Difference between revisions of "2020 AMC 8 Problems/Problem 20"
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\end{tabular} | \end{tabular} | ||
\endgroup</cmath> | \endgroup</cmath> | ||
− | <math> | + | <math>\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2</math> |
==Solution 1== | ==Solution 1== | ||
Line 22: | Line 22: | ||
==Solution 2== | ==Solution 2== | ||
− | Notice the average height of the trees ends with <math>0.2</math> therefore the sum of all five heights of the trees must end with <math>1</math>. (<math>0.2</math> | + | Notice the average height of the trees ends with <math>0.2</math>; therefore, the sum of all five heights of the trees must end with <math>1</math> or <math>6</math>. (<math>0.2</math> <math>\cdot</math> <math>5</math> = <math>1</math>) |
We already know Tree <math>2</math> is <math>11</math> meters tall. Both Tree <math>1</math> and Tree <math>3</math> must <math>22</math> meters tall - since neither can be <math>5.5</math>. | We already know Tree <math>2</math> is <math>11</math> meters tall. Both Tree <math>1</math> and Tree <math>3</math> must <math>22</math> meters tall - since neither can be <math>5.5</math>. | ||
Once again, apply our observation for solving for the Tree <math>4</math>'s height. Tree <math>4</math> can't be <math>11</math> meters for the sum of the five tree heights to still end with <math>1</math>. Therefore, the Tree <math>4</math> is <math>44</math> meters tall. | Once again, apply our observation for solving for the Tree <math>4</math>'s height. Tree <math>4</math> can't be <math>11</math> meters for the sum of the five tree heights to still end with <math>1</math>. Therefore, the Tree <math>4</math> is <math>44</math> meters tall. | ||
− | Now | + | Now, Tree <math>5</math> can either be <math>22</math> or <math>88</math>. Find the average height for both cases of Tree <math>5</math>. Doing this, we realize the Tree <math>5</math> must be <math>22</math> for the average height to end with <math>0.2</math> and that the average height is <math>\boxed{\textbf{(B)}\ 24.2}</math>. |
==Solution 3== | ==Solution 3== | ||
Line 45: | Line 45: | ||
If Tree 4 now has a height of <math>11</math>, then Tree 5 would need to have height <math>22</math>, but in that case <math>S</math> would equal <math>88</math>, which is not <math>1</math> more than a multiple of <math>5</math>. So we instead take Tree 4 to have height <math>44</math>. Then the sum of the heights of the first 4 trees is <math>22+11+22+44 = 99</math>, so using a height of <math>22</math> for Tree 5 gives <math>S=121</math>, which is <math>1</math> more than a multiple of <math>5</math> (whereas <math>88</math> gives <math>S = 187</math>, which is not). Thus the average height of the trees is <math>\frac{121}{5} = \boxed{\textbf{(B) }24.2}</math> meters. | If Tree 4 now has a height of <math>11</math>, then Tree 5 would need to have height <math>22</math>, but in that case <math>S</math> would equal <math>88</math>, which is not <math>1</math> more than a multiple of <math>5</math>. So we instead take Tree 4 to have height <math>44</math>. Then the sum of the heights of the first 4 trees is <math>22+11+22+44 = 99</math>, so using a height of <math>22</math> for Tree 5 gives <math>S=121</math>, which is <math>1</math> more than a multiple of <math>5</math> (whereas <math>88</math> gives <math>S = 187</math>, which is not). Thus the average height of the trees is <math>\frac{121}{5} = \boxed{\textbf{(B) }24.2}</math> meters. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Since we know that the tree heights have to be integers, then it is immediate that Tree 1 and 3 have a height of <math>22</math>. Now using the information given by the last column (that the average of the heights of the trees ends in <math>.2</math>), we can tell that the sum of all the heights of the trees ends in either <math>1</math> or <math>6</math>, because those are the only numbers from <math>0</math> to <math>9</math> that are congruent to <math>1</math> after taking modulo <math>5</math>. The two multiples of eleven (eleven because all of the tree heights have to be a multiple of eleven if they are integers) that come to mind are <math>66</math> and <math>121</math>. Since the sum of the heights of Tree 1, 2, and 3 is already <math>55</math>, we know that <math>66</math> is impossible to obtain. Then, we can decide with relative confidence that the answer should be <math>\frac{121}{5} = \boxed{\textbf{(B) }24.2}</math>. | ||
+ | |||
+ | |||
+ | P.S. : We can check our solution by trying to obtain <math>121</math>. Surely enough, we have Tree 1 = <math>22</math>, Tree 2 = <math>11</math>, Tree 3 = <math>22</math>, Tree 4 = <math>44</math>, and Tree 5 = <math>22</math>. These all add up to get <math>121</math>. | ||
+ | |||
+ | ==Solution 5 (basic modular arithmetic)== | ||
+ | Since her average ends with 0.2, the sum of her tree heights must be <math>\equiv 1 \pmod {5}</math>. | ||
+ | If Tree 1's height is <math>a</math>, Tree 3's is <math>b</math>, Tree 4's is <math>c</math>, and Tree 5's is <math>d</math>, then we <math>a</math> and <math>b</math> are both 22, since all of the tree heights are integers. | ||
+ | Now we have <math>22+11+22+c+d \equiv 1 \pmod {5}</math>. Simplifying, we get <math>c+d \equiv 1 \pmod {5}</math>. | ||
+ | The only possible combination of <math>c</math> and <math>d</math> that abides the condition are 44 and 22, respectively. | ||
+ | Adding these up, we get 121, and <math>\frac{121}{5}</math> is <math>\boxed{\textbf{(B) }24.2}</math>. | ||
+ | |||
+ | -kn07 | ||
+ | |||
+ | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=bHNrBwwUCMI | ||
+ | |||
+ | ~NiuniuMaths | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/UnVo6jZ3Wnk?si=1OAhRlAjzYP_EzBa&t=3750 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (🚀 Very fast 🚀)== | ||
+ | https://youtu.be/swfsLlZVGbM | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
− | https://youtu.be/ | + | https://youtu.be/PCb5vOIioJA |
+ | |||
+ | Please like and subscribe! | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/uCV9-WpaIjw | ||
+ | ~savannahsolver | ||
+ | ==Video Solution== | ||
https://youtu.be/VnOecUiP-SA | https://youtu.be/VnOecUiP-SA | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=1045 | ||
+ | |||
+ | ~Interstigation | ||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=19|num-a=21}} | {{AMC8 box|year=2020|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:36, 26 January 2024
Contents
- 1 Problem 20
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5 (basic modular arithmetic)
- 7 Video Solution by NiuniuMaths (Easy to understand!)
- 8 Video Solution by Math-X (First understand the problem!!!)
- 9 Video Solution (🚀 Very fast 🚀)
- 10 Video Solution
- 11 Video Solution by WhyMath
- 12 Video Solution
- 13 Video Solution by Interstigation
- 14 See also
Problem 20
A scientist walking through a forest recorded as integers the heights of trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
Solution 1
We will show that ,
,
,
, and
meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height
meters, we can deduce that Trees 1 and 3 both have a height of
meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of
and
,
and
, or
and
. Checking each of these, in the first case, the average is
meters, which doesn't end in
as the problem requires. Therefore, we consider the other cases. With
and
, the average is
meters, which again does not end in
, but with
and
, the average is
meters, which does. Consequently, the answer is
.
Solution 2
Notice the average height of the trees ends with ; therefore, the sum of all five heights of the trees must end with
or
. (
=
)
We already know Tree
is
meters tall. Both Tree
and Tree
must
meters tall - since neither can be
.
Once again, apply our observation for solving for the Tree
's height. Tree
can't be
meters for the sum of the five tree heights to still end with
. Therefore, the Tree
is
meters tall.
Now, Tree
can either be
or
. Find the average height for both cases of Tree
. Doing this, we realize the Tree
must be
for the average height to end with
and that the average height is
.
Solution 3
As in Solution 1, we shall show that the heights of the trees are ,
,
,
, and
meters. Let
be the sum of the heights, so that the average height will be
meters. We note that
, so in order for
to end in
,
must be one more than a multiple of
. Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both
meters. At this point, our table looks as follows:
If Tree 4 now has a height of , then Tree 5 would need to have height
, but in that case
would equal
, which is not
more than a multiple of
. So we instead take Tree 4 to have height
. Then the sum of the heights of the first 4 trees is
, so using a height of
for Tree 5 gives
, which is
more than a multiple of
(whereas
gives
, which is not). Thus the average height of the trees is
meters.
Solution 4
Since we know that the tree heights have to be integers, then it is immediate that Tree 1 and 3 have a height of . Now using the information given by the last column (that the average of the heights of the trees ends in
), we can tell that the sum of all the heights of the trees ends in either
or
, because those are the only numbers from
to
that are congruent to
after taking modulo
. The two multiples of eleven (eleven because all of the tree heights have to be a multiple of eleven if they are integers) that come to mind are
and
. Since the sum of the heights of Tree 1, 2, and 3 is already
, we know that
is impossible to obtain. Then, we can decide with relative confidence that the answer should be
.
P.S. : We can check our solution by trying to obtain . Surely enough, we have Tree 1 =
, Tree 2 =
, Tree 3 =
, Tree 4 =
, and Tree 5 =
. These all add up to get
.
Solution 5 (basic modular arithmetic)
Since her average ends with 0.2, the sum of her tree heights must be .
If Tree 1's height is
, Tree 3's is
, Tree 4's is
, and Tree 5's is
, then we
and
are both 22, since all of the tree heights are integers.
Now we have
. Simplifying, we get
.
The only possible combination of
and
that abides the condition are 44 and 22, respectively.
Adding these up, we get 121, and
is
.
-kn07
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=bHNrBwwUCMI
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=1OAhRlAjzYP_EzBa&t=3750
~Math-X
Video Solution (🚀 Very fast 🚀)
~Education, the Study of Everything
Video Solution
Please like and subscribe!
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1045
~Interstigation
See also
2020 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.