AMC 10 · 2002 · #1
Grade 4 arithmeticPick an answer.
Adding nine large numbers and long-dividing by 9 is slow and error-prone. Instead, notice the shared structure with Tool #5 (Look for a Pattern): every number is a run of 9s, so each equals 9 times a run of 1s. Tool #7 (Identify Subproblems) then splits the work — factor the 9 out so it cancels the divide-by-9, then add the leftover runs of 1s column by column. Finally Tool #3 (Eliminate Possibilities) reads off which listed digit is missing, since four of the five choices actually appear in M.
Read the pattern of the set
Every member is a run of nines, the k-th holding k nines, so there are exactly nine numbers.
Spotting that the numbers are just growing runs of the same digit turns a messy list into one clean rule.
4.OA.C.5Look For A PatternRewrite each run of 9s
A run of nines is nine times the matching run of ones, so 999 = 9 × 111 and so on.
A block of nines is exactly nine copies of the same block of ones.
4.NBT.B.5Identify SubproblemsFactor out the 9 and cancel
Factoring the 9 out of the sum cancels the divide-by-9, leaving just the runs of ones.
Pulling out the shared factor 9 lets it wipe out the divide-by-9, so the hard arithmetic disappears.
Pulling the shared factor 9 out front lets it cancel against the division by 9.
▸ Why?
A factor shared by every term can be lifted out of the whole sum at once.
▸ Why?
Dividing by 9 undoes multiplying by 9, so the two operations wipe each other out.
Add the runs of 1s by columns
Adding those by columns with no carrying gives 123456789, which never contains 0, choice (A).
Each column just counts how many of the numbers are long enough to reach that place, so the totals march 9,8,7,…,1.
4.NBT.B.4Look For A PatternA row of nines is just nine times a row of ones, so factoring out the 9 cancels the divide-by-9 and the answer 123456789 falls out with no messy arithmetic.
- Read the pattern of the set
- Rewrite each run of 9s
- Factor out the 9 and cancel
- Add the runs of 1s by columns