AMC 10 · 2005 · #20
Grade 6 arithmeticWhat is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
Pick an answer.
AMC 10 2005 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: We form 5-digit numbers by arranging the digits $1, 3, 5, 7, 8$, using each digit exactly once in each number. There are many such numbers. We want the average (mean) of all of them — the sum of every number divided by how many numbers there are.
Givens: The available digits are $1, 3, 5, 7, 8$, and each number uses all five exactly once; Every digit is nonzero, so any arrangement is a genuine 5-digit number (no leading-zero worry); Answer choices: (A) $48000$, (B) $49999.5$, (C) $53332.8$, (D) $55555$, (E) $56432.8$
Unknowns: The mean of the whole collection of these 5-digit numbers
Understand
Restated: We form 5-digit numbers by arranging the digits $1, 3, 5, 7, 8$, using each digit exactly once in each number. There are many such numbers. We want the average (mean) of all of them — the sum of every number divided by how many numbers there are.
Givens: The available digits are $1, 3, 5, 7, 8$, and each number uses all five exactly once; Every digit is nonzero, so any arrangement is a genuine 5-digit number (no leading-zero worry); Answer choices: (A) $48000$, (B) $49999.5$, (C) $53332.8$, (D) $55555$, (E) $56432.8$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #16 Change Focus / Count the Complement, #15 Organize Information in More Ways, #7 Identify Subproblems
Listing all the numbers and adding them is hopeless by hand, so the winning idea is Tool #5 (Look for a Pattern): every digit is treated the same way, so across all the arrangements each digit lands in each place-value column equally often. Tool #16 (Change Focus) reframes the average — instead of averaging whole numbers, we average one place-value column at a time. Tool #15 (Organize Information) writes each number in place-value form so those columns are visible. Tool #7 (Identify Subproblems) then splits the work into 'find the average digit' and 'multiply by the place values', which are both easy.
Execute — Answer: C
5.NBT.A.1 Step 1 Write a number by place value
- Any of these numbers is built from its five digits weighted by place value: the ten-thousands digit is worth $10000$, the thousands digit $1000$, and so on down to the ones digit worth $1$.
- So a number is completely determined by which digit sits in each of the five columns.
💡 A number is just its digits, each multiplied by what its column is worth.
6.EE.A.3 Step 2 Average each place on its own
- To average all the numbers, we can average each place-value column separately and then add the columns back up, because the average of a sum equals the sum of the averages.
- So the mean number equals the average ten-thousands digit times $10000$, plus the average thousands digit times $1000$, and so on.
- Write $\overline{d_k}$ for the average digit in place $k$.
💡 Averaging each column first gives the same total as averaging whole numbers, so we can handle one column at a time.
6.SP.B.5 Step 3 Every place has the same average digit
- Look at one column, say the ones place.
- As we run through all the arrangements, the five digits take turns sitting there, and by symmetry each digit lands in that place exactly the same number of times.
- So the average digit in the ones place is just the ordinary average of $1, 3, 5, 7, 8$.
- Nothing about the ones place is special, so every column has this same average digit.
💡 No digit prefers any column, so on average every column holds the same typical digit — the mean of the five.
6.EE.A.3 Step 4 Combine the columns
- Since each column contributes $4.8$ times its own place value, factor the $4.8$ out.
- What is left inside is the sum of the five place values, which is $10000+1000+100+10+1 = 11111$.
💡 Every place carries the same average digit, so the place values simply add up to $11111$.
5.NBT.B.7 Step 5 Multiply to finish
- Compute $4.8 \times 11111$.
- As whole numbers, $48 \times 11111 = 533328$; dividing by $10$ puts the one decimal place back, giving $4.8 \times 11111 = 53332.8$.
- That matches choice (C).
💡 Multiply as whole numbers first, then restore the single decimal place.
5.NBT.A.1 Any of these numbers is built from its five digits weighted by place value: the 6.EE.A.3 To average all the numbers, we can average each place-value column separately an 6.SP.B.5 Look at one column, say the ones place. As we run through all the arrangements, 6.EE.A.3 Since each column contributes $4.8$ times its own place value, factor the $4.8$ 5.NBT.B.7 Compute $4.8 \times 11111$. As whole numbers, $48 \times 11111 = 533328$; dividi Review
Reasonableness: The average digit $4.8$ is just under $5$, and multiplying by $11111$ should land the answer just under the all-fives number $5 \times 11111 = 55555$. Our result $53332.8$ sits exactly there — a little below $55555$ — which is the right size for a typical digit slightly below $5$. This also exposes the traps: (D) $55555$ is what you get if you wrongly use an average digit of $5$; (A) $48000$ and (E) $56432.8$ are far from the estimate. So (C) is the only reasonable value.
Alternative: Count and sum directly. There are $5! = 120$ numbers. Fixing one digit in one place leaves $4! = 24$ ways to arrange the rest, so each digit appears in each place exactly $24$ times. The digits total $1+3+5+7+8 = 24$, so every place's digits sum to $24 \times 24 = 576$. The grand total of all $120$ numbers is $576 \times 11111 = 6{,}399{,}936$, and the mean is $6{,}399{,}936 \div 120 = 53{,}332.8$, confirming (C).
CCSS standards used (min grade 6)
5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Writing each 5-digit number as its digits weighted by place value ($10000, 1000, 100, 10, 1$).)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Averaging each place-value column separately, then factoring the common average digit $4.8$ out of the place values.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Finding the mean of the five digits, $(1+3+5+7+8)/5 = 4.8$, as the average digit in every place.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Evaluating $4.8 \times 11111 = 53332.8$ to get the final average.)
⭐ When every arrangement is equally likely, each place-value column just shows the average digit, so the average number is that one average digit times $11111$.
⭐ When every arrangement is equally likely, each place-value column just shows the average digit, so the average number is that one average digit times $11111$.
More like this
Same archetype — closest grade level first.
