AMC 10 · 2021 · #1
Easy mode Grade 4Here are four numbers: 1234, 2341, 3412, and 4123.
Add all four together. What is the total?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Add the four four-digit numbers $1234$, $2341$, $3412$, and $4123$, and match the total to one of the five listed choices.
Givens: The four numbers to add are $1234$, $2341$, $3412$, and $4123$; Every one of them is built from exactly the digits $1$, $2$, $3$, $4$, each used once; Each number is the previous one with its leading digit moved to the end (a cyclic shift); Answer choices: (A) $10{,}000$, (B) $10{,}010$, (C) $10{,}110$, (D) $11{,}000$, (E) $11{,}110$
Unknowns: The value of the sum $1234 + 2341 + 3412 + 4123$
Understand
Restated: Add the four four-digit numbers $1234$, $2341$, $3412$, and $4123$, and match the total to one of the five listed choices.
Givens: The four numbers to add are $1234$, $2341$, $3412$, and $4123$; Every one of them is built from exactly the digits $1$, $2$, $3$, $4$, each used once; Each number is the previous one with its leading digit moved to the end (a cyclic shift); Answer choices: (A) $10{,}000$, (B) $10{,}010$, (C) $10{,}110$, (D) $11{,}000$, (E) $11{,}110$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #15 Organize Information in More Ways, #7 Identify Subproblems
Straight column addition works, but the four addends are cyclic shifts of the same four digits, and Tool #5 (Look for a Pattern) turns that observation into the whole solution. Tool #15 (Organize Information in More Ways) is what makes the pattern visible: stack the numbers and read down each place-value column instead of reading across each number. Tool #7 (Identify Subproblems) then splits the total into four independent column sums, each of which is the same easy sum $1 + 2 + 3 + 4$.
Execute — Answer: E
4.NBT.A.2 Step 1 Stack the numbers and read down
- Write the four numbers in a column, lined up by place value.
- Reading across a row gives you one number; reading down a column gives you the four digits that share a place value.
- The thousands column reads $1, 2, 3, 4$.
- The hundreds column reads $2, 3, 4, 1$.
- The tens column reads $3, 4, 1, 2$.
- The ones column reads $4, 1, 2, 3$.
💡 Reading a column of stacked numbers tells you which digits are competing for the same place value.
4.NBT.A.1 Step 2 Every column holds 1, 2, 3, 4
- Look at what each column contains.
- Each column has the digits $1$, $2$, $3$, and $4$ in some order, and never a repeat.
- That happens because each number is the one above it shifted one place, so as you go down the stack a given digit slides from column to column and lands in each column exactly once.
- So every single column adds to $1 + 2 + 3 + 4 = 10$.
💡 A cyclic shift moves every digit through every place, so no place gets favored.
4.NBT.B.4 Step 3 Rebuild the total from the columns
- Each column contributes $10$ of its own place value: $10$ thousands, $10$ hundreds, $10$ tens, and $10$ ones.
- That is $10 \times 1000 + 10 \times 100 + 10 \times 10 + 10 \times 1$, and pulling out the common factor $10$ gives $10 \times 1111 = 11110$.
- Written with a comma, the sum is $11{,}110$, which is choice (E).
💡 If every place is worth the same count, factor that count out and multiply once instead of adding four times.
4.NBT.A.2 Write the four numbers in a column, lined up by place value. Reading across a ro 4.NBT.A.1 Look at what each column contains. Each column has the digits $1$, $2$, $3$, and 4.NBT.B.4 Each column contributes $10$ of its own place value: $10$ thousands, $10$ hundre Review
Reasonableness: Add the numbers directly as a check: $1234 + 2341 = 3575$ and $3412 + 4123 = 7535$, and $3575 + 7535 = 11110$. That confirms $11{,}110$. A size estimate agrees too: each addend is a bit under $2500$ on average, and four of them land near $10{,}000$, so a total just past $11{,}000$ is the right magnitude. The competing choices are all too small to hold four numbers averaging roughly $2778$.
Alternative: Track only the ones digit and the carrying. The ones column adds to $10$, so the sum ends in $0$ and carries $1$ into the tens. The tens column also adds to $10$, plus the carried $1$, giving $11$: write $1$, carry $1$. The same thing repeats in the hundreds and thousands columns, so every remaining digit is $1$ and a final $1$ is carried out the front. The digits come out $11110$ without ever computing the full total, and only choice (E) ends in $0$ with $1$s in front.
CCSS standards used (min grade 4)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Lining the four addends up by place value and reading each place-value column to see which digits share a place.)4.NBT.A.1Recognize that a digit represents ten times what it represents in place to its right (Knowing that a column of digits is worth its digit sum times that column's place value, so a column summing to $10$ contributes $10$ thousands, $10$ hundreds, and so on.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Combining the four column contributions into $10 \times 1111 = 11{,}110$ and verifying with a direct multi-digit addition.)
⭐ When numbers are made of the same digits shuffled around, add down the columns instead of across the rows: every column carries the same digit sum, and the total falls out at once.
⭐ When numbers are made of the same digits shuffled around, add down the columns instead of across the rows: every column carries the same digit sum, and the total falls out at once.
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