2020 AMC 10B Problems/Problem 25
- The following problem is from both the 2020 AMC 10B #25 and 2020 AMC 12B #24, so both problems redirect to this page.
Contents
Problem
Let denote the number of ways of writing the positive integer as a productwhere , 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
Note that . Since there are at most six not necessarily 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 in a manner that is , or . A split has ways of happening ( and symmetry; and symmetry), a split has ways of happening (due to all being distinct) and a split has ways of happening ( and symmetry) so in this case we have ways.
: We have as our baseline, and for the two other 's, we have a or split. The former grants us ways ( and symmetry and and symmetry) and the latter grants us also ways ( and symmetry and 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
Solution 2
As before, note that , and we need to consider 6 different cases, one for each possible value of , 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 factors. First, the factorization needs to contain one factor that is itself a multiple of . There are to choose from. That leaves slots left to fill, each of which must contain at least one factor of . Once we have filled in a to each of the remaining slots, we're left with twos.
Consider the remaining factors of left to assign to the factors. Using stars and bars, the number of ways to do this is: This makes possibilities for each k.
To obtain the total number of factorizations, add all possible values for k:
Solution 3
Begin by examining . can take on any value that is a factor of except . For each choice of , the resulting must have a product of . This means the number of ways the rest , can be written by the scheme stated in the problem for each is equal to , since the product of is counted as one valid product if and only if , the product has the properties that factors are greater than , and differently ordered products are counted separately.
For example, say the first factor is . Then, the remaining numbers must multiply to , so the number of ways the product can be written beginning with is . To add up all the number of solutions for every possible starting factor, must be calculated and summed for all possible , except and , since a single is not counted according to the problem statement. The however, is counted, but only results in possibility, the first and only factor being . This means
.
Instead of calculating D for the larger factors first, reduce , , and into sums of where to ease calculation. Following the recursive definition sums of where c takes on every divisor of n except for 1 and itself, the sum simplifies to
+
, so the sum further simplifies to
, after combining terms. From quick casework,
and . Substituting these values into the expression above,
.
~monmath a.k.a Fmirza
Solution 4
Note that , and that of a perfect power of a prime is relatively easy to calculate. Also note that you can find from by simply totaling the number of ways there are to insert a into a set of numbers that multiply to .
First, calculate . Since , all you have to do was find the number of ways to divide the 's into groups, such that each group has at least one . By stars and bars, this results in way with five terms, ways with four terms, ways with three terms, ways with two terms, and way with one term. (The total, , is not needed for the remaining calculations.)
Then, to get , in each possible sequence, insert a somewhere in it, either by placing it somewhere next to the original numbers (in one of ways, where is the number of terms in the sequence), or by multiplying one of the numbers by (in one of ways). There are ways to do this with one term, with two, with three, with four, and with five.
The resulting number of possible sequences is . ~emerald_block
Solution 5 (Minimal Casework)
Consider the arrangement of the prime factors of 96 in a line . An arrangement of factors can be created by placing "dividers" to group primes. For example, is equivalent to the arrangement . Because there are ways to order the prime factors, and ways to place dividers, this gives us an initial ways to arrange divisors.
However, through this method, we overcount cases where is combined with another factor. For example, the arrangement can be written as or . Precisely, we double count any case with as a factor, triple count any case with , quadruple count any case with , etc.
Now, consider all cases where must be grouped with at least one . This can be expressed in the same "line" format as , where dividers can again be placed to group divisors. In this case, there are ways to order divisors, and ways to place dividers, so we have an possible sequences for this case. Notice that in this format, we double count cases where is a factor, we triple count cases where is a factor, etc. Precisely, for any case counted times in the first step, it is counted times in this step. Thus, if we subtract, we count each case exactly once.
So, we get:
. ~hdai1122
Solution 6 (Another Fast Way)
First we factor into numbers where . By applying stars and bars there are ways. Then we can either insert into each of the spaces between (or beyond) 's, or multiply it to one of the 's, a total of ways. Hence the answer to the problem is
~ asops
Solution 7 (Integer Partition)
Note that . depends on dividing into different terms, which is the integer partition of .
Divide into term:
There is only one way.
Divide into terms:
Case : is alone has different arrangements.
Case : is with . For and , can be with any term from the tuples, and the number of arrangement of the terms is .
Divide into terms:
Case : is alone . For and , there are arrangements each.
Case : is with . For and , can be with any term from the tuples. If is with for the first tuple, or for the second tuple, the number of arrangements is for each. If is with for the first tuple, or for the second tuple, the number of arrangements is for each.
Divide into terms:
Case : is alone . For and , there are arrangements each.
Case : is with . For , can be with any term from the tuple. If is with , the number of arrangements is . If is with , the number of arrangements is .
Divide into terms:
When dividing into parts there are cases.
Case : is alone . For , there are arrangements.
Case : is with . For , can only be with . The number of arrangements is
Divide into terms:
. The number of arrangements of is
Solution 8
Ignore the first and first count which . This implies that is less than or equal to . Now, we can see that can lie between the two , or contribute to one of them. This gives if . Now, just sum up gives .
~Hayabusa1
Video Solution by OmegaLearn
https://youtu.be/8WrdYLw9_ns?t=995
~ pi_is_3.14
Video Solution by MathEx
https://www.youtube.com/watch?v=977F9lBb37E
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 |
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.