AMC 10 · 2007 · #15

Grade 7 arithmetic
mean-median-mode-rangelinear-equations-one-varcasework caseworkconvert-to-algebra ↑ Prerequisites: mean-median-mode-range
📏 Long solution 💡 3 insights
Problem
One new number is added to a four-number set so the five-number set has median equal to mean. Find the total of every value that works.

Pick an answer.

(A)
7
(B)
9
(C)
19
(D)
24
(E)
26
How to solve
Strategy Make a Systematic List

The mean is easy — it is (28+n)/5 for every n. The median is the hard half, because it is a different expression depending on whether n lands below the old middle, inside it, or above it. So the spine of the problem is Tool #2 (Make a Systematic List): carve the number line into stretches on which the median is a fixed expression, and make sure the carving leaves no gap and no overlap, because a missed stretch is a missed value of n and the question asks for the sum of all of them. Inside each stretch Tool #4 (Introduce a Variable) and Tool #13 (Convert to Algebra) reduce the condition to one linear equation with exactly one root. Tool #6 (Guess and Check) then does the step most quick write-ups skip. Each case assumed n lies in some range and used that assumption to compute the median; the number it spits out is only a real solution if it actually lands back inside the range it assumed. Without that return trip the casework proves necessity and never sufficiency.

1STEP 1

Write the mean once, for good

The mean can be written once and reused.

mean=(3+6+9+10+n)/5=(28+n)/5
2STEP 2

The median is the middle of 6, n, 9

The median takes one of three values depending on where the number lands.

median= 6,& n < 6 ; n,& 6 < n < 9 ; 9,& n > 9 n ≠ 6, n ≠ 9 (forbidden)
3STEP 3

Case n < 6: the median is 6

The low case gives 2.

(28+n)/5=6 ⟹ 28+n=30 ⟹ n=2
4STEP 4

Case 6 < n < 9: the median is n

The middle case gives 7.

(28+n)/5=n ⟹ 28+n=5n ⟹ 4n=28 ⟹ n=7
5STEP 5

Case n > 9: the median is 9

The high case gives 17.

(28+n)/5=9 ⟹ 28+n=45 ⟹ n=17
6STEP 6

Send each answer back to its own case

Each value falls inside its own case, so all three survive.

2 < 6 ✓ 6 < 7 < 9 ✓ 17 > 9 ✓
7STEP 7

Rebuild the sets, then add

Adding gives 26, choice (E).

{2,3,6,9,10}, {3,6,7,9,10}, {3,6,9,10,17} → 2+7+17=26 → (E)
Answer
26
The answer list itself is evidence. Every wrong choice is the sum of a proper subset of {2,7,17}: (A) 7 is the middle case alone, (B) 9=2+7 drops the large solution, (C) 19=2+17 drops the middle solution, (D) 24=7+17 drops the small solution, and only (E) 26=2+7+17 uses all three. That pattern is a strong signal that exactly three cases exist and that the trap being tested is exactly the one this solution guards against — losing a case, not miscalculating one. Two more checks pass. Direction: each solution pulls the mean toward the median it was supposed to match — n=2 drags the mean of a set skewed high back down to 6, and n=17 drags it up to 9, which is why the small solution sits far below the set and the large one far above. Symmetry: the mean of the original four is 28/4=7, which is exactly the middle solution — that is no coincidence, since the middle case demanded precisely that the four old numbers average to n.
💡Key takeaway

When the median depends on where the new number lands, split the number line so no stretch is missed — then send every answer back to the case that produced it, because a case that names a number outside its own window has named nothing.

  • Write the mean once, for good
  • The median is the middle of 6, n, 9
  • Case n < 6: the median is 6
  • Case 6 < n < 9: the median is n
  • Case n > 9: the median is 9
  • Send each answer back to its own case
  • Rebuild the sets, then add