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 | + | 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 == | ||
− | + | 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: | |
+ | :<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> | ||
+ | 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 == | ||
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{{AIME box|year=1987|num-b=2|num-a=4}} | {{AIME box|year=1987|num-b=2|num-a=4}} |
Revision as of 17:27, 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 and itself, note that squares like don't work, an example would be ), or (2) it the cube of a prime number (an example would be ). Thus, listing them all out, we get:
Directly summing them up yields .
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 |