AMC 10 · 2015 · #7
Grade 4 arithmeticPick an answer.
AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Counting an unfamiliar list like 13, 16, 19, …c is hard to eyeball, so Tool #9 (Solve an Easier Related Problem) reshapes it into a list anyone can count at a glance. First Tool #5 (Look for a Pattern) reveals the constant step of 3; then shifting every term down and dividing by that step turns the sequence into 0, 1, 2, …c, 20, where the count is obvious. Reshaping never changes how many terms there are.
Find the constant step
The terms grow by a constant step: 16 - 13 = 3 and 19 - 16 = 3.
A sequence is easiest to handle once you know the single fixed amount it jumps by each time.
4.OA.C.5Look For A PatternShift the list to start at 0
Subtract 13 from every term so the list runs 0, 3, 6, …, 60.
Shifting every number by the same amount keeps the count the same, just like moving a row of chairs sideways.
4.NBT.B.4Solve An Easier Related ProblemDivide every term by 3
Divide every term by 3, giving the simple list 0, 1, 2, …, 20.
Squeezing the gaps to size 1 makes the terms line up with the counting numbers you already know.
4.NBT.B.6Solve An Easier Related ProblemCount the whole numbers from 0 to 20
Both endpoints count, so terms = 20 + 1 = 21 — choice (B).
Counting fence posts, not gaps: a row with 20 gaps between posts has 21 posts because both ends count.
4.NBT.B.4Look For A PatternTo count an evenly-spaced list, divide the gap from first to last by the step, then add one for the term you started on.
- Find the constant step
- Shift the list to start at 0
- Divide every term by 3
- Count the whole numbers from 0 to 20