Difference between revisions of "2020 AMC 10B Problems/Problem 25"
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+ | {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #25]] and [[2020 AMC 12B Problems|2020 AMC 12B #24]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
Line 29: | Line 31: | ||
==Solution 2== | ==Solution 2== | ||
− | As before, note that <math>96=2^5\cdot3</math>, and we need to consider 6 different cases, one for each possible value of <math>k</math>, 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 <math>k</math> factors. First, the factorization needs to contain one factor that is itself a multiple of <math>3</math> | + | As before, note that <math>96=2^5\cdot3</math>, and we need to consider 6 different cases, one for each possible value of <math>k</math>, 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 <math>k</math> factors. First, the factorization needs to contain one factor that is itself a multiple of <math>3</math>. There are <math>k</math> to choose from. That leaves <math>k-1</math> slots left to fill, each of which must contain at least one factor of <math>2</math>. Once we have filled in a <math>2</math> to each of the remaining slots, we're left with <math>5-(k-1)=6-k</math> twos. |
Consider the remaining <math>6-k</math> factors of <math>2</math> left to assign to the <math>k</math> factors. Using stars and bars, the number of ways to do this is: | Consider the remaining <math>6-k</math> factors of <math>2</math> left to assign to the <math>k</math> factors. Using stars and bars, the number of ways to do this is: | ||
− | <cmath>{{(6-k)+ | + | <cmath>{{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}</cmath> This makes <math>k{5\choose{6-k}}</math> possibilities for each k. |
To obtain the total number of factorizations, add all possible values for k: <cmath>\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{\textbf{(A) } \text{112}}.</cmath> | To obtain the total number of factorizations, add all possible values for k: <cmath>\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{\textbf{(A) } \text{112}}.</cmath> | ||
+ | |||
+ | ~Continuous_Pi | ||
==Solution 3== | ==Solution 3== | ||
Line 68: | Line 72: | ||
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]] | 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]] | ||
− | == Video Solution == | + | ==Solution 5 (Minimal Casework)== |
− | https://youtu.be/8WrdYLw9_ns?t= | + | Consider the arrangement of the prime factors of 96 in a line <math>(2,2, 2, 2, 2, 3)</math>. An arrangement of factors can be created by placing "dividers" to group primes. For example, <math>(2, 2, |, 2, 2, 2, |, 3)</math> is equivalent to the arrangement <math>4 \cdot 8 \cdot 3</math>. Because there are <math>6</math> ways to order the prime factors, and <math>2^5</math> ways to place dividers, this gives us an initial <math>6 \cdot 2^5</math> ways to arrange divisors. |
+ | |||
+ | However, through this method, we overcount cases where <math>3</math> is combined with another factor. For example, the arrangement <math>4 \cdot 6 \cdot 4</math> can be written as <math>(2, 2, |, 2, 3, |, 2, 2)</math> or <math>(2, 2, |, 3, 2, |, 2, 2)</math>. Precisely, we double count any case with <math>6</math> as a factor, triple count any case with <math>12</math>, quadruple count any case with <math>24</math>, etc. | ||
+ | |||
+ | Now, consider all cases where <math>3</math> must be grouped with at least one <math>2</math>. This can be expressed in the same "line" format as <math>(2, 2, 2, 2, 6)</math>, where dividers can again be placed to group divisors. In this case, there are <math>5</math> ways to order divisors, and <math>2^4</math> ways to place dividers, so we have an <math>5 \cdot 2^4</math> possible sequences for this case. Notice that in this format, we double count cases where <math>12</math> is a factor, we triple count cases where <math>24</math> is a factor, etc. Precisely, for any case counted <math>n</math> times in the first step, it is counted <math>n - 1</math> times in this step. Thus, if we subtract, we count each case exactly once. | ||
+ | |||
+ | So, we get: | ||
+ | |||
+ | <math>6 \cdot 2^5 - 5 \cdot 2^4 = \boxed{\textbf{(A) }112}</math>. ~[[User:hdai1122|hdai1122]] | ||
+ | |||
+ | ==Solution 6 (Another Fast Way)== | ||
+ | |||
+ | First we factor <math>32</math> into <math>m</math> numbers <math>g_1, \cdots, g_m</math> where <math>g_i>1,i=1,\ldots,m</math>. By applying stars and bars there are <math>\binom{5-1}{m-1}</math> ways. Then we can either insert <math>3</math> into each of the <math>m+1</math> spaces between (or beyond) <math>g_i</math>'s, or multiply it to one of the <math>g_i</math>'s, a total of <math>2m+1</math> ways. Hence the answer to the problem is | ||
+ | |||
+ | <cmath> | ||
+ | \sum_{m=1}^5 (2m+1)\binom{5-1}{m-1} = \sum_{n=0}^4 (2n+3) \binom{4}{n} = 8\sum_{n=0}^4 \frac{n}{4} \binom{4}{n} + 3 \sum_{n=0}^4 \binom{4}{n} = 8 \sum_{n=0}^4 \binom{3}{n-1} + 3\sum_{n=0}^4 \binom{4}{n} = 8 \cdot 2^3 + 3\cdot 2^4 = \boxed{\textbf{(A) }112}. | ||
+ | </cmath> | ||
+ | |||
+ | ~ asops | ||
+ | |||
+ | ==Solution 7 (Integer Partition)== | ||
+ | |||
+ | Note that <math>96 = 2^5 \cdot 3</math>. <math>D(n)</math> depends on dividing <math>2^5</math> into different terms, which is the [https://en.wikipedia.org/wiki/Partition_(number_theory) integer partition] of <math>5</math>. | ||
+ | |||
+ | Divide <math>96</math> into <math>1</math> term: | ||
+ | |||
+ | There is only one way. <math>\underline{\textbf{1}}</math> | ||
+ | |||
+ | |||
+ | Divide <math>96</math> into <math>2</math> terms: | ||
+ | |||
+ | Case <math>1</math>: <math>3</math> is alone | ||
+ | <math>(2^5,3)</math> has <math>{\textbf{2}}</math> different arrangements. | ||
+ | |||
+ | Case <math>2</math>: <math>3</math> is with <math>2^n</math> | ||
+ | <math>5 = 1 + 4 = 2 + 3</math>. For <math>(2, 2^4)</math> and <math>(2^2, 2^3)</math>, <math>3</math> can be with any term from the <math>2</math> tuples, and the number of arrangement of the <math>2</math> terms is <math>2</math>. <math>2 \cdot 2 \cdot 2 = {\textbf{8}}</math> | ||
+ | |||
+ | <math>2 + 8 = \underline{\textbf{10}}</math> | ||
+ | |||
+ | |||
+ | Divide <math>96</math> into <math>3</math> terms: | ||
+ | |||
+ | Case <math>1</math>: <math>3</math> is alone | ||
+ | <math>5 = 2 + 3 = 1 + 4</math>. For <math>(2^2, 2^3, 3)</math> and <math>(2, 2^4, 3)</math>, there are <math>3!</math> arrangements each. <math>2 \cdot 3! = {\textbf{12}}</math> | ||
+ | |||
+ | Case <math>2</math>: <math>3</math> is with <math>2^n</math> | ||
+ | <math>5 = 1 + 1 + 3 = 1 + 2 + 2</math>. For <math>(2, 2, 2^3)</math> and <math>(2, 2^2, 2^2)</math>, <math>3</math> can be with any term from the <math>2</math> tuples. If <math>3</math> is with <math>2^3</math> for the first tuple, or <math>2</math> for the second tuple, the number of arrangements is <math>3</math> for each. If <math>3</math> is with <math>2</math> for the first tuple, or <math>2^2</math> for the second tuple, the number of arrangements is <math>6</math> for each. <math>2 \cdot 3 + 2 \cdot 6 = {\textbf{18}}</math> | ||
+ | |||
+ | <math>12 + 18 = \underline{\textbf{30}}</math> | ||
+ | |||
+ | |||
+ | Divide <math>96</math> into <math>4</math> terms: | ||
+ | |||
+ | Case <math>1</math>: <math>3</math> is alone | ||
+ | <math>5 = 1 + 1 + 3 = 1 + 2 + 2</math>. For <math>(2, 2, 2^3, 3)</math> and <math>(2, 2^2, 2^2, 3)</math>, there are <math>\frac{4!}{2!}</math> arrangements each. <math>2 \cdot \frac{4!}{2!} = {\textbf{24}}</math> | ||
+ | |||
+ | Case <math>2</math>: <math>3</math> is with <math>2^n</math> | ||
+ | <math>5 = 1 + 1 + 1 + 2</math>. For <math>(2, 2, 2, 2^2)</math>, <math>3</math> can be with any term from the tuple. If <math>3</math> is with <math>2^2</math>, the number of arrangements is <math>\frac{4!}{3!}</math>. If <math>3</math> is with <math>2</math>, the number of arrangements is <math>\frac{4!}{2!}</math>. | ||
+ | <math>\frac{4!}{3!} + \frac{4!}{2!} = {\textbf{16}}</math> | ||
+ | |||
+ | <math>24 + 16 = \underline{\textbf{40}}</math> | ||
+ | |||
+ | |||
+ | Divide <math>96</math> into <math>5</math> terms: | ||
+ | |||
+ | When dividing <math>5</math> into <math>5</math> parts there are <math>2</math> cases. | ||
+ | |||
+ | Case <math>1</math>: <math>3</math> is alone | ||
+ | <math>5 = 1 + 1 + 1 + 2</math>. For <math>(2, 2, 2, 2^2, 3)</math>, there are <math>\frac{5!}{3!}</math> arrangements. <math>\frac{5!}{3!} = {\textbf{20}}</math> | ||
+ | |||
+ | Case <math>2</math>: <math>3</math> is with <math>2^n</math> | ||
+ | <math>5 = 1 + 1 + 1 + 1 + 1</math>. For <math>(2, 2, 2, 2, 2)</math>, <math>3</math> can only be with <math>2</math>. The number of arrangements is <math>\frac{5!}{4!} = \textbf{5}</math> | ||
+ | |||
+ | <math>20 + 5 = \underline{\textbf{25}}</math> | ||
+ | |||
+ | |||
+ | Divide <math>96</math> into <math>6</math> terms: | ||
+ | |||
+ | <math>5 = 1 + 1 + 1 + 1 + 1</math>. The number of arrangements of <math>(2, 2, 2, 2, 2, 3)</math> is <math>\frac{6!}{5!} = \underline{\textbf{6}}</math> | ||
+ | |||
+ | |||
+ | <math> 1 + 10 + 30 + 40 + 25 + 6 = \boxed{\textbf{(A) }112}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 8== | ||
+ | |||
+ | Ignore the <math>3</math> first and first count <math>2^{x_1+x_2+...+x^n}=32</math> which <math>x_1+x_2+...+x_n=5</math>. This implies that <math>n</math> is less than or equal to <math>5</math>. Now, we can see that <math>3</math> can lie between the two <math>x_i, x_{i+1}</math>, or contribute to one of them. This gives <math>2k+1</math> if <math>x_1+...+x_k=5</math>. Now, just sum up gives <math>\binom{5-1}{5-1}\cdot 11+\binom{5-1}{4-1}\cdot 9+\binom{5-1}{3-1}\cdot 7+\binom{5-1}{2-1}\cdot 5+\binom{5-1}{1-1}\cdot 3=\boxed{112}</math>. | ||
+ | |||
+ | ==Solution 9 (Cleanest Algebra)== | ||
+ | Consider how we partition the factors of <math>96 = 2^5\cdot 3</math>. For each <math>k</math>, there are two cases. Either we can put the <math>2</math>s into <math>k</math> nonzero parts, so that the <math>3</math> shares a partition with some <math>2</math>s, which can be done in <math>k\binom{5-(k-1)+(k-1)-1}{k-2} = k\binom{4}{k-2}</math> ways, or we can put the <math>2</math>s into <math>k-1</math> nonzero parts and put the <math>3</math> in its own partition, which can be done in <math>k\binom{5-k+k-1}{k-1} = k\binom{4}{k-1}</math> ways. Summing over all <math>k</math>, we have | ||
+ | <cmath>\sum_{k=1}^6 \binom{4}{k-2} + \sum_{k=1}^6 \binom{4}{k-1}.</cmath> | ||
+ | |||
+ | But <math>\sum_{k=1}^6 \binom{4}{k-2} = 2\binom{4}{0}+3\binom{4}{1} + \dots + 6\binom{4}{4} = 4 \cdot \sum_{i=0}^4 \binom{4}{i} = 4 \cdot 2^4 = 64</math>. Similarly, <math>\sum_{k=1}^6 \binom{4}{k-1} = 3\cdot 2^4 = 48</math>. So our answer is <math>64+48 = \boxed{\textbf{(A) }112}</math>. | ||
+ | |||
+ | ~InsetIowa9 | ||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/8WrdYLw9_ns?t=995 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
− | |||
− | |||
− | + | == Video Solution by MathEx == | |
+ | https://www.youtube.com/watch?v=977F9lBb37E | ||
+ | |||
− | ==Video Solution== | + | == Video Solution == |
− | https:// | + | Video is 4:29 long Uses casework. https://youtu.be/fR3VL7g90QM |
+ | Mark888 | ||
==See Also== | ==See Also== | ||
− | |||
{{AMC10 box|year=2020|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2020|ab=B|num-b=24|after=Last Problem}} | ||
+ | {{AMC12 box|year=2020|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:12, 11 July 2024
- 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 product
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 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:
~Continuous_Pi
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
.
Solution 9 (Cleanest Algebra)
Consider how we partition the factors of . For each
, there are two cases. Either we can put the
s into
nonzero parts, so that the
shares a partition with some
s, which can be done in
ways, or we can put the
s into
nonzero parts and put the
in its own partition, which can be done in
ways. Summing over all
, we have
But . Similarly,
. So our answer is
.
~InsetIowa9
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
Video Solution
Video is 4:29 long Uses casework. https://youtu.be/fR3VL7g90QM Mark888
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.