Competition · AMC preparation · step 4 of 4
AMC 8 · 2016 · #5
Grade 4 number-theoryPick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The "divide by 10, remainder 3" clue is very tight: it forces the units digit to be 3, so only nine candidates exist (13, 23, 33, …, 93). Tool #2 (Systematic List) writes them out in order. Tool #3 (Eliminate) then applies the second condition — divide by 9, remainder 1 — to throw out the candidates that fail. To check divisibility-by-9 quickly, we use Tool #5 (Look for a Pattern): a number's remainder mod 9 equals the digit-sum's remainder mod 9, so we just need the digit sum to leave remainder 1. One candidate survives. Then a single division by 11 finishes the problem — no algebra needed.
List the candidates ending in 3
Remainder 3 when divided by 10 forces the units digit to be 3, so N is one of the nine two-digit numbers ending in 3.
Reading the units digit straight off a number is a Grade 4 place-value move — the units digit IS the remainder when you divide by 10.
4.NBT.A.2Make A Systematic ListRecall the digit-sum rule
Use the digit-sum rule: any number's remainder mod 9 equals its digit sum taken mod 9, so just add the digits and look for remainder 1.
Spotting the digit-sum shortcut is a Grade 4 multiples-and-divisibility pattern that saves nine long divisions.
When a whole number is divided by 9, the remainder it leaves is the same as the remainder its digit sum leaves.
▸ Why?
A two-digit number is built from its digits by place value, and each ten is just one nine plus one extra, so the whole number equals its digit sum plus a whole pile of nines.
▸ Why?
Writing the number by place value splits it into tens and ones: the tens digit stands for that many tens, and the units digit for that many ones.
▸ Why?
Ten is nine and one, so every ten breaks into a nine plus a spare one; a batch of tens therefore becomes that same count of nines plus that same count of spare ones.
▸ Why?
Write the digit sum as some whole nines plus a leftover smaller than nine; the extra pile of nines only raises the count of nines while that same small leftover stays behind, and a leftover smaller than nine is exactly the remainder when dividing by nine.
Test each candidate
Add the digits of each candidate: only 73 gives digit sum 10, which leaves remainder 1 when divided by 9, so N = 73.
Sweeping through a short candidate list and crossing off the failures is the textbook Grade 4 "reason about results" elimination move.
4.OA.A.3Eliminate PossibilitiesDivide 73 by 11
Divide N = 73 by 11: since 11 × 6 = 66 and 11 × 7 exceeds 73, the remainder is 73 - 66 = 7.
Dividing a two-digit number by a one- or two-digit number with a remainder is exactly the Grade 4 long-division standard.
4.NBT.B.6Make A Systematic ListThis AMC 8 problem only needs Grade 4 ideas — units-digit place value, the digit-sum rule for 9, and basic division with remainders — that you already know!
- List the candidates ending in 3
- Recall the digit-sum rule
- Test each candidate
- Divide 73 by 11
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