1984 AIME Problems/Problem 4: Difference between revisions

Revision as of 23:43, 21 June 2021

Problem

Let $S$ be a list of positive integers - not necessarily distinct - in which the number $68$ appears. The arithmetic mean of the numbers in $S$ is $56$ . However, if $68$ is removed, the arithmetic mean of the numbers is $55$ . What's the largest number that can appear in $S$ ?

Solution 1 (Two Variables)

Suppose that $S$ has $n$ numbers other than $68,$ and the sum of these numbers is $s.$

We are given that $\begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*}$ Clearing denominators, we have $\begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*}$ Subtracting the equations, we get $68=n+56,$ from which $n=12$ and $s=660.$

The sum of the twelve remaining numbers is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$ s and one $660-11\cdot1=\boxed{649}.$

~JBL (Solution)

~MRENTHUSIASM (Reconstruction)

@@ Line 2: / Line 2: @@
 Let <math>S</math> be a list of positive integers - not necessarily distinct - in which the number <math>68</math> appears. The arithmetic mean of the numbers in <math>S</math> is <math>56</math>. However, if <math>68</math> is removed, the arithmetic mean of the numbers is <math>55</math>. What's the largest number that can appear in <math>S</math>?
-== Solution ==
+== Solution 1 (Two Variables) ==
-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
+Suppose that <math>S</math> has <math>n</math> numbers other than <math>68,</math> and the sum of these numbers is <math>s.</math>
-<math>s + 68 = 56n + 56</math> and
+We are given that
+<cmath>\begin{align*}
+\frac{s+68}{n+1}&=56, \\
+\frac{s}{n}&=55.
+\end{align*}</cmath>
+Clearing denominators, we have
+<cmath>\begin{align*}
+s+68&=56n+56, \\
+s&=55n.
+\end{align*}</cmath>
+Subtracting the equations, we get <math>68=n+56,</math> from which <math>n=12</math> and <math>s=660.</math>
-<math>s = 55n</math> so
+The sum of the twelve remaining numbers is <math>660.</math> To maximize the largest number, we minimize the other eleven numbers: We can have eleven <math>1</math>s and one <math>660-11\cdot1=\boxed{649}.</math>
-<math>68 = n + 56</math>,
+~JBL (Solution)
-<math>n = 12</math> and
+~MRENTHUSIASM (Reconstruction)
-<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 ==

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1984 AIME Problems/Problem 4: Difference between revisions

Revision as of 23:43, 21 June 2021

Problem

Solution 1 (Two Variables)

See also