Difference between revisions of "1997 AHSME Problems/Problem 22"
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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math> | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math> | ||
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+ | ==Solution== | ||
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+ | Working backwards, if <math>6 \le E \le 10</math>, then <math>6 \pm 11 \le A \le 10 \pm 11</math>. Since <math>A</math> is a positive integer, <math>17 \le A \le 21</math>. | ||
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+ | Since <math>17 \le A \le 21</math>, we know that <math>17 \pm 19 \le B \le 21 \pm 19</math>. But if <math>B=36</math>, which is the smallest possible "plus" value, then <math>E + A + B = 6 + 17 + 36 = 59</math>, which is too much money. | ||
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+ | Hence, <math>17 - 19 \le B \le 21 - 19</math>. But since <math>B</math> must be a positive integer, that leaves only two possibilities: <math>B = 1</math> or <math>B=2</math>, which correspond with <math>E = 9</math> and <math>E = 10</math>. | ||
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+ | Concentrating only on <math>E=9</math>, we have <math>E=9</math> leading to <math>A = 9 + 11 = 20</math>, which leads to <math>B = 20 - 19 = 1</math>, which leads to <math>C = 1 + 7 = 8</math>. Thus far we have given out <math>9 + 20 + 1 + 8 = 38</math> dollars. This means that Dick must have <math>56 - 38 = 18</math> dollars. However, the difference between Carlos and Dick is not <math>5</math> dollars. | ||
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+ | Thus, the right answer must be <math>\boxed{\textbf{(E)}}</math>. Verifying, if <math>E = 10</math>, then <math>A = 10 + 11 = 21</math>, <math>B = 21 - 19 = 2</math>, which leads to <math>C = 2 + 7 = 9</math>. Thus far, we have given out <math>10 + 21 + 2 + 9 = 42</math> dollars, leaving <math>56 - 42 = 14</math> dollars for Dick. Dick does indeed have <math>5</math> dollars more than Carlos, and <math>4</math> dollars more than Elgin. | ||
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=21|num-a=23}} | {{AHSME box|year=1997|num-b=21|num-a=23}} | ||
+ | {{MAA Notice}} |
Latest revision as of 17:46, 9 January 2015
Problem
Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had dollars. The absolute difference between the amounts Ashley and Betty had to spend was dollars. The absolute difference between the amounts Betty and Carlos had was dollars, between Carlos and Dick was dollars, between Dick and Elgin was dollars, and between Elgin and Ashley was dollars. How many dollars did Elgin have?
Solution
Working backwards, if , then . Since is a positive integer, .
Since , we know that . But if , which is the smallest possible "plus" value, then , which is too much money.
Hence, . But since must be a positive integer, that leaves only two possibilities: or , which correspond with and .
Concentrating only on , we have leading to , which leads to , which leads to . Thus far we have given out dollars. This means that Dick must have dollars. However, the difference between Carlos and Dick is not dollars.
Thus, the right answer must be . Verifying, if , then , , which leads to . Thus far, we have given out dollars, leaving dollars for Dick. Dick does indeed have dollars more than Carlos, and dollars more than Elgin.
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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