Competition · AMC preparation · step 4 of 4
AMC 8 · 2020 · #19
Grade 4 number-theorycountingPick an answer.
AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The condition "divisible by 15" splits cleanly into two smaller, very different sub-conditions: divisible by 5 (a last-digit rule) and divisible by 3 (a digit-sum rule). Tool #7 (Identify Subproblems) lets us pin down one digit at a time — the 5-rule forces the value of a, then the 3-rule narrows down b. Once both digits are constrained, Tool #2 (Make a Systematic List) finishes the job: write the few remaining candidate numbers in order and count them. We deliberately avoid Tool #13 (Algebra) — divisibility rules and listing handle this with no equations needed.
Write the flippy digit pattern
A five-digit flippy number must read ababa — positions 1, 3, 5 are a and positions 2, 4 are b — with a ≠ 0 and a ≠ b.
Reading a multi-digit number by its place-value positions is exactly the Grade 4 "read and write multi-digit whole numbers" skill.
4.NBT.A.2Identify SubproblemsUse the divisibility rule for 5
Divisibility by 5 needs the last digit 0 or 5; that last digit is a and a ≠ 0, so a = 5 and the number is 5b5b5 (b ≠ 5).
Recognizing multiples of 5 by the last digit is part of the Grade 4 "factors and multiples" cluster.
4.OA.B.4Identify SubproblemsUse the divisibility rule for 3
The digits of 5b5b5 sum to 15 + 2b; since 15 is a multiple of 3, we only need 2b to be a multiple of 3 too.
The "sum-of-digits divisible by 3" rule is a standard Grade 4 divisibility test that lives in the factors-and-multiples standard.
The number 5b5b5 is a multiple of 3 exactly when the sum of its digits, 15 + 2b, is a multiple of 3.
▸ Why?
Written by place value, 5b5b5 is 5 · 10000 + b · 1000 + 5 · 100 + b · 10 + 5, and this whole number leaves the same remainder when divided by 3 as the plain sum of its digits, 15 + 2b.
▸ Why?
The number really is each digit times its place value added up, because every higher place bundles ten of the place just below it.
▸ Why?
Every place value is one more than a multiple of 3: 10 = 9 + 1, 100 = 99 + 1, 10000 = 9999 + 1, since 9, 99, and 9999 are each a whole number of equal groups of 3 (for example 99 is 33 groups of 3).
▸ Why?
So each digit times its place splits into the digit times a multiple of 3 plus the digit times one, which turns the number into one big multiple of 3 plus the plain digit sum 15 + 2b.
▸ Why?
That big multiple of 3 drops out: the number is the multiple-of-3 part joined to the digit-sum part with no gap or overlap, so the whole is a multiple of 3 exactly when the digit-sum part is.
Find the allowed digits b
2b is a multiple of 3 exactly when b is, so b ∈ {0, 3, 6, 9}, and every one of these already satisfies b ≠ 5.
Listing the one-digit multiples of 3 is direct Grade 4 multiples reasoning.
4.OA.B.4Identify SubproblemsList every valid number
List them in order — 50505, 53535, 56565, 59595 — giving 4 flippy multiples of 15, which is choice (B).
Generating every number from a clear rule ("5b5b5 for each allowed b") and counting matches the Grade 4 "generate a pattern following a rule" standard.
4.OA.C.5Make A Systematic ListThis AMC 8 problem only needs Grade 4 divisibility rules — "ends in 0 or 5" for 5, and "digits add up to a multiple of 3" for 3 — that you already know!
- Write the flippy digit pattern
- Use the divisibility rule for 5
- Use the divisibility rule for 3
- Find the allowed digits b
- List every valid number
A parent dashboard for the family lives at sensimlab.com.