AMC 10 · 2007 · #8
Grade 7 arithmeticOn the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form bbcac, where 0≤a<b<c≤9, and b was the average of a and c. How many different five-digit numbers satisfy all these properties?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A five-digit number has digits of the form $bbcac$, where the digits satisfy $0 \le a < b < c \le 9$ and $b$ is the average of $a$ and $c$. Count how many different five-digit numbers fit all of these conditions.
Givens: The number's digits, in order, are $b,\,b,\,c,\,a,\,c$ — the pattern $bbcac$; The three digit-values satisfy the strict chain $0 \le a < b < c \le 9$; $b$ is the average of $a$ and $c$, that is $b = \dfrac{a+c}{2}$; Answer choices: (A) $12$, (B) $16$, (C) $18$, (D) $20$, (E) $25$
Unknowns: The count of distinct five-digit numbers $bbcac$ that meet every condition
Understand
Restated: A five-digit number has digits of the form $bbcac$, where the digits satisfy $0 \le a < b < c \le 9$ and $b$ is the average of $a$ and $c$. Count how many different five-digit numbers fit all of these conditions.
Givens: The number's digits, in order, are $b,\,b,\,c,\,a,\,c$ — the pattern $bbcac$; The three digit-values satisfy the strict chain $0 \le a < b < c \le 9$; $b$ is the average of $a$ and $c$, that is $b = \dfrac{a+c}{2}$; Answer choices: (A) $12$, (B) $16$, (C) $18$, (D) $20$, (E) $25$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #16 Change Focus / Count the Complement, #5 Look for a Pattern
The question is a pure 'how many' count, so Tool #2 (Make a Systematic List) is the engine: once the real freedom in the problem is found, list and count the valid cases. Tool #4 (Introduce a Variable) is already handed to us as $a,b,c$; the key is reading $b = \tfrac{a+c}{2}$ as a rule that ties the three together. Tool #16 (Change Focus) is the crux move: instead of hunting for triples $(a,b,c)$, notice that $b$ is completely decided by $a$ and $c$, so the whole problem collapses to counting the allowed pairs $(a,c)$ — and the average condition quietly forces $a$ and $c$ to share the same parity.
Execute — Answer: D
6.EE.B.6 Step 1 Read the digit pattern and its rules
- The receipt reads $bbcac$, so only three distinct values appear: $a$, $b$, and $c$.
- They must climb strictly, $a < b < c$, and $b$ equals the average of $a$ and $c$.
- So the whole puzzle lives in choosing the three digit-values, not in arranging them — the arrangement $bbcac$ is fixed.
💡 Name what actually varies — here it is just the three digit-values, and the pattern around them never changes.
6.SP.A.3 Step 2 See b as the midpoint of a and c
- The average of $a$ and $c$ is $\dfrac{a+c}{2}$, the value sitting exactly halfway between them.
- So $b$ is pinned to the midpoint of $a$ and $c$.
- That means $b$ is not a free choice at all: pick $a$ and $c$, and $b$ is forced.
💡 The average of two numbers is their midpoint, so knowing the two ends fixes the middle.
2.OA.C.3 Step 3 The midpoint forces a and c to match in parity
- Since $a + c = 2b$, the sum $a+c$ must be even.
- A sum of two whole numbers is even only when both are even or both are odd, so $a$ and $c$ must have the same parity.
- Bonus: whenever $a < c$ share a parity, the midpoint $b = \tfrac{a+c}{2}$ is a whole digit strictly between them, so $a < b < c$ holds by itself and $b \ge 1$ — a genuine five-digit number every time.
💡 Two numbers add to an even total exactly when they are both even or both odd.
7.SP.C.8 Step 4 Reframe: just count valid pairs (a, c)
- Because $b$ is decided by $a$ and $c$, and $a<b<c$ takes care of itself, the entire count equals the number of pairs $(a,c)$ with $a < c$ that share a parity.
- So I only need to count same-parity pairs among the digits $0$–$9$: first the even ones, then the odd ones.
💡 When one quantity is forced by the others, count only the free choices.
7.SP.C.8 Step 5 Count the both-even pairs
- The even digits are $0, 2, 4, 6, 8$ — five of them.
- I need pairs with $a < c$.
- Fixing the smaller digit: $0$ pairs with $\{2,4,6,8\}$ (4 ways), $2$ with $\{4,6,8\}$ (3), $4$ with $\{6,8\}$ (2), $6$ with $\{8\}$ (1).
- That is $4+3+2+1 = 10$ pairs.
💡 List pairs by their smaller member so nothing is counted twice.
7.SP.C.8 Step 6 Count the both-odd pairs and add
- The odd digits are $1, 3, 5, 7, 9$ — also five, so by the exact same tally there are $4+3+2+1 = 10$ odd pairs.
- Adding the two cases gives $10 + 10 = 20$ valid pairs, hence $20$ valid numbers.
- That is choice (D).
- (A tempting $25$ appears if you forget the parity rule and count opposite-parity pairs instead.)
💡 The two parity cases are separate and cover everything, so just add their counts.
6.EE.B.6 The receipt reads $bbcac$, so only three distinct values appear: $a$, $b$, and $ 6.SP.A.3 The average of $a$ and $c$ is $\dfrac{a+c}{2}$, the value sitting exactly halfwa 2.OA.C.3 Since $a + c = 2b$, the sum $a+c$ must be even. A sum of two whole numbers is ev 7.SP.C.8 Because $b$ is decided by $a$ and $c$, and $a<b<c$ takes care of itself, the ent 7.SP.C.8 The even digits are $0, 2, 4, 6, 8$ — five of them. I need pairs with $a < c$. F 7.SP.C.8 The odd digits are $1, 3, 5, 7, 9$ — also five, so by the exact same tally there Review
Reasonableness: Cross-check the total a different way: among all digits $0$–$9$ there are $\binom{10}{2}=45$ pairs $a<c$. A pair is same-parity ($a+c$ even) or opposite-parity ($a+c$ odd). Same-parity pairs are exactly the $20$ we counted, leaving $45-20=25$ opposite-parity pairs. Only the same-parity pairs give a whole-number average $b$, so $20$ is right — and the leftover $25$ is precisely the trap choice (E) you'd land on by ignoring the even-sum requirement. The count is comfortably in the range of the answer choices, confirming (D).
Alternative: Loop on $b$ instead of on the pair. For each middle value $b$, the equal spacing means $a = b-d$ and $c = b+d$ for some step $d \ge 1$, and both must stay digits $0$–$9$. Counting the allowed $d$ for each $b$ from $1$ to $8$ gives $1,2,3,4,4,3,2,1$, which again sums to $20$.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Treating the digits as variables $a,b,c$ and reading the pattern $bbcac$ with its rule $b=\tfrac{a+c}{2}$.)6.SP.A.3Recognize a measure of center (mean) for a data set (Interpreting 'the average of $a$ and $c$' as the midpoint $\tfrac{a+c}{2}$, so $a+c=2b$.)2.OA.C.3Determine whether a group has an odd or even number of members (Concluding that $a+c$ being even forces $a$ and $c$ to share the same parity.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and counting (Systematically listing and counting the same-parity digit pairs $(a,c)$ to reach the total of $20$.)
⭐ When one value is forced to be the average of two others, their sum must be even, so both are even or both are odd — then you just count the pairs.
⭐ When one value is forced to be the average of two others, their sum must be even, so both are even or both are odd — then you just count the pairs.
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