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

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(solution, there probably is a better way to do it though)
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== Problem ==
 
== Problem ==
By a proper divisior 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?
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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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== Solution ==
 
== 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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:<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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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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== See also ==
 
== See also ==
* [[1987 AIME Problems]]
 
 
 
{{AIME box|year=1987|num-b=2|num-a=4}}
 
{{AIME box|year=1987|num-b=2|num-a=4}}

Revision as of 18:27, 15 February 2007

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

A number is nice in one of two instances: (1) it has exactly two distinct prime divisors (excluding $1$ and itself, note that squares like $3^2 = 9$ don't work, an example would be $3 \cdot 5 = 15$), or (2) it the cube of a prime number (an example would be $2^3 = 8$). Thus, listing them all out, we get:

$2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,\ 3^3 = 27,\ 3 \cdot 11 = 33$

Directly summing them up yields $2(3 + 4 + 5 + 7 + 11 + 13) + 3(5 + 7 + 9 + 11) = 182$.

See also

1987 AIME (ProblemsAnswer KeyResources)
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