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

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#It is the cube of a prime number.  
 
#It is the cube of a prime number.  
 
#:If we let <math>n=p^3</math> with <math>p</math> prime, then its proper divisors are <math>p</math> and <math>p^2</math>, and <math>p(n) = p \cdot p^2 =n</math>.  
 
#:If we let <math>n=p^3</math> with <math>p</math> prime, then its proper divisors are <math>p</math> and <math>p^2</math>, and <math>p(n) = p \cdot p^2 =n</math>.  
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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>).  
 
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 former case, it suffices to note that <math>p(n) \ge (pr) \cdot (qr) = pqr^2 > pqr = n</math>.  
  
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Since <math>e \neq 3</math>, in the case <math>e = 0 \Longrightarrow n = 1</math> does not work.  
 
Since <math>e \neq 3</math>, in the case <math>e = 0 \Longrightarrow n = 1</math> does not work.  
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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>.
 
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>.
 
[[Sum]]ming these yields <math>\boxed{182}</math>.
 
[[Sum]]ming these yields <math>\boxed{182}</math>.
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Alternatively, we could note that <math>n</math> is only nice when it only has two divisors, which, when multiplied, clearly yield <math>n</math>. We know that when the [[prime factorization]] of <math>n = a_1^{b_1} \cdot a_2^{b_2} \cdot a_3^{b_3} . . . \cdot a_m^{b_m}</math>, the number of factors <math>f(n)</math> of <math>n</math> is <cmath>f(n) = (b_1 + 1)(b_2 +1)(b_3 +1) . . . (b_m +1).</cmath>
 
Alternatively, we could note that <math>n</math> is only nice when it only has two divisors, which, when multiplied, clearly yield <math>n</math>. We know that when the [[prime factorization]] of <math>n = a_1^{b_1} \cdot a_2^{b_2} \cdot a_3^{b_3} . . . \cdot a_m^{b_m}</math>, the number of factors <math>f(n)</math> of <math>n</math> is <cmath>f(n) = (b_1 + 1)(b_2 +1)(b_3 +1) . . . (b_m +1).</cmath>

Revision as of 23:38, 12 May 2020

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

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 $p(n) = p \cdot p^2 \cdots p^{(e-1)} = p^{(e-1)e/2}$.

For $p(n) = n$, we need $p^{(e-1)e/2} = p^e \Longrightarrow e^2 - e = 2e \Longrightarrow$ $e = 0 or e = 3$.

Since $e \neq 3$, in 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 $\boxed{182}$.

Alternatively, we could note that $n$ is only nice when it only has two divisors, which, when multiplied, clearly yield $n$. We know that when the prime factorization of $n = a_1^{b_1} \cdot a_2^{b_2} \cdot a_3^{b_3} . . . \cdot a_m^{b_m}$, the number of factors $f(n)$ of $n$ is \[f(n) = (b_1 + 1)(b_2 +1)(b_3 +1) . . . (b_m +1).\]

Since $n$ is nice, it may only have $4$ factors ($1$, $n$, $p$, and $q$). This means that $f(n) = 4$. The number $4$ can only be factored into $(2)(2)$ or $(4)(1)$, which means that either $b_1 = 1$ and $b_2 = 1$, or $b_1 = 3$. Therefore the only two cases are $n = pq$, or $n = p^3$.

See also

1987 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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