AMC 8 · 2014 · #10

Grade 4 arithmetic
sequences-arithmeticmulti-digit-arithmetic identify-subproblems ↑ Prerequisites: sequences-arithmeticmulti-digit-arithmetic
📏 Short solution 💡 2 insights
📘 View easy version →
Problem
The first AMC 8 took place in 1985 and one has been held every year since. Samantha was 12 years old in the year she sat the seventh AMC 8. In what year was she born?

Pick an answer.

(A)
1979
(B)
1980
(C)
1981
(D)
1982
(E)
1983

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

The years 1985, 1986, 1987, … form a simple pattern — one contest per year, going up by 1 each time. Tool #5 (Look for a Pattern) turns "the seventh AMC 8" into a year by counting six steps forward from 1985. Tool #7 (Identify Subproblems) keeps the work tidy by splitting the question into two clean pieces: (a) find the year of the seventh contest, then (b) subtract Samantha's age to get her birth year. Algebra (tool #13) is overkill here — a Grade 4 pattern + subtraction is enough.

1STEP 1

List the contest years — each goes up by 1, so the n-th AMC 8 lands in 1985 + (n - 1).

1985, 1986, 1987, 1988, 1989, 1990, 1991
2STEP 2

Put n = 7: the seventh contest is 6 years after the first, so it falls in 1991.

1985 + (7 - 1) = 1985 + 6 = 1991
3STEP 3

Samantha turned 12 in 1991, so subtract her age: 1991 - 12 gives her birth year.

1991 - 12 = 1979
4STEP 4

Match the birth year to the answer choices.

1979 → (A)
Answer
1979
Sanity-check the pattern: the 1st contest is 1985, so the 2nd is 1986, the 3rd is 1987, … — six steps later the 7th must be 1991. Then a 12-year-old in 1991 was born in 1991 - 12 = 1979, which lines up with choice (A). The other choices (1980–1983) would mean Samantha was 11, 10, 9, or 8 in 1991 — none match the "turned 12" clue.
💡Key takeaway

This AMC 8 problem only needs Grade 4 skills — counting up by 1 to spot a year pattern, then a single subtraction — that you already know!