AMC 10 · 2019 · #12

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

base-conversionplace-valuedigit-sumbound-inequality-then-enumerate bound-inequality-then-enumerateidentify-subproblems ↑ Prerequisites: base-conversionplace-value

📏 Medium solution 💡 3 insights

Problem

Among positive integers less than 2019, what is the greatest possible sum of digits when the integer is written in base 7?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

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 666₇ (all 6s), so try thousands digits 0, 1, 2, … in front of 666₇ until the value crosses 2018. Tool #3 matches 22 to the choices. Algebra is unnecessary; this is base-conversion plus a small directional search.

1STEP 1

Convert the cap 2018 to base 7 by repeated division to see the digit ceiling: 2018 = 5612₇.

2018 = 5 · 7³ + 6 · 7² + 1 · 7 + 2 = 5612₇

2STEP 2

A 5th base-7 digit needs ≥ 7⁴ = 2401 ＞ 2018, so n has at most 4 digits; the ceiling 24 is out of reach since 6666₇ = 2400 ＞ 2018.

6666₇ = 6(343+49+7+1) = 6 · 400 = 2400 ＞ 2018

3STEP 3

Fix the bottom three digits as 666₇ (digit sum 18); with thousands digit t the number is 343t + 342.

t · 343 + 342 ＜ 2019 → t ＜

\frac{2019 - 342}{343}

\frac{1677}{343}

≈ 4.89

4STEP 4

The largest integer with t ＜ 4.89 is t = 4, giving 4666₇ = 1714 ＜ 2019 with digit sum 4 + 6 + 6 + 6 = 22.

4666₇ = 1714 ＜ 2019, 4 + 6 + 6 + 6 = 22

5STEP 5

With t = 5 the lower digits are forced small: the best is 5612₇ = 2018 with digit sum only 5 + 6 + 1 + 2 = 14 ＜ 22, so 22 stands.

5666₇ = 2057 ＞ 2018; 5612₇ = 2018, digit sum 5+6+1+2 = 14

6STEP 6

Match 22 to the answer choices: (C).

22 → (C)

Answer

Sanity check 4666₇: 4 · 343 + 6 · 49 + 6 · 7 + 6 = 1372 + 294 + 42 + 6 = 1714 ＜ 2019 ✓ and digit sum 22. The absolute ceiling 4 · 6 = 24 is unreachable because 6666₇ = 2400 ＞ 2018; 23 would need digits like 5,6,6,6 (sum 23), but the smallest such is 5666₇ = 2057 ＞ 2018. So 22 is genuinely the best — matches choice (C).

💡Key takeaway

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 = 5612₇ uses 4 digits, your best bet is to load three 6s in the bottom and the largest legal digit on top — that is 4666₇ = 1714, with digit sum 4+6+6+6 = 22, answer (C).

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