Difference between revisions of "1984 AIME Problems/Problem 4"
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s&=55n. | s&=55n. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Subtracting the equations, we get <math>68=n+56,</math> from which <math>n=12</math> | + | Subtracting the equations, we get <math>68=n+56,</math> from which <math>n=12.</math> It follows that <math>s=660.</math> |
− | The sum of the twelve remaining numbers is <math>660.</math> To maximize the largest number, we minimize the other eleven numbers: | + | The sum of the twelve remaining numbers is <math>660.</math> To maximize the largest number, we must minimize the other eleven numbers: We can have eleven <math>1</math>s and one <math>660-11\cdot1=\boxed{649}.</math> |
− | |||
− | We can have eleven <math>1</math>s and one <math>660-11\cdot1=\boxed{649}.</math> | ||
~JBL (Solution) | ~JBL (Solution) | ||
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\textbf{Final} & n & 55 & 55n | \textbf{Final} & n & 55 & 55n | ||
\end{array}</cmath> | \end{array}</cmath> | ||
+ | We are given that <cmath>56(n+1)-68=55n,</cmath> from which <math>n=12.</math> It follows that the sum of the remaining numbers is <math>55n=660.</math> We continue with the last paragraph of Solution 1 to get the answer <math>\boxed{649}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 01:31, 22 June 2021
Problem
Let be a list of positive integers--not necessarily distinct--in which the number
appears. The average (arithmetic mean) of the numbers in
is
. However, if
is removed, the average of the remaining numbers drops to
. What is the largest number that can appear in
?
Solution 1 (Two Variables)
Suppose that has
numbers other than
and the sum of these numbers is
We are given that
Clearing denominators, we have
Subtracting the equations, we get
from which
It follows that
The sum of the twelve remaining numbers is To maximize the largest number, we must minimize the other eleven numbers: We can have eleven
s and one
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
Solution 2 (One Variable)
Suppose that has
numbers other than
We have the following table:
We are given that
from which
It follows that the sum of the remaining numbers is
We continue with the last paragraph of Solution 1 to get the answer
~MRENTHUSIASM
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 |