AMC 10 · 2024 · #15

• Translate the conditions into algebra.
• Sum of the six fixed numbers: $1 + 2.2 + 3.2 + 5.2 + 6.2 + 7 = 24.8$.
• For the mean to be an integer $M$, we need $24.8 + x + y + z = 9M$.
• So $x + y + z \in \{2.2, 11.2, 20.2, 29.2, \dots\}$.
• Also from the range argument above, $x \geq 0$.

💡 The mean condition fixes the decimal part of $x+y+z$ to $.2$ — a one-line filter on the sum.

• Split on $(\min, \max)$ of the full list.
• Four cases, but $\min = 1$ AND $\max = 7$ gives range $6 \neq 7$, so three live cases: (I) $\min = 1$, $\max = z$ ($z > 7$, $x \geq 1$, so $z = 8$); (II) $\min = x$, $\max = 7$ ($x < 1$, $z \leq 7$, so $x = 0$); (III) $\min = x$, $\max = z$ ($0 \leq x < 1$, $z > 7$, $z = x + 7$).

💡 Range $= 7$ has only three structural ways to happen — list them and treat each cleanly.

• Case I: $z = 8$, $1 \leq x \leq y \leq 8$.
• The sum $x + y + z = x + y + 8 \geq 10$, so $x + y + 8 \in \{11.2, 20.2\}$ (the only feasible mean values in range; $29.2$ would force $x+y > 21$, impossible with $y \leq 8$).
• Sub-case $x + y = 3.2$ forces $1 \leq x \leq 1.6$, but then the sorted list begins $1, x, y, 2.2, 3.2, \dots$ so the median is $3.2$ — not integer.
• Discard.
• Sub-case $x + y = 12.2$ with $1 \leq x \leq y \leq 8$ gives $4.2 \leq x \leq 6.1$.

💡 Walk both feasible sums; one is killed by the median, the other narrows $x$ to a short interval.

• Continue Case I.
• The sorted list begins $1, 2.2, 3.2, \ldots$.
• The 4th term is $\min(x, 5.2)$, the 5th (median) follows.
• If $4.2 \leq x \leq 5.2$ the median is $5.2$ (not integer).
• If $5.2 < x \leq 6.1$ the 5th term is $x$, so $x$ itself must be a positive integer in $(5.2, 6.1]$: only $x = 6$.
• Then $y = 12.2 - 6 = 6.2$.
• Triple: $(6, 6.2, 8)$.
• Verify: sorted $\{1, 2.2, 3.2, 5.2, 6, 6.2, 6.2, 7, 8\}$, median $6$, mean $45/9 = 5$, range $7$.
• All four conditions hold.

💡 Push the median condition through the sorted list; in this sub-case only $x = 6$ survives.

• Case II: $x = 0$, $0 \leq y \leq z \leq 7$.
• The sum $y + z \in \{2.2, 11.2\}$ (higher values impossible since $y, z \leq 7$).
• The sub-case $y + z = 2.2$ leaves the sorted list beginning $0, y, 1, \dots$ with median $2.2$ — kill.
• The sub-case $y + z = 11.2$ gives $4.2 \leq y \leq 5.6$.
• The sorted list begins $0, 1, 2.2, 3.2, \dots$, and the 5th term is $\min(y, 5.2)$.
• For it to be an integer with $y \geq 4.2$, only $y = 5$ works.
• Then $z = 6.2$.
• Triple: $(0, 5, 6.2)$.
• Verify: sorted $\{0, 1, 2.2, 3.2, 5, 5.2, 6.2, 6.2, 7\}$, median $5$, mean $36/9 = 4$, range $7$.
• Valid.

💡 Same machinery as Case I — the median equation forces $y$ to be an integer in a narrow window.

• Case III: $z = x + 7$, $0 \leq x < 1$.
• The sorted list begins $x, 1, 2.2, 3.2, \dots$, so the 5th term (median) is $\min(y, 5.2)$ provided $y > 3.2$ (otherwise the median is $3.2$, not integer).
• So $y$ must be an integer in $(3.2, 5.2]$ — meaning $y \in \{4, 5\}$.
• Mean condition: $24.8 + 2x + y + 7 = 9M$, i.e.
• $31.8 + 2x + y = 9M$.
• If $y = 4$: $35.8 + 2x = 9M$, only $M = 4$ works ($x = 0.1$).
• If $y = 5$: $36.8 + 2x = 9M$, $M = 4$ needs $x = -0.4 < 0$, $M = 5$ needs $x = 4.1 \not< 1$ — both fail.

💡 Two integer choices of $y$, each plugged into the mean equation — one yields a legal $x$, the other doesn't.

• Verify the Case III candidate.
• Triple: $(0.1, 4, 7.1)$.
• Sorted: $\{0.1, 1, 2.2, 3.2, 4, 5.2, 6.2, 7, 7.1\}$, median $4$, mean $(24.8 + 11.2)/9 = 36/9 = 4$, range $7.1 - 0.1 = 7$.
• All four conditions hold.
• Total over all cases: $3$ valid triples.

💡 Three cases, three survivors — the answer count drops cleanly out of the casework.