AMC 8 · 2020 · #7

Grade 4 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

digit-constraintscombinations-basicsystematic-enumeration digit-constraintscaseworksystematic-enumeration ↑ Prerequisites: systematic-enumeration

📏 Medium solution 💡 3 insights

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Problem

Count the four-digit integers strictly between 2020 and 2400 whose four digits are all different and appear in strictly increasing order from left to right (so each digit is smaller than the next one).

Pick an answer.

(A)

$text{9}$

(B)

$text{10}$

(C)

$text{15}$

(D)

$text{21}$

(E)

$text{28}$

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The problem says 'how many integers', so reach for Tool #2 (Systematic List). But before listing four-digit numbers blindly, use Tool #3 (Eliminate Possibilities) on the two leading digits: the range 2020 ＜ N ＜ 2400 together with the strict-increase rule forces a and b to specific values, leaving only the last two digits free. Once a and b are pinned down, we just list all increasing pairs (c, d) chosen from the digits larger than b — a short, fully bounded list.

1STEP 1

N is between 2020 and 2400, so the thousands digit must be a = 2 (1 is too small, 3 or more too big).

2020 ＜ N ＜ 2400 ⟹ a = 2

2STEP 2

Since a = 2 and digits climb, b must exceed 2; but b = 4 forces N ≥ 2456 ＞ 2400, so b = 3 is the only fit.

a = 2, a ＜ b, N ＜ 2400 ⟹ b = 3

3STEP 3

Now every valid N is 23cd, so just count increasing pairs (c, d) chosen from {4, 5, 6, 7, 8, 9}.

N = 23cd, c, d ∈ {4, 5, 6, 7, 8, 9}, c ＜ d

4STEP 4

List by the smaller digit c: 5, 4, 3, 2, 1, 0 pairs for c = 4, 5, 6, 7, 8, 9 — adding gives 15 pairs.

5 + 4 + 3 + 2 + 1 + 0 = 15

5STEP 5

Each pair (c, d) makes exactly one number 23cd, so the integer count equals the pair count: 15.

count = 15 ⟹ (C)

Answer

text{15}

Spot-check three of the 15 integers: 2345 (digits 2,3,4,5, strictly increasing, in range) — valid. 2378 (digits 2,3,7,8) — valid. 2389 (digits 2,3,8,9) — valid. The largest one we built is 2389 ＜ 2400, and the smallest is 2345 ＞ 2020, so the entire family sits inside the required range. The count 15 matches answer (C), and the other choices (9, 10, 21, 28) don't arise from any natural mis-count here.

💡Key takeaway

This AMC 8 problem only needs Grade 4 place-value comparison plus a tidy list you can add up — no fancy combinations formula required!

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