AMC 10 · 2011 · #13
Easy mode Grade 3You are making three-digit numbers with these rules:
- The number is between 200 and 700.
- The number is even.
- The three digits are all different.
- Every digit must come from the list 1,2,5,7,8,9.
How many such numbers can you make?
Pick an answer.
AMC 10 2011 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: Count the whole numbers between 200 and 700 that are even, whose three digits are all different, and where every digit is taken from the set $\{1,2,5,7,8,9\}$.
Givens: The number is between 200 and 700; The number is even; All three digits are different; Each digit must come from $\{1,2,5,7,8,9\}$
Unknowns: How many such three-digit numbers exist
Understand
Restated: Count the whole numbers between 200 and 700 that are even, whose three digits are all different, and where every digit is taken from the set $\{1,2,5,7,8,9\}$.
Givens: The number is between 200 and 700; The number is even; All three digits are different; Each digit must come from $\{1,2,5,7,8,9\}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The question asks "how many," so I build the numbers digit by digit and count the valid choices at each place. The units digit is the tightest constraint (it decides evenness), so I split into cases by the hundreds digit, count each case as choices multiplied together, and add the cases.
Execute — Answer: A
2.NBT.A.1 Step 1 Pin down the hundreds digit
- A number between 200 and 700 has hundreds digit $2, 3, 4, 5,$ or $6$.
- But every digit must come from $\{1,2,5,7,8,9\}$, and only $2$ and $5$ appear in both lists.
- So the hundreds digit is either $2$ or $5$.
💡 The hundreds digit alone decides which 100-block the number sits in, so start by keeping only the ones that fit the range.
2.OA.C.3 Step 2 Pin down the units digit
- A number is even exactly when its last digit is even.
- The only even digits in $\{1,2,5,7,8,9\}$ are $2$ and $8$.
- So the units digit must be $2$ or $8$.
💡 Even or odd is decided by the ones digit, so filtering the last spot handles the whole evenness rule.
3.OA.A.1 Step 3 Case 1: hundreds digit is 2
- The units digit must be even and different from the hundreds digit $2$, so the only choice left is $8$.
- The tens digit can be any allowed digit except $2$ and $8$, leaving $\{1,5,7,9\}$ — that is $4$ choices.
- This case gives $1 \times 4 = 4$ numbers.
💡 Once the hundreds and units are locked, the count is just how many digits are still free for the tens place.
3.OA.A.1 Step 4 Case 2: hundreds digit is 5
- The units digit is even, so it is $2$ or $8$; neither equals $5$, so both work — that is $2$ choices.
- After the hundreds and units are used, $6 - 2 = 4$ allowed digits remain for the tens place.
- This case gives $2 \times 4 = 8$ numbers.
💡 Count each place's free choices and multiply, because every units choice pairs with every tens choice.
2.OA.A.1 Step 5 Add the cases
- The two cases cannot overlap because they use different hundreds digits, so add them: $4 + 8 = 12$.
- There are $12$ such numbers, which is choice (A).
💡 Separate, non-overlapping cases can simply be added to get the whole count.
2.NBT.A.1 A number between 200 and 700 has hundreds digit $2, 3, 4, 5,$ or $6$. But every 2.OA.C.3 A number is even exactly when its last digit is even. The only even digits in $\ 3.OA.A.1 The units digit must be even and different from the hundreds digit $2$, so the o 3.OA.A.1 The units digit is even, so it is $2$ or $8$; neither equals $5$, so both work — 2.OA.A.1 The two cases cannot overlap because they use different hundreds digits, so add Review
Reasonableness: The total $12$ is small, which fits: the digit set is tight, the number must be even, and no digit may repeat, so most combinations are ruled out. A quick sanity check on Case 1 — $2\_8$ with tens from $\{1,5,7,9\}$ gives $218, 258, 278, 298$, exactly $4$ numbers — confirms the counting method, and it matches choice (A).
Alternative: List by units digit instead. Numbers ending in $2$ need hundreds digit $5$ (since $2$ is taken) with $4$ tens choices: $4$ numbers. Numbers ending in $8$ can have hundreds digit $2$ or $5$, each with $4$ tens choices: $8$ numbers. Total $4 + 8 = 12$, the same answer.
CCSS standards used (min grade 3)
2.NBT.A.1Understand that the three digits of a three-digit number represent hundreds, tens, and ones (Reading the hundreds digit to decide which numbers fall between 200 and 700.)2.OA.C.3Determine whether a group of objects has an odd or even number (Using the last digit to keep only the even numbers.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Multiplying the number of choices for each digit place to count the numbers in each case.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Adding the two non-overlapping case totals into the final count.)
⭐ Lock the digits that face hard rules first (range and even), then count the free choices left and multiply, adding up separate cases at the end.
⭐ Lock the digits that face hard rules first (range and even), then count the free choices left and multiply, adding up separate cases at the end.
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