AMC 10 · 2012 · #12
Easy mode Grade 4Here is the rule for a leap year (a year with 366 days): a year is a leap year if it divides evenly by 400 (like 2000), or if it divides evenly by 4 but not by 100 (like 2012). February 7, 2012 was a Tuesday, and it was exactly 200 years after the writer Charles Dickens was born. What day of the week was Dickens born on?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: February 7, 2012 (a Tuesday) is the 200th anniversary of Charles Dickens' birth. Find the day of the week on which he was born, given the leap-year rule: a year is a leap year if it is divisible by 400, or divisible by 4 but not by 100.
Givens: February 7, 2012 was a Tuesday.; That date is exactly the 200th anniversary of the birth.; Leap year rule: divisible by 400, or divisible by 4 but not by 100.; A common year has 365 days; a leap year has 366 days.
Unknowns: The day of the week on which Charles Dickens was born.
Understand
Restated: February 7, 2012 (a Tuesday) is the 200th anniversary of Charles Dickens' birth. Find the day of the week on which he was born, given the leap-year rule: a year is a leap year if it is divisible by 400, or divisible by 4 but not by 100.
Givens: February 7, 2012 was a Tuesday.; That date is exactly the 200th anniversary of the birth.; Leap year rule: divisible by 400, or divisible by 4 but not by 100.; A common year has 365 days; a leap year has 366 days.
Plan
Primary tool: #11 Work Backwards
Secondary: #5 Look for a Pattern, #7 Identify Subproblems, #3 Eliminate Possibilities
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.
Execute — Answer: A
4.OA.A.3 Step 1 Pin down the birth year
- An anniversary counts whole years since the event.
- The 200th anniversary falls on February 7, 2012, so the birth happened 200 years earlier on the same month and day: February 7, 1812.
💡 Going back 200 anniversaries just means subtracting 200 from the year while keeping February 7.
4.NBT.B.6 Step 2 How a date shifts weekday each year
- A week is 7 days.
- A common year has 365 days, and 365 = 52 x 7 + 1, so the same calendar date lands 1 weekday later the next year.
- A leap year has 366 days, and 366 = 52 x 7 + 2, so a leap year pushes the date 2 weekdays later.
- Only the leftover days past whole weeks matter.
💡 Whole weeks return you to the same weekday, so only the remainder after dividing by 7 moves the day.
4.OA.B.4 Step 3 Count the leap years crossed
- Moving from February 7, 1812 forward to February 7, 2012 crosses the years 1812 through 2011.
- Count the years divisible by 4: from 1812 to 2008 that is (2008 - 1812) / 4 + 1 = 50 years.
- Now apply the rule: 1900 is divisible by 100 but not 400, so it is not a leap year and must be removed; 2000 is divisible by 400 and stays.
- That leaves 49 leap years and 200 - 49 = 151 common years.
💡 Only leap days add the extra shift, and the divisible-by-100 rule quietly removes 1900.
4.OA.A.3 Step 4 Add the shifts and step back
- Each common year shifts the weekday by 1 and each leap year by 2, so across the 200 years the total forward shift is 151 x 1 + 49 x 2 = 249 days of weekday movement.
- Dividing by 7 leaves a remainder of 4, since 249 = 7 x 35 + 4, so February 7, 2012 lands 4 weekdays after February 7, 1812.
- To undo that, step 4 days backward from Tuesday: Tuesday, Monday, Sunday, Saturday, Friday.
- Dickens was born on a Friday, which is answer (A).
💡 The birthday is 4 weekdays before 2012's Tuesday, so count 4 days backward to reach it.
4.OA.A.3 An anniversary counts whole years since the event. The 200th anniversary falls o 4.NBT.B.6 A week is 7 days. A common year has 365 days, and 365 = 52 x 7 + 1, so the same 4.OA.B.4 Moving from February 7, 1812 forward to February 7, 2012 crosses the years 1812 4.OA.A.3 Each common year shifts the weekday by 1 and each leap year by 2, so across the Review
Reasonableness: The total shift 249 is close to 250, and 250 = 7 x 35 + 5 would give 5; our exact count gives remainder 4, matching Tuesday minus 4 = Friday. A quick sanity check: 200 years is a little over 28 whole weeks per year cycle, and the leftover 4 days is small and plausible. Friday is one of the listed choices, so the answer is consistent.
Alternative: Instead of counting common and leap years separately, note every year adds at least 1 weekday, so the base shift is 200, and each of the 49 leap years adds 1 more: 200 + 49 = 249. This reaches the same remainder 4 with less arithmetic.
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Locating the birth year and stepping backward the right number of weekdays.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing day counts by 7 to find how many weekdays a date shifts.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Counting multiples of 4 and applying the divisibility-based leap-year rule.)
⭐ To 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.
⭐ To 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.
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