Difference between revisions of "2020 AMC 10B Problems/Problem 25"
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| − | Thus, adding, we have <math>1+10+30+40+25+6=\textbf{(A) } 112</math>. | + | Thus, adding, we have <math>1+10+30+40+25+6=\boxed{\textbf{(A) } 112}</math>. |
~kevinmathz | ~kevinmathz | ||
Revision as of 22:37, 7 February 2020
Problem
Let 
denote the number of ways of writing the positive integer 
as a product![\[n = f_1\cdot f_2\cdots f_k,\]](http://latex.artofproblemsolving.com/7/f/5/7f52b53b0b17822b9bde3837f27c952c28b38559.png)
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
Note that 
. Since there are at most six not nexxessarily distinct factors 
multiplying to 
, we have six cases: 
Now we look at each of the six cases.
: We see that there is way, merely .
: This way, we have the 
in one slot and 
in another, and symmetry. The four other 's leave us with 
ways and symmetry doubles us so we have 
.
: We have 
as our baseline. We need to multiply by in places, and see that we can split the remaining three powers of 2 in a manner that is 3-0-0, 2-1-0 or 1-1-1. A 3-0-0 split has 
ways of happening (24-2-2 and symmetry; 2-3-16 and symmetry), a 2-1-0 split has 
ways of happening (due to all being distinct) and a 1-1-1 split has ways of happening (6-4-4 and symmetry) so in this case we have 
ways.
: We have 
as our baseline, and for the two other 's, we have a 2-0-0-0 or 1-1-0-0 split. The former grants us 
ways (12-2-2-2 and symmetry and 3-8-2-2 and symmetry) and the latter grants us also 
ways (6-4-2-2 and symmetry and 3-4-4-2 and symmetry) for a total of 
ways.
: We have 
as our baseline and one place to put the last two: on another two or on the three. On the three gives us ways due to symmetry and on another two gives us 
ways due to symmetry. Thus, we have 
ways.
: We have 
and symmetry and no more twos to multiply, so by symmetry, we have ways.
Thus, adding, we have 
.
~kevinmathz
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
