Somewhat similar to the first solution, we see that the number <math>n</math> has two even factors for every odd factor.  Thus, if <math>x</math> is an odd factor of <math>n</math>, then <math>2x</math> and <math>4x</math> must be the two corresponding even factors.  So, the prime factorization of <math>n</math> is <math>2^2 3^a 5^b 7^c...</math> for some set of integers <math>a, b, c, ...</math>
  
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<math>3^3*5=135</math>
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Since there are <math>18</math> factors of <math>n</math>, we can write:
  
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<math>3^5=243</math>
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<math>(2+1)(a+1)(b+1)(c+1)... = 18</math>
  
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Therefore, the smallest value of <math>whatever</math> is 135.
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<math>(a+1)(b+1)(c+1)... = 6</math>
  
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<math>135*4=\boxed{540}</math>
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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>.
  
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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>. Finally, you get <math>180</math>
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-dragoon
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2000|n=II|num-b=3|num-a=5}}
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 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 14:55, 11 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 $2^2n$. Finally, you 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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