Difference between revisions of "2020 AMC 8 Problems/Problem 20"
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==Problem== | ==Problem== | ||
Let <math>a_{1}, a_{2}, a_{3}, a_{4},</math> and <math>a_{5}</math> be five positive integers such that <math>a_{2}=11</math> and <math>\frac{a_{i}}{a_{i+1}}\in\{\frac{1}{2}, 2\}</math> for <math>1\leq n\leq 4</math>. If <math>\{\frac{a_{1}+a_{2}+a_{3}+a_{4}+a_{5}}{5}\}=0.2</math>, compute <math>\frac{a_{1}+a_{2}+a_{3}+a_{4}+a_{5}}{5}</math>. | Let <math>a_{1}, a_{2}, a_{3}, a_{4},</math> and <math>a_{5}</math> be five positive integers such that <math>a_{2}=11</math> and <math>\frac{a_{i}}{a_{i+1}}\in\{\frac{1}{2}, 2\}</math> for <math>1\leq n\leq 4</math>. If <math>\{\frac{a_{1}+a_{2}+a_{3}+a_{4}+a_{5}}{5}\}=0.2</math>, compute <math>\frac{a_{1}+a_{2}+a_{3}+a_{4}+a_{5}}{5}</math>. | ||
+ | |||
+ | Note: <math>\{x\}</math> denotes the fractional part of <math>x</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> | <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> |
Revision as of 23:02, 9 December 2020
Problem
Let and
be five positive integers such that
and
for
. If
, compute
.
Note: denotes the fractional part of
.
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
. (
*
=
)
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 the 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.
Video Solution
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.