Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2020 AMC 10B Problems/Problem 25

Revision as of 00:50, 8 February 2020 by Monmath (talk | contribs) (→‎Solution)

Problem

Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product\[n = f_1\cdot f_2\cdots f_k,\]where $k\ge1$, the $f_i$ are integers strictly greater than $1$, 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 $6$ can be written as $6$, $2\cdot 3$, and $3\cdot2$, so $D(6) = 3$. What is $D(96)$?

$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

Solution

Note that $96 = 2^5 \cdot 3$. Since there are at most six not nexxessarily distinct factors $>1$ multiplying to $96$, we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.


$k=1$: We see that there is $1$ way, merely $96$.

$k=2$: This way, we have the $3$ in one slot and $2$ in another, and symmetry. The four other $2$'s leave us with $5$ ways and symmetry doubles us so we have $10$.

$k=3$: We have $3, 2, 2$ as our baseline. We need to multiply by $2$ in $3$ 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 $6 + 3 = 9$ ways of happening (24-2-2 and symmetry; 2-3-16 and symmetry), a 2-1-0 split has $6 \cdot 3 = 18$ ways of happening (due to all being distinct) and a 1-1-1 split has $3$ ways of happening (6-4-4 and symmetry) so in this case we have $9+18+3=30$ ways.

$k=4$: We have $3, 2, 2, 2$ as our baseline, and for the two other $2$'s, we have a 2-0-0-0 or 1-1-0-0 split. The former grants us $4+12=16$ ways (12-2-2-2 and symmetry and 3-8-2-2 and symmetry) and the latter grants us also $12+12=24$ ways (6-4-2-2 and symmetry and 3-4-4-2 and symmetry) for a total of $16+24=40$ ways.

$k=5$: We have $3, 2, 2, 2, 2$ as our baseline and one place to put the last two: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5 \cdot 4 = 20$ ways due to symmetry. Thus, we have $5+20=25$ ways.

$k=6$: We have $3, 2, 2, 2, 2, 2$ and symmetry and no more twos to multiply, so by symmetry, we have $6$ ways.


Thus, adding, we have $1+10+30+40+25+6=\boxed{\textbf{(A) } 112}$.

~kevinmathz

Solution 2

As before, note that $96=2^5\cdot3$, and we need to consider 6 different cases, one for each possible value of $k$, the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization needs to contain one factor that is itself a multiple of $3$, and there are $k$ to choose from, and the rest must contain at least one factor of $2$. Next, consider the remaining $6-n$ factors of $2$ left to assign to the $k$ factors. Using stars and bars, the number of ways to do this is \[{{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}\] This makes $k{5\choose{6-k}}$ possibilities for each k.

To obtain the total number of factorizations, add all possible values for k: \[\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{\textbf{(A) } \text{112}}\].

Solution 3

Begin examining $f_1$. $f_1$ can take on any value that is a factor of $96$ except $1$. For each choice of $f_1$, the resulting $f_2...f_k$ must have a product of $96/f_1$. This means the number of ways the rest $f_a$ $1<a<=k$ can be written by the scheme stated in the problem for each $f_1$ is $D(96/f_1)$, since the product of $f_1 \cdot f_2$...$\cdot f_k$ is 96 if and only if $f_1 \cdot x=96$ and the product $x$ has the properties that factors are greater than $1$, and order matters in counting the solutions, which is how $D$ is defined.

For example, say the first factor is $2$, the remaining numbers must multiply to $48$, so the number of ways the product can be written beginning with $2$ is $D(48)$. To add up all possible starting factors, $D(96/f_a)$ must be calculated and summed for all $f_a$ except $96$ and $1$, since a single $1$ is not counted according to the problem statement. The $96$ however, is counted, but only results in $1$ possibility, the first and only factor being $96$. This means $D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(4)+D(2)+1$. Instead of calculating D for the larger factors first, reduce $D(48)$, $D(32)$, and $D(24)$ into sums of $D(m)$ where $m<=16$ to ease calculation.

Following the definition $D(n)=($sums of $D(c))+1$ where c takes on every divisor of n except for 1 and itself, the sum simplifies to $(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)+ (D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(4)+D(2)+1=\boxed{\textbf{(A) } 112}$.

~monmath a.k.a Fmirza

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_25&oldid=117311"