Difference between revisions of "2020 AMC 8 Problems/Problem 4"
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==Solution 2== | ==Solution 2== | ||
The first hexagon has <math>1</math> dot. The second hexagon has <math>1+6</math> dots. The third hexagon <math>1+6+12</math> dots. Following this pattern, we predict that the fourth hexagon will have <math>1+6+12+18=37</math> dots <math>\implies\boxed{\textbf{(B) }37}</math>.<br> | The first hexagon has <math>1</math> dot. The second hexagon has <math>1+6</math> dots. The third hexagon <math>1+6+12</math> dots. Following this pattern, we predict that the fourth hexagon will have <math>1+6+12+18=37</math> dots <math>\implies\boxed{\textbf{(B) }37}</math>.<br> | ||
− | ~ | + | ~ junaidmansuri |
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=3|num-a=5}} | {{AMC8 box|year=2020|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 06:18, 18 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
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 |
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