Question

The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the largest of the six integers is

• A. 26
• B. 28
• C. 30
• D. 32
• E. 34

Hint:

To maximize the largest term with mean/median/mode constraints, minimize the other terms: push the lower ones down, keep the median pair as low as allowed, and ensure the mode condition (uniqueness and its value) isn’t violated.

Answer 1

The Correct Option is D

Let the ordered integers be \(a_1\le a_2\le a_3\le a_4\le a_5\le a_6\).
Step 1: Use mean and median.
Mean \(=15⇒ a_1+\cdots+a_6=90\).
Median \(=18⇒ \dfrac{a_3+a_4}{2}=18⇒ a_3+a_4=36.\)
Step 2: Enforce the “only mode \(<18\)” condition.
Exactly one value occurring most often must be \(<18\). To maximize \(a_6\), minimize the other five while keeping a unique mode \(<18\) and avoiding any other value repeating as often.
If \(a_3=a_4=18\), then 18 would repeat and could become (or tie for) the mode, which is forbidden. Hence choose the smallest unequal pair summing to 36:
\[ (a_3,a_4)=(17,19). \] Step 3: Create the unique mode \(<18\) and minimize the first five.
Make the smallest two equal to create the mode and keep them minimal:
\[ a_1=a_2=1 (\text{mode }=1<18). \] Now \(a_5\ge a_4=19\). If \(a_5=19\), then 19 would occur twice, tying the mode (two 1's vs. two 19's), which violates “only mode”. Therefore take the least allowed: \[ a_5=20. \] Step 4: Compute \(a_6\) and conclude maximality.
\[ a_6=90-(a_1+a_2+a_3+a_4+a_5)=90-(1+1+17+19+20)=32. \] Any other valid choice has \(a_5\ge 20\) and/or larger earlier terms, which only decreases \(a_6\). Hence the maximum possible largest integer is \[ \boxed{32}. \]