Difference between revisions of "2015 AMC 10A Problems/Problem 3"
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− | + | ==Problem== | |
+ | Ann made a <math>3</math>-step staircase using <math>18</math> toothpicks as shown in the figure. How many toothpicks does she need to add to complete a <math>5</math>-step staircase? | ||
+ | |||
+ | <math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}}\ 22\qquad\textbf{(E)}\ 24</math> | ||
+ | |||
+ | ==Solution== | ||
+ | We can see that a <math>1</math>-step staircase requires <math>4</math> toothpicks and a <math>2</math>-step staircase requires <math>10</math> toothpicks. Thus, to go from a <math>1</math>-step to <math>2</math>-step staircase, <math>6</math> additional toothpicks are needed and to go from a <math>2</math>-step to <math>3</math>-step staircase, <math>8</math> additional toothpicks are needed. Applying this pattern, to go from a <math>3</math>-step to <math>4</math>-step staircase, <math>10</math> additional toothpicks are needed and to go from a <math>4</math>-step to <math>5</math>-step staircase, <math>12</math> additional toothpicks are needed. Our answer is <math>10+12=\boxed{\textbf{(D)}\ 22}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2015|ab=A|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Revision as of 16:42, 4 February 2015
Problem
Ann made a -step staircase using toothpicks as shown in the figure. How many toothpicks does she need to add to complete a -step staircase?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}}\ 22\qquad\textbf{(E)}\ 24$ (Error compiling LaTeX. Unknown error_msg)
Solution
We can see that a -step staircase requires toothpicks and a -step staircase requires toothpicks. Thus, to go from a -step to -step staircase, additional toothpicks are needed and to go from a -step to -step staircase, additional toothpicks are needed. Applying this pattern, to go from a -step to -step staircase, additional toothpicks are needed and to go from a -step to -step staircase, additional toothpicks are needed. Our answer is .
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.