Difference between revisions of "2020 AMC 10B Problems/Problem 25"
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~monmath a.k.a Fmirza | ~monmath a.k.a Fmirza | ||
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+ | ==Solution 4== | ||
+ | Note that <math>96 = 3 \cdot 2^5</math>, and that <math>D</math> of a perfect power of a prime is relatively easy to calculate. Also note that you can find <math>D(96)</math> from <math>D(32)</math> by simply totaling the number of ways there are to insert a <math>3</math> into a set of numbers that multiply to <math>32</math>. | ||
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+ | First, calculate <math>D(32)</math>. Since <math>32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2</math>, all you have to do was find the number of ways to divide the <math>2</math>'s into groups, such that each group has at least one <math>2</math>. By stars and bars, this results in <math>1</math> way with five terms, <math>4</math> ways with four terms, <math>6</math> ways with three terms, <math>4</math> ways with two terms, and <math>1</math> way with one term. (The total, <math>16</math>, is not needed for the remaining calculations.) | ||
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+ | Then, to get <math>D(96)</math>, in each possible <math>D(32)</math> sequence, insert a <math>3</math> somewhere in it, either by placing it somewhere next to the original numbers (in one of <math>n+1</math> ways, where <math>n</math> is the number of terms in the <math>D(32)</math> sequence), or by multiplying one of the numbers by <math>3</math> (in one of <math>n</math> ways). There are <math>2+1=3</math> ways to do this with one term, <math>3+2=5</math> with two, <math>7</math> with three, <math>9</math> with four, and <math>11</math> with five. | ||
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+ | The resulting number of possible sequences is <math>3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{\textbf{(A) }112}</math>. ~[[User:emerald_block|emerald_block]] | ||
==See Also== | ==See Also== |
Revision as of 23:51, 7 February 2020
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
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
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 , and there are to choose from, and the rest must contain at least one factor of . Next, 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 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 , since the product of ... is 96 if and only if and the product has the properties that factors are greater than , and order matters in counting the solutions, which is how is defined.
For example, say the first factor is , the remaining numbers must multiply to , so the number of ways the product can be written beginning with is . To add up all possible starting factors, must be calculated and summed for all 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 definition sums of where c takes on every divisor of n except for 1 and itself, the sum simplifies to .
~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
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.