AMC 10 · 2019 · #12
Grade 6 arithmeticProblem
What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than 2019?
Pick an answer.
AMC 10 2019 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: Among positive integers less than $2019$, what is the greatest possible sum of digits when the integer is written in base $7$?
Givens: The number $n$ is a positive integer with $n < 2019$; We write $n$ in base $7$ (digits are $0,1,2,3,4,5,6$); We sum those base-$7$ digits; Answer choices: $11,\, 14,\, 22,\, 23,\, 27$
Unknowns: The maximum possible base-$7$ digit sum
Understand
Restated: Among positive integers less than $2019$, what is the greatest possible sum of digits when the integer is written in base $7$?
Givens: The number $n$ is a positive integer with $n < 2019$; We write $n$ in base $7$ (digits are $0,1,2,3,4,5,6$); We sum those base-$7$ digits; Answer choices: $11,\, 14,\, 22,\, 23,\, 27$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #6 Guess and Check, #3 Eliminate Possibilities
Tool #9 (Easier Problem): first compute the base-$7$ form of $2018$ (the largest allowed $n$) — that tells us the digit budget. Tool #6 (Guess & Check): the digit-sum hero is the number $666_7$ (all $6$s), so try thousands digits $0, 1, 2, \dots$ in front of $666_7$ until the value crosses $2018$. Tool #3 matches $22$ to the choices. Algebra is unnecessary; this is base-conversion plus a small directional search.
Execute — Answer: C
5.NBT.A.1 Step 1 - Find the base-$7$ form of $2018$ to see the digit ceiling.
- Divide repeatedly by $7$: $2018 = 5\cdot343 + 303$; $303 = 6\cdot49 + 9$; $9 = 1\cdot7 + 2$.
- So $2018 = 5612_7$.
💡 Grade 5 place value: dividing by $7$ peels off one base-$7$ digit at a time.
5.NBT.A.1 Step 2 - Note any number with $5$ or more base-$7$ digits is $\ge 7^4 = 2401 > 2018$.
- So $n$ has at most $4$ base-$7$ digits, and the theoretical max digit sum is $4 \cdot 6 = 24$.
- The number $6666_7 = 2400 > 2018$, so that ceiling is not reachable.
💡 Grade 5: a $4$-digit base-$7$ number caps out at $6666_7 = 2400$ — close to $2018$ but over.
6.EE.B.8 Step 3 - Try fixing the bottom three digits as $666_7$ (digit sum $18$) and check what thousands digit keeps the value under $2019$.
- With thousands digit $t$, the number is $t\cdot 343 + 666_7 = 343t + 342$.
💡 Grade 6 inequalities: directly compare the candidate to the cap $2018$.
4.NBT.B.4 Step 4 - Largest integer $t$ with $t < 4.89$ is $t = 4$.
- That gives the number $4666_7 = 4\cdot 343 + 342 = 1372 + 342 = 1714$, which is under $2019$.
- Its digit sum is $4 + 6 + 6 + 6 = 22$.
💡 Grade 4 addition: stuffing the lower three slots with $6$s and putting $4$ on top gives a legal candidate.
5.NBT.A.1 Step 5 - Confirm no larger digit sum is possible.
- With $t = 5$, the candidate $5666_7 = 5 \cdot 343 + 342 = 2057 > 2018$ — illegal.
- Pushing one of the lower $6$s higher than $6$ is impossible (digit cap).
- So any legal $4$-digit number with thousands digit $\le 4$ has digit sum $\le 4 + 6 + 6 + 6 = 22$, and any number with thousands digit $5$ has $5\cdot 7^3 = 1715$ used in the top slot, leaving room $2018 - 1715 = 303 = 612_7$, whose lower digit sum is $6+1+2 = 9$; total $5 + 9 = 14 < 22$.
💡 Grade 5: once thousands digit reaches $5$, the lower digits are forced small to stay under $2019$.
4.NBT.A.2 Step 6 Match $22$ to the choices.
💡 Grade 4: pick the matching value from the answer list.
5.NBT.A.1 Find the base-$7$ form of $2018$ to see the digit ceiling. Divide repeatedly by 5.NBT.A.1 Note any number with $5$ or more base-$7$ digits is $\ge 7^4 = 2401 > 2018$. So 6.EE.B.8 Try fixing the bottom three digits as $666_7$ (digit sum $18$) and check what th 4.NBT.B.4 Largest integer $t$ with $t < 4.89$ is $t = 4$. That gives the number $4666_7 = 5.NBT.A.1 Confirm no larger digit sum is possible. With $t = 5$, the candidate $5666_7 = 5 4.NBT.A.2 Match $22$ to the choices. Review
Reasonableness: Sanity check $4666_7$: $4 \cdot 343 + 6 \cdot 49 + 6 \cdot 7 + 6 = 1372 + 294 + 42 + 6 = 1714 < 2019$ ✓ and digit sum $22$. The absolute ceiling $4 \cdot 6 = 24$ is unreachable because $6666_7 = 2400 > 2018$; $23$ would need digits like $5,6,6,6$ (sum $23$), but the smallest such is $5666_7 = 2057 > 2018$. So $22$ is genuinely the best — matches choice $(C)$.
Alternative: Tool #16 (Change focus / Complement): instead of maximizing the digit sum, minimize the "deficit" $24 - \text{digit sum}$. The deficit equals the sum of $(6 - d_i)$ across the four base-$7$ digits. The minimum deficit is achieved at $4666_7$ with deficit $2$, giving digit sum $24 - 2 = 22$. Same answer $(C)$.
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the digit sum $22$ to the multiple-choice value.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the digits $4 + 6 + 6 + 6 = 22$.)5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Reading base-$7$ place value (digits times $7^0, 7^1, 7^2, 7^3$) for $2018$ and $4666_7$.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Setting $343t + 342 < 2019$ to find the largest legal thousands digit $t = 4$.)
⭐ This AMC 10 problem only needs Grade 6 place-value reasoning you already know! In base $7$ each digit is at most $6$, and since $2018 = 5612_7$ uses $4$ digits, your best bet is to load three $6$s in the bottom and the largest legal digit on top — that is $4666_7 = 1714$, with digit sum $4+6+6+6 = \mathbf{22}$, answer $(C)$.
⭐ This AMC 10 problem only needs Grade 6 place-value reasoning you already know! In base $7$ each digit is at most $6$, and since $2018 = 5612_7$ uses $4$ digits, your best bet is to load three $6$s in the bottom and the largest legal digit on top — that is $4666_7 = 1714$, with digit sum $4+6+6+6 = \mathbf{22}$, answer $(C)$.
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