AMC 10 · 2016 · #5
Grade 4 arithmeticPick an answer.
The problem looks like a calendar problem, with two and a half years, three different Decembers and a leap year to worry about. It is not. The names of the weekdays repeat in a cycle of length 7, and the problem already hands over the total number of days elapsed, 919. So the pattern (Tool #5) is the whole engine: after any multiple of 7 days you are back on the same weekday name. That turns the messy calendar question into a far easier one (Tool #9): what is left over when 919 is split into whole weeks? The one piece of real computation, dividing 919 by 7 and keeping the remainder, is the subproblem to isolate (Tool #7). Everything else is counting two steps forward from Thursday.
Keep only the day count
Only the day count matters.
Every day on the calendar is one day long, so once you know how many days passed, the calendar has nothing left to tell you.
4.MD.A.2Solve An Easier Related ProblemWeekdays repeat every 7 days
Weekdays repeat every seven days.
A week is a loop of length seven; going around it a whole number of times brings you back to the exact same spot.
A week is a loop of length seven, so going around it a whole number of times lands on the same day.
▸ Why?
The weekdays repeat exactly every seven days, so a full week changes nothing at all.
▸ Why?
Any number of days splits into whole weeks plus one remainder, and only that remainder moves the day.
Divide 919 by 7
Dividing leaves a remainder of 2.
Division with remainder is exactly the tool for 'how many whole weeks fit, and what is left over'.
4.NBT.B.6Identify SubproblemsDiscard the 131 whole weeks
The whole weeks change nothing.
Throwing away complete loops costs nothing, so a two-and-a-half-year problem collapses to a two-day problem.
4.OA.C.5Solve An Easier Related ProblemStep 2 days forward from Thursday
Stepping forward gives Saturday, choice (B).
After the whole weeks are gone, all that is left is walking two names forward around the loop.
4.OA.A.3Look For A PatternWeekdays run in a loop of 7, so divide the number of days by 7 and only the remainder moves you.
- Keep only the day count
- Weekdays repeat every 7 days
- Divide 919 by 7
- Discard the 131 whole weeks
- Step 2 days forward from Thursday