Difference between revisions of "2020 AMC 12B Problems/Problem 24"
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− | To make a product of <math>n</math>, we can either have just <math>n</math>, or we can have a divisor <math>d</math> of <math>n</math>, <math>1 < d < n</math>, followed by a way to make a product of <math>\frac{n}{d}</math>. Thus, <math>D(n) = 1 + \sum_{d | n} D \left(\frac{n}{d} \right)</math> for all possible <math>d</math>. It's easy to calculate <math>D(n)</math> for all powers of <math>2</math>, since powers of <math>2</math> only have powers of <math>2</math> as divisors. We have | + | To make a product of <math>n</math>, we can either have just <math>n</math>, or we can have a divisor <math>d</math> of <math>n</math>, <math>1 < d < n</math>, followed by a way to make a product of <math>\frac{n}{d}</math>. Thus, <math>D(n) = 1 + \sum_{d | n, d > 1} D \left(\frac{n}{d} \right)</math> for all possible <math>d</math>. It's easy to calculate <math>D(n)</math> for all powers of <math>2</math>, since powers of <math>2</math> only have powers of <math>2</math> as divisors. We have |
<cmath>D(2) = 1</cmath> | <cmath>D(2) = 1</cmath> | ||
<cmath>D(4) = 2</cmath> | <cmath>D(4) = 2</cmath> |
Revision as of 18:30, 9 February 2020
Let denote the number of ways of writing the positive integer
as a product
where
, the
are integers strictly greater than
, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number
can be written as
,
, and
, so
. What is
?
Solution 1
To make a product of , we can either have just
, or we can have a divisor
of
,
, followed by a way to make a product of
. Thus,
for all possible
. It's easy to calculate
for all powers of
, since powers of
only have powers of
as divisors. We have
Now we will calculate
for all
in the form
, for
. Note that each divisor of
is either of the form
or
. Therefore, to calculate each
, we will sum the first
values in both the tables we created, and add
. We have
Solution 2
Bash.
Since , for the number of
, we have the following cases:
Case 1: , we have
, only 1 case.
Case 2: , we have
, totally
cases.
Case 3: , we have
, totally
cases.
Case 4: , we have
, totally
cases.
Case 5: , we have
, totally
cases.
Case 6: , we have
, totally
cases.
Thus, add all of them together, we have cases. Put
.
~FANYUCHEN20020715
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.