AMC 10 · 2003 · #25
Grade 6 number-theoryPick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The clean way in is to name the pieces: Tool #4 (Introduce a Variable) writes the split as n = 100q + r, turning the word 'quotient and remainder' into one equation. The magic is what that equation reveals: rewrite 100q + r as 99q + (q + r). Since 99 is a multiple of 11, the messy target q + r and the number n itself always leave the same remainder when divided by 11 (Tool #5, Look for a Pattern — the pattern is that 100 is just 1 more than a multiple of 11). So asking '11 divides q+r' is exactly the same as asking '11 divides n.' That swap is Tool #9 (Solve an Easier Related Problem): instead of chasing two variables, we just count the 5-digit multiples of 11, which is a short, direct count.
Write the division as one equation
Dividing n by 100 means n = 100q + r with 0 ≤ r ≤ 99: q is the first three digits (100 to 999), r the last two (0 to 99).
Dividing by 100 just slices a number's last two digits off as the remainder and keeps the front as the quotient.
4.NBT.B.6Introduce A VariableRewrite so q+r appears
We want q + r, but the equation shows 100q + r. Split 100q into 99q + q, and the sum we want appears: n = 99q + (q + r).
Peeling 99q off of 100q leaves exactly one extra q to pair with r, surfacing the sum we care about.
6.EE.A.3Introduce A VariableSee that q+r and n share divisibility by 11
Since 99 = 9 × 11, the term 99q is always a multiple of 11, so q + r is divisible by 11 exactly when n itself is.
Two numbers that differ by a multiple of 11 always give the same answer to 'is it divisible by 11?'
Two numbers that differ by a multiple of eleven always give the same answer to being divisible by eleven.
▸ Why?
Only the remainder after dividing survives, so a shift by a whole multiple leaves it untouched.
▸ Why?
Shifting both quantities by the same amount leaves their difference exactly as it was.
Turn it into counting multiples of 11
So count the 5-digit multiples of 11: the smallest is 11 × 910 = 10010 and the largest is 11 × 9090 = 99990.
The valid values of n are an evenly spaced list of multiples of 11, pinned between the first and last one that still has five digits.
4.OA.B.4Solve An Easier Related ProblemCount the list
Counting 11 × 910 through 11 × 9090 is just counting 910 to 9090: 9090 - 910 + 1 = 8181, choice (B).
Counting evenly spaced numbers means counting their multiplier labels: last minus first, plus one for the endpoint.
4.OA.C.5Solve An Easier Related ProblemSince 100 is one more than a multiple of 11, the number n and the sum q+r always leave the same remainder when divided by 11, so the question is just 'how many 5-digit multiples of 11 are there?' — and that count is 8181.
- Write the division as one equation
- Rewrite so q+r appears
- See that q+r and n share divisibility by 11
- Turn it into counting multiples of 11
- Count the list