AMC 10 · 2003 · #18
Grade 6 number-theoryPick an answer.
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
The division reads n = 100q + r with the remainder under one hundred.
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
Splitting the hundred makes the wanted sum appear: 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 is a multiple of eleven, the condition becomes simply 11 divides n.
Two numbers that differ by a multiple of 11 always give the same answer to 'is it divisible by 11?'
Two numbers differing by a multiple of 11 always give the same answer to the question of divisibility by 11.
▸ Why?
Splitting a number into whole elevens plus a leftover is possible in exactly one way, and adding a whole eleven leaves that leftover alone.
▸ Why?
The split into quotient and remainder came from place value, since dividing by 100 slices the last two digits off.
Turn it into counting multiples of 11
The five-digit multiples of eleven run from 10010 to 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 those gives 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