Difference between revisions of "1984 AIME Problems/Problem 4"
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== Solution == | == Solution == | ||
− | Suppose <math>S</math> has <math>n</math> members other than 68, and the sum of these members is <math>s</math>. Then we're given that <math>\frac{s + 68}{n + 1} = 56</math> and <math>\frac{s}{n} = 55</math>. Multiplying to clear [[denominator]]s, we have <math>s + 68 = 56n + 56</math> and <math>s = 55n</math> so <math>68 = n + 56</math>, <math>n = 12</math> and <math>s = 12\cdot 55 = 660</math>. Because the sum and number of the elements of <math>S</math> are fixed, if we want to maximize the largest number in <math>S</math>, we should take all but one member of <math>S</math> to be as small as possible. Since all members of <math>S</math> are positive integers, the smallest possible value of a member is 1. Thus the largest possible element is <math>660 - 11 | + | Suppose <math>S</math> has <math>n</math> members other than 68, and the sum of these members is <math>s</math>. Then we're given that <math>\frac{s + 68}{n + 1} = 56</math> and <math>\frac{s}{n} = 55</math>. Multiplying to clear [[denominator]]s, we have <math>s + 68 = 56n + 56</math> and <math>s = 55n</math> so <math>68 = n + 56</math>, <math>n = 12</math> and <math>s = 12\cdot 55 = 660</math>. Because the sum and number of the elements of <math>S</math> are fixed, if we want to maximize the largest number in <math>S</math>, we should take all but one member of <math>S</math> to be as small as possible. Since all members of <math>S</math> are positive integers, the smallest possible value of a member is 1. Thus the largest possible element is <math>660 - 11 = \boxed{649}</math>. |
== See also == | == See also == |
Revision as of 18:59, 17 January 2011
Problem
Let be a list of positive integers - not necessarily distinct - in which the number appears. The arithmetic mean of the numbers in is . However, if is removed, the arithmetic mean of the numbers is . What's the largest number that can appear in ?
Solution
Suppose has members other than 68, and the sum of these members is . Then we're given that and . Multiplying to clear denominators, we have and so , and . Because the sum and number of the elements of are fixed, if we want to maximize the largest number in , we should take all but one member of to be as small as possible. Since all members of are positive integers, the smallest possible value of a member is 1. Thus the largest possible element is .
See also
1984 AIME (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 | ||
All AIME Problems and Solutions |