Competition · AMC preparation · step 4 of 4
AMC 8 · 2020 · #7
Grade 4 countingPick an answer.
AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Pin down the thousands digit
N is between 2020 and 2400, so the thousands digit must be a = 2 (1 is too small, 3 or more too big).
Comparing four-digit numbers by their leading digit is exactly Grade 4 place value: only numbers starting with 2 land in the 2,xxx family.
4.NBT.A.2Eliminate PossibilitiesPin down the hundreds digit
Since a = 2 and digits climb, b must exceed 2; but b = 4 forces N ≥ 2456 ＞ 2400, so b = 3 is the only fit.
Squeezing b between two inequalities is the same multi-digit comparison move, just applied to the hundreds place.
4.NBT.A.2Eliminate PossibilitiesReduce to the last two digits
Now every valid N is 23cd, so just count increasing pairs (c, d) chosen from {4, 5, 6, 7, 8, 9}.
Stripping the problem down to 'count pairs from a small set' is a Grade 4 multi-step word-problem move.
4.OA.A.3Make A Systematic ListList the digit pairs
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.
Counting items grouped into rows of 5, 4, 3, 2, 1 is the same as a Grade 2 rectangular-array total: just add the row sizes.
Grouping the increasing pairs (c, d) from {4, 5, 6, 7, 8, 9} by their smaller digit c gives group sizes 5, 4, 3, 2, 1, 0, which add up to 15.
▸ Why?
Every increasing pair has exactly one smaller digit c, so filing each pair under its own c sorts all the pairs into separate groups with none left out and none placed in two groups at once.
▸ Why?
When a collection is split into groups that overlap nowhere and miss nothing, the sizes of the groups add back up to the size of the whole collection.
▸ Why?
The group with smaller digit c holds one pair for each digit d larger than c up to 9, so its size is just how many digits sit above c: 5 when c = 4, then 4, 3, 2, 1, and 0 when c = 9.
▸ Why?
Matching each pair in the group with its own larger digit d links them one for one, so the group holds exactly as many pairs as there are digits above c.
Count the numbers
Each pair (c, d) makes exactly one number 23cd, so the integer count equals the pair count: 15.
One-to-one correspondence between pairs and numbers means the totals match — Grade 4 multi-step reasoning.
4.OA.A.3Make A Systematic ListThis AMC 8 problem only needs Grade 4 place-value comparison plus a tidy list you can add up — no fancy combinations formula required!
- Pin down the thousands digit
- Pin down the hundreds digit
- Reduce to the last two digits
- List the digit pairs
- Count the numbers
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