Difference between revisions of "1994 AJHSME Problems/Problem 21"
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A gumball machine contains <math>9</math> red, <math>7</math> white, and <math>8</math> blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is | A gumball machine contains <math>9</math> red, <math>7</math> white, and <math>8</math> blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is | ||
| − | <math>\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18</math> | + | <math>\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18</math> |
| + | |||
| + | ==Solution== | ||
| + | If a person gets three gumballs of each of the three colors, that is, <math>9</math> gumballs, then the <math>10^{\text{th}}</math> gumball must be the fourth one for one of the colors. Therefore, the person must buy <math>\boxed{\text{(C)}\ 10}</math> gumballs. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=20|num-a=22}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:10, 7 November 2018
Problem
A gumball machine contains 
red, 
white, and 
blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is
Solution
If a person gets three gumballs of each of the three colors, that is, gumballs, then the 
gumball must be the fourth one for one of the colors. Therefore, the person must buy 
gumballs.
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
