Difference between revisions of "2016 AMC 10A Problems/Problem 20"
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+ | ==Problem== | ||
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>? | For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>? | ||
<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math> | <math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math> | ||
− | ==Solution== | + | ==Solution 1== |
All the desired terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math> (the <math>1^t</math> part is necessary to make stars and bars work better.) | All the desired terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math> (the <math>1^t</math> part is necessary to make stars and bars work better.) | ||
− | Since <math>x</math>, <math>y</math>, <math>z</math>, and <math>w</math> must be at least <math>1</math> (<math>t</math> can be 0), let <math>x' = x - 1</math>, <math>y' = y - 1</math>, <math>z' = z - 1</math>, and <math>w' = w - 1</math>, so <math>x' + y' + z' + w' + t = N - 4</math>. Now, we use stars and bars to see that there are <math>\binom{N}{4}</math> solutions to this equation. We notice that <math>\binom{14}{4} = 1001</math>, | + | Since <math>x</math>, <math>y</math>, <math>z</math>, and <math>w</math> must be at least <math>1</math> (<math>t</math> can be <math>0</math>), let <math>x' = x - 1</math>, <math>y' = y - 1</math>, <math>z' = z - 1</math>, and <math>w' = w - 1</math>, so <math>x' + y' + z' + w' + t = N - 4</math>. Now, we use [[stars and bars]] (also known as ball and urn) to see that there are <math>\binom{(N-4)+4}{4}</math> or <math>\binom{N}{4}</math> solutions to this equation. We notice that <math>1001=7\cdot11\cdot13</math>, which leads us to guess that <math>N</math> is around these numbers. This suspicion proves to be correct, as we see that <math>\binom{14}{4} = 1001</math>, giving us our answer of <math>\boxed{\textbf{(B) }14.}</math> |
− | + | Note: An alternative is instead of making the transformation, we "give" the variables <math>x, y, z, w</math> 1, and then proceed as above. | |
− | + | ~ Mathkiddie(minor edits by vadava_lx) | |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | By the [[Hockey Stick Identity]], the number of terms that have all <math>a,b,c,d</math> raised to a positive power is <math>\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}</math>. We now want to find some <math>N</math> such that <math>\binom{N}{4} = 1001</math>. As mentioned above, after noticing that <math>1001 = 7\cdot11\cdot13</math>, and some trial and error, we find that <math>\binom{14}{4} = 1001</math>, giving us our answer of <math>\boxed{\textbf{(B) }14.}</math> | ||
+ | |||
+ | ~minor edits by vadava_lx | ||
+ | |||
+ | ==Solution 3 (Casework)== | ||
+ | |||
+ | The terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math>. The problem becomes distributing <math>N</math> identical balls to <math>5</math> different boxes <math>(x, y, z, w, t)</math> such that each of the boxes <math>(x, y, z, w)</math> has at least <math>1</math> ball. The <math>N</math> balls in a row have <math>N-1</math> gaps among them. We are going to put <math>4</math> or <math>3</math> divisors into those <math>N-1</math> gaps. There are <math>2</math> cases of how to put the divisors. | ||
+ | |||
+ | Case <math>1</math>: | ||
+ | Put 4 divisors into <math>N-1</math> gaps. It corresponds to each of <math>(a, b, c, d, 1)</math> has at least one term. There are <math>\binom{N-1}{4}</math> terms. | ||
+ | |||
+ | Case <math>2</math>: | ||
+ | Put 3 divisors into <math>N-1</math> gaps. It corresponds to each of <math>(a, b, c, d)</math> has at least one term. There are <math>\binom{N-1}{3}</math> terms. | ||
+ | |||
+ | So, there are <math>\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}</math> terms. <math>\binom{N}{4} = 1001</math>, and since we have <math>\binom{14}{4} = 1001, N=\boxed{\textbf{(B) }14.}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/yGJwp72qPzk?t=88 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=R3eJW3PCYMs | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/TpG8wlj4eRA with 5 Stars and Bars examples preceding the solution. Time stamps in description to skip straight to solution. | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
+ | |||
{{AMC10 box|year=2016|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2016|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:13, 8 October 2023
Contents
Problem
For some particular value of , when
is expanded and like terms are combined, the resulting expression contains exactly
terms that include all four variables
and
, each to some positive power. What is
?
Solution 1
All the desired terms are in the form , where
(the
part is necessary to make stars and bars work better.)
Since
,
,
, and
must be at least
(
can be
), let
,
,
, and
, so
. Now, we use stars and bars (also known as ball and urn) to see that there are
or
solutions to this equation. We notice that
, which leads us to guess that
is around these numbers. This suspicion proves to be correct, as we see that
, giving us our answer of
Note: An alternative is instead of making the transformation, we "give" the variables 1, and then proceed as above.
~ Mathkiddie(minor edits by vadava_lx)
Solution 2
By the Hockey Stick Identity, the number of terms that have all raised to a positive power is
. We now want to find some
such that
. As mentioned above, after noticing that
, and some trial and error, we find that
, giving us our answer of
~minor edits by vadava_lx
Solution 3 (Casework)
The terms are in the form , where
. The problem becomes distributing
identical balls to
different boxes
such that each of the boxes
has at least
ball. The
balls in a row have
gaps among them. We are going to put
or
divisors into those
gaps. There are
cases of how to put the divisors.
Case :
Put 4 divisors into
gaps. It corresponds to each of
has at least one term. There are
terms.
Case :
Put 3 divisors into
gaps. It corresponds to each of
has at least one term. There are
terms.
So, there are terms.
, and since we have
Video Solution by OmegaLearn
https://youtu.be/yGJwp72qPzk?t=88
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=R3eJW3PCYMs
Video Solution 2
https://youtu.be/TpG8wlj4eRA with 5 Stars and Bars examples preceding the solution. Time stamps in description to skip straight to solution.
~IceMatrix
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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