AMC 8 · 2014 · #10

Easy mode Grade 4
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Problem

The very first AMC 88 contest was held in 19851985. There has been one AMC 88 every year since then, with no skipping.

Samantha took the 77th AMC 88. That same year, she turned 1212 years old.

What year was Samantha born?

(A) 1979(B) 1980(C) 1981(D) 1982(E) 1983\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983

Pick an answer.

(A)
1979
(B)
1980
(C)
1981
(D)
1982
(E)
1983
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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: AMC $8$ #$1$ was held in $1985$; The contest has been held every year since (one per year); Samantha turned $12$ the year of the seventh AMC $8$; Answer choices: (A) $1979$, (B) $1980$, (C) $1981$, (D) $1982$, (E) $1983$

Unknowns: Samantha's birth year

Understand

Restated: 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?

Givens: AMC $8$ #$1$ was held in $1985$; The contest has been held every year since (one per year); Samantha turned $12$ the year of the seventh AMC $8$; Answer choices: (A) $1979$, (B) $1980$, (C) $1981$, (D) $1982$, (E) $1983$

Plan

Primary tool: #5 Look for a Pattern

Secondary: #7 Identify Subproblems

The years $1985, 1986, 1987, \dots$ 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.

Execute — Answer: A

#5 Look for a Pattern 4.OA.C.5 Step 1

List the AMC $8$ years until the seventh one to see the pattern: each year goes up by $1$, so the $n$-th contest is in year $1985 + (n - 1)$.

$$1985, 1986, 1987, 1988, 1989, 1990, \underline{1991}$$

💡 Writing out the first few terms makes the rule obvious: add $1$ each step. This is the Grade $4$ "generate and analyze patterns" move.

#5 Look for a Pattern 4.NBT.B.4 Step 2

Confirm the pattern with the formula: the seventh contest is $6$ years after the first.

$$1985 + (7 - 1) = 1985 + 6 = 1991$$

💡 Adding $6$ to a four-digit year is the Grade $4$ "fluently add multi-digit whole numbers" skill.

#7 Identify Subproblems 4.NBT.B.4 Step 3

Now solve the second subproblem: given that Samantha was $12$ in $1991$, subtract to find her birth year.

$$1991 - 12 = 1979$$

💡 Splitting the question into "find the year, then find the age" is the Tool #$7$ subproblems move; the subtraction itself is standard Grade $4$ arithmetic.

#7 Identify Subproblems 4.NBT.B.4 Step 4

Match $1979$ to the answer choices.

$$1979 \;\Rightarrow\; \textbf{(A)}$$

💡 Reading the choice list is the final "close the subproblem" check.

[1] #5 4.OA.C.5 List the AMC $8$ years until the seventh one to see the pattern: each year goes
[2] #5 4.NBT.B.4 Confirm the pattern with the formula: the seventh contest is $6$ years after the
[3] #7 4.NBT.B.4 Now solve the second subproblem: given that Samantha was $12$ in $1991$, subtrac
[4] #7 4.NBT.B.4 Match $1979$ to the answer choices.

Review

Reasonableness: Sanity-check the pattern: the $1$st contest is $1985$, so the $2$nd is $1986$, the $3$rd is $1987$, $\dots$ — six steps later the $7$th 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.

Alternative: Tool #$11$ (Work Backwards) gives the same answer in one chain: start from "age $12$ at the seventh AMC $8$," rewind $12$ years to the birth year, and substitute the contest year $1991$ — so birth year $= 1991 - 12 = 1979$. Or tool #$6$ (Guess and Check) on the choices: only $1979 + 12 = 1991$ matches the seventh-contest year; $1980 + 12 = 1992$, $1981 + 12 = 1993$, etc., do not.

CCSS standards used (min grade 4)

  • 4.OA.C.5 Generate and analyze patterns (Recognizing that AMC $8$ years form the arithmetic pattern $1985, 1986, 1987, \dots$ (step $+ 1$) and using the rule "$n$-th term $= 1985 + (n - 1)$" to land on $1991$ for the seventh contest.)
  • 4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Performing the four-digit arithmetic $1985 + 6 = 1991$ and then $1991 - 12 = 1979$ to get Samantha's birth year.)

⭐ 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!

⭐ 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!