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.
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 PatternOnly 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 CheckSub-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.
4.OA.B.4Look For A PatternThe 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 SubproblemsCombining 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!