AMC 10 · 2012 · #12
Grade 4 number-theoryPick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The end state is known (Tuesday, February 7, 2012) and we want the start state 200 years earlier, so Work Backwards fits exactly. The engine is a weekly pattern in how a fixed date shifts weekday from year to year; counting how many of those 200 years are leap years is a clean subproblem, and the answer is one of five listed days.
Pin down the birth year
An anniversary counts whole years, so the birth was 200 years earlier on the same date: February 7, 1812.
Going back 200 anniversaries just means subtracting 200 from the year while keeping February 7.
4.OA.A.3Work BackwardsHow a date shifts weekday each year
Since 365 = 52 x 7 + 1 and 366 = 52 x 7 + 2, a common year pushes a fixed date 1 weekday later, a leap year 2.
Whole weeks return you to the same weekday, so only the remainder after dividing by 7 moves the day.
Whole weeks return you to the same weekday, so only the remainder after dividing moves the day.
▸ Why?
After seven days everything returns to where it began, so full weeks leave no trace.
▸ Why?
What is left after removing those full weeks is exactly the remainder.
Count the leap years crossed
From 1812 to 2011, 50 years are divisible by 4; drop 1900 to get 49 leap years and 151 common years.
Only leap days add the extra shift, and the divisible-by-100 rule quietly removes 1900.
4.OA.B.4Identify SubproblemsAdd the shifts and step back
Total shift 151 x 1 + 49 x 2 = 249 and 249 = 7 x 35 + 4, so 4 days back from Tuesday gives Friday.
The birthday is 4 weekdays before 2012's Tuesday, so count 4 days backward to reach it.
4.OA.A.3Work BackwardsTo find a weekday far in the past, count how many days the date drifts each year, keep only the leftover after dividing by 7, and step that many days backward.
- Pin down the birth year
- How a date shifts weekday each year
- Count the leap years crossed
- Add the shifts and step back