AMC 10 · 2012 · #9

Grade 4 number-theory
modular-arithmeticpattern-recognition work-backwardscasework ↑ Prerequisites: modular-arithmetic
📏 Medium solution 💡 2 insights
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Problem
A known weekday sits a whole number of years after a birth date. Find the weekday of the birth.

Pick an answer.

(A)
Friday
(B)
Saturday
(C)
Sunday
(D)
Monday
(E)
Tuesday
How to solve
Strategy Work Backwards

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.

1STEP 1

Pin down the birth year

Subtracting gives the birth year.

2012 - 200 = 1812
2STEP 2

How a date shifts weekday each year

An ordinary year pushes the weekday by one.

365 = 52 × 7 + 1, 366 = 52 × 7 + 2
3STEP 3

Count the leap years crossed

Counting the leap years takes the century rule.

(2008 - 1812)/4 + 1 = 50, 50 - 1 = 49 leap years, 200 - 49 = 151 common years
4STEP 4

Add the shifts and step back

Stepping back gives Friday.

151 × 1 + 49 × 2 = 249, 249 = 7 × 35 + 4, Tuesday - 4 = Friday
Answer
Friday
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.
💡Key takeaway

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.

  • Pin down the birth year
  • How a date shifts weekday each year
  • Count the leap years crossed
  • Add the shifts and step back