2020 AMC 10B Problem 25

Problem 25 of 25HarderNumber Theory Combinatorics

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,

D(6) = 3.

What is

D(96)?

Show solution

Solution

Write

96=2^5\cdot3.

Suppose an ordered factorization has

k

factors. Exactly one factor contains the single prime factor

3

; choose its position in

k

ways. The other

k-1

factors must each contain at least one factor of

2,

while the factor containing

3

may contain any number of factors of

2.

Distributing the five factors of

2

under these conditions can be done in

\binom{5}{k-1}

ways. Therefore the number of ordered factorizations with

k

factors is

k\binom{5}{k-1},

where

1\le k\le6.

Thus

D(96)=\sum_{k=1}^6 k\binom{5}{k-1}.

Letting

j=k-1,

this becomes

\begin{aligned} &\sum_{j=0}^5 (j+1)\binom5j \\ &\quad =\sum_{j=0}^5 j\binom5j+\sum_{j=0}^5\binom5j \\ &\quad =5\cdot2^4+2^5 \\ &\quad =80+32=112. \end{aligned}

Thus, A is the correct answer.

AoPS wiki

Tagged: prime factorization · combinations

Answer choices

Solution

More practice