Difference between revisions of "2020 AMC 8 Problems/Problem 4"
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==Solution 5== | ==Solution 5== | ||
− | Adding up the dots by rows in each hexagon, we see that the first hexagon has <math>1</math> dot, the second has <math>2+3+2</math> dots and the third has <math>3+4+5+4+3</math> dots. Following the pattern, the fourth hexagon has <math>4+5+6+7+6+5+4=\boxed{\textbf{(B) }37}</math>. | + | Adding up the dots by rows in each hexagon, we see that the first hexagon has <math>1</math> dot, the second has <math>2+3+2</math> dots and the third has <math>3+4+5+4+3</math> dots. Following the pattern, the fourth hexagon has <math>4+5+6+7+6+5+4=\boxed{\textbf{(B) }37}</math>. Note yes this is what you do and if you don't undertstand just add them up to get 37$. |
-vaporwave | -vaporwave |
Revision as of 18:59, 19 November 2020
Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
Diagram by sircalcsalot
Solution
We find the pattern . The sum of the first four numbers in this sequence is
.
Solution 2
The first hexagon has dot. The second hexagon has
dots. The third hexagon
dots. Following this pattern, we predict that the fourth hexagon will have
dots
.
~junaidmansuri
Solution 3
Each band adds so we have
[pog]
Solution 4
Let the hexagon with dot be
. Notice that the rest of the terms are generated by the recurrence relation
for
. Using this, we find that
and
.
-franzliszt
Solution 5
Adding up the dots by rows in each hexagon, we see that the first hexagon has dot, the second has
dots and the third has
dots. Following the pattern, the fourth hexagon has
. Note yes this is what you do and if you don't undertstand just add them up to get 37$.
-vaporwave
Video Solution
~savannahsolver
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.