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Difference between revisions of "1987 AIME Problems/Problem 3"

(solution, there probably is a better way to do it though)
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== Problem ==
 
== Problem ==
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By a proper [[divisor]] of a [[natural number]] we mean a [[positive]] [[integer|integral]] divisor other than 1 and the number itself. A natural number greater than 1 will be called "nice" if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?
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By a proper [[divisor]] of a [[natural number]] we mean a [[positive]] [[integer|integral]] divisor other than 1 and the number itself. A natural number greater than 1 will be called ''nice'' if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?
  
 
== Solution ==
 
== Solution ==
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A number is nice in one of two instances: (1) it has exactly two distinct [[prime]] divisors (excluding <math>1</math> and itself, note that squares like <math>3^2 = 9</math> don't work, an example would be <math>3 \cdot 5 = 15</math>), or (2) it the cube of a [[prime]] number (an example would be <math>2^3 = 8</math>). Thus, listing them all out, we get:
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Let <math>p(n)</math> denote the product of the distinct proper divisors of <math>n</math>. A number <math>n</math> is ''nice'' in one of two instances:  
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:<math>2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,\ 3^3 = 27,\ 3 \cdot 11 = 33</math>
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#It has exactly two distinct [[prime]] divisors.
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Directly [[sum]]ming them up yields <math>2(3 + 4 + 5 + 7 + 11 + 13) + 3(5 + 7 + 9 + 11) = 182</math>.<!--Is there an easier way to manipulate?-->
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#:If we let <math>n = pq</math>, where <math>p,q</math> are the prime factors, then its proper divisors are <math>1,\,p,\,q</math>, and <math>p(n) = 1 \cdot p \cdot q = n</math>.
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#It is the cube of a prime number.
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#:If we let <math>n=p^3</math> with <math>p</math> prime, then its proper divisors are <math>1,\,p,\,p^2</math>, and <math>p(n) = 1 \cdot p \cdot p^2 n</math>.
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----
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We now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form <math>n = pqr</math> (with <math>p,q</math> prime and <math>r > 1</math>) or <math>n = p^e</math> (with <math>e \neq 3</math>).
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 +
In the former case, it suffices to note that <math>p(n) \ge (pr) \cdot (qr) = pqr^2 > pqr = n</math>. In the latter case, then <math>p(n) = 1 \cdot p \cdot p^2 \cdots p^e = p^{e(e+1)/2}</math>. For <math>p(n) = n</math>, we need <math>p^{e(e+1)/2} = p^e \Longrightarrow e^2 + e = 2e \Longrightarrow e = 0,3</math> (the case <math>e = 0 \Longrightarrow n = 1</math> does not work).
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----
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Thus, listing out the first ten numbers to fit this form, <math>2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,</math> <math>\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,</math> <math>\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,</math> <math>\ 3^3 = 27,\ 3 \cdot 11 = 33</math>.
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[[Sum]]ming these yields <math>\boxed{182}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1987|num-b=2|num-a=4}}
 
{{AIME box|year=1987|num-b=2|num-a=4}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 20:31, 9 April 2008

Problem

By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called nice if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?

Solution

Let $p(n)$ denote the product of the distinct proper divisors of $n$. A number $n$ is nice in one of two instances:

  1. It has exactly two distinct prime divisors.
    If we let $n = pq$, where $p,q$ are the prime factors, then its proper divisors are $1,\,p,\,q$, and $p(n) = 1 \cdot p \cdot q = n$.
  2. It is the cube of a prime number.
    If we let $n=p^3$ with $p$ prime, then its proper divisors are $1,\,p,\,p^2$, and $p(n) = 1 \cdot p \cdot p^2 n$.

We now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form $n = pqr$ (with $p,q$ prime and $r > 1$) or $n = p^e$ (with $e \neq 3$).

In the former case, it suffices to note that $p(n) \ge (pr) \cdot (qr) = pqr^2 > pqr = n$. In the latter case, then $p(n) = 1 \cdot p \cdot p^2 \cdots p^e = p^{e(e+1)/2}$. For $p(n) = n$, we need $p^{e(e+1)/2} = p^e \Longrightarrow e^2 + e = 2e \Longrightarrow e = 0,3$ (the case $e = 0 \Longrightarrow n = 1$ does not work).

Thus, listing out the first ten numbers to fit this form, $2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,$ $\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,$ $\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,$ $\ 3^3 = 27,\ 3 \cdot 11 = 33$. Summing these yields $\boxed{182}$.

See also

1987 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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