Competition · AMC preparation · step 4 of 4
AMC 8 · 2018 · #7
Grade 4 number-theoryPick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question hides two clean sub-questions (Tool #7): (a) use the divisibility-by-9 clue to figure out the digit U, and (b) once the number is known, compute its remainder when divided by 8. Sub-question (a) is exactly the digit-sum pattern for 9 (Tool #5: a number is divisible by 9 when its digits add to a multiple of 9); since U ranges over only 10 values, Tool #6 (Guess and Check) also works in seconds. Sub-question (b) is a single division with remainder, no algebra needed.
Add the known digits
Sub-problem A: by the divisibility rule for 9, sum the known digits — 2 + 0 + 1 + 8 + U = 11 + U.
Adding the four known digits is a Grade 4 multi-digit addition fact (2+0+1+8 = 11).
4.NBT.B.4Look For A PatternFind the missing digit
Only U = 7 makes 11 + U = 18 a multiple of 9 (11 ≤ 11 + U ≤ 20), so the number is 20187.
Checking which sum is a multiple of 9 is a Grade 4 multiples-of-a-number task.
4.OA.B.4Guess And CheckSplit off a multiple of 8
Sub-problem B: since 1000 (hence 20000) is a multiple of 8, 20187 mod 8 equals just 187 mod 8.
Spotting that multiples of 1000 are also multiples of 8 is a Grade 4 multiples observation.
To find the remainder when 20187 is divided by 8, we only need the remainder of its last three digits, 187, divided by 8.
▸ Why?
20187 breaks into 20000 plus 187, and 20000 is an exact number of 8s, so all of the leftover after dividing by 8 comes from the 187 part.
▸ Why?
20000 is a whole number of 8s because 20000 = 20 × 1000 and 1000 = 8 × 125, so 20000 = 8 × 2500.
▸ Why?
Regrouping 20 × 1000 as 20 × (8 × 125) and then as (20 × 125) × 8 keeps the same product, so 20000 equals 8 × 2500.
▸ Why?
Since 20000 = 8 × 2500, it is 2500 equal groups of 8 with nothing left over, which is exactly what being a multiple of 8 means.
▸ Why?
Because 20000 is complete groups of 8, the only part that can leave a leftover after dividing by 8 is the remaining 187.
▸ Why?
20187 is just its two parts, 20000 and 187, added back together with nothing missing or counted twice.
▸ Why?
Splitting 20187 = 8 × 2500 + 187 shows 20000 adds only whole groups of 8 (leftover 0), so writing the division as 20187 = q × 8 + r with 0 ≤ r ＜ 8 forces that small leftover r to come entirely from the 187.
Divide 187 by 8
The biggest multiple of 8 up to 187 is 8 × 23 = 184, so 187 = 8 × 23 + 3, leaving remainder 3.
Finding the quotient 23 and remainder 3 from a three-digit dividend is exactly the Grade 4 division-with-remainder standard.
4.NBT.B.6Identify SubproblemsRead off the remainder
Combining both sub-problems, 20187 mod 8 = 187 mod 8 = 3, which matches choice (B).
Stitching the two sub-answers together is the Tool #7 "combine subproblems" move.
4.NBT.B.6Identify SubproblemsThis AMC 8 problem only needs Grade 4 divisibility rules and division-with-remainder you already know!
- Add the known digits
- Find the missing digit
- Split off a multiple of 8
- Divide 187 by 8
- Read off the remainder
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