AMC 8 · 2022 · #13

Grade 4 algebranumber-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
systems-of-equationslinear-diophantinesystematic-enumeration convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: linear-equations-two-varsystematic-enumeration
📏 Medium solution 💡 3 insights
📘 View easy version →
Problem
We have two positive whole numbers. The first one equals twice the second one, plus some extra positive whole number k (the value in the blank). The two numbers also add up to 28. The question is: how many different positive whole numbers k can go in the blank?

Pick an answer.

(A)
6
(B)
7
(C)
8
(D)
9
(E)
10

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The question "how many values of k work?" begs for a Tool #2 systematic list — pick an ordering for b (the smaller of the two numbers) and list every b that produces a valid positive k. To set up the list cleanly we first use Tool #7 (Subproblems): combine the two facts (a = 2b + k and a + b = 28) into a single arithmetic relationship 3b + k = 28, so k = 28 - 3b. Then Tool #6 (Guess and Check) lets us walk b = 1, 2, 3, … and stop the moment k would drop below 1. This avoids reaching for Tool #13 (algebra) when ordinary subtraction and a list do the job.

1STEP 1

"a is k more than twice b" means a = 2b + k; substitute into a + b = 28 to get 3b + k = 28.

a = 2b + k, a + b = 28 → 3b + k = 28
2STEP 2

Solve for k: k = 28 - 3b. Now ask which positive integers b keep 28 - 3b positive.

k = 28 - 3b
3STEP 3

k ≥ 1 needs 28 - 3b ≥ 1, so 3b ≤ 27 and b ≤ 9; b runs over the whole numbers 1 to 9.

28 - 3b ≥ 1 ⇔ 3b ≤ 27 ⇔ b ≤ 9
4STEP 4

List (b, k) for b = 1 to 9: k = 25, 22, 19, …, 1, and a = 28 - b stays positive throughout.

b & k = 28-3b & a = 28-b ; 1 & 25 & 27 ; 2 & 22 & 26 ; 3 & 19 & 25 ; 4 & 16 & 24 ; 5 & 13 & 23 ; 6 & 10 & 22 ; 7 & 7 & 21 ; 8 & 4 & 20 ; 9 & 1 & 19
5STEP 5

Count the distinct k values 25, 22, 19, 16, 13, 10, 7, 4, 1 — that is 9 different numbers, choice (D).

#{25, 22, 19, 16, 13, 10, 7, 4, 1} = 9 → (D)
Answer
9
Sanity-check the two extreme rows of the table. When b = 1: the bigger number is a = 27, and 2b + k = 2 + 25 = 27. Yes — a + b = 28 and a is 25 more than twice b. When b = 9: a = 19, and 2b + k = 18 + 1 = 19. Yes — a + b = 28 and a is just 1 more than twice b. Trying b = 10 would force k = 28 - 30 = -2, which is not a positive integer, so the list really does stop at b = 9. Nine values is consistent, and choice (D) 9 is the answer.
💡Key takeaway

This AMC 8 problem only needs Grade 4 "twice as many" thinking and a careful list you already know how to make!

More like this

Same archetype — closest grade level first.