Difference between revisions of "2000 AIME II Problems/Problem 4"
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<math>(a+1)(b+1)(c+1)... = 6</math> | <math>(a+1)(b+1)(c+1)... = 6</math> | ||
− | 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 | + | 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>. |
== See also == | == See also == |
Revision as of 08:39, 30 April 2019
Contents
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 is
. If a number has
factors, then it can have at most
distinct primes in its factorization.
Dividing the greatest power of from
, we have an odd integer with six positive divisors, which indicates that it either is (
) a prime raised to the
th power, or two primes, one of which is squared. The smallest example of the former is
, while the smallest example of the latter is
.
Suppose we now divide all of the odd factors from ; then we require a power of
with
factors, namely
. Thus, our answer is
.
Solution 2
Somewhat similar to the first solution, we see that the number has two even factors for every odd factor. Thus, if
is an odd factor of
, then
and
must be the two corresponding even factors. So, the prime factorization of
is
for some set of integers
Since there are factors of
, we can write:
Since only has factors from the set
, either
and all other variables are
, or
and
, with again all other variables equal to
. This gives the two numbers
and
. The latter number is smaller, and is equal to
.
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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