Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #13
Grade 4 number-theory
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Dividing 2, 4, 6, … by 7 produces a repeating cycle of remainders, so Tool #5 (Look for a Pattern) is the natural primary tool — compute the first several remainders, spot the cycle, then count efficiently. Tool #2 (Systematic List) supports this by listing the first seven remainders in order so the cycle is visible. Tool #3 (Eliminate Possibilities) is the multiple-choice closer: once we have the seven counts, only one of (A)-(E) matches and the rest can be crossed out.
Count the even numbers
The even numbers 2, 4, …, 50 are 2·1 through 2·25, so there are exactly 25 of them — the seven bars must sum to 25.
Pairing each even number with a counting number 1-25 is a Grade 4 multi-step word-problem move.
4.OA.A.3Make A Systematic ListList the remainders
Listing the remainders of the first even numbers when divided by 7 exposes a length-7 cycle: 2, 4, 6, 1, 3, 5, 0.
Finding whole-number quotients and remainders is exactly the Grade 4 division-with-remainder skill.
4.NBT.B.6Make A Systematic ListConfirm the cycle repeats
16 leaves remainder 2, restarting the cycle; adding 14 (= 2·7) never changes the remainder mod 7, so the 7-block repeats forever.
Spotting a repeating cycle in a generated number list is the Grade 4 pattern-rule standard in action.
4.OA.C.5Look For A PatternCount the complete cycles
Dividing 25 by 7 gives 25 = 7·3 + 4, so the cycle runs 3 full times (21 numbers) with 4 leftover entries.
Splitting 25 into 3 full groups of 7 plus 4 leftovers is the same Grade 4 division-with-remainder idea, used at the cycle level.
4.NBT.B.6Look For A PatternTally each remainder
After 3 cycles every remainder sits at 3; the leftovers 44, 46, 48, 50 give remainders 2, 4, 6, 1, so bump those four bins by 1.
Combining "3 from each cycle" with "+1 for the leftovers" is a Grade 4 two-step word-problem combination.
When all 25 even numbers are sorted by their remainder on division by 7, the seven bins (remainders 0 through 6) end up holding 3, 4, 4, 3, 4, 3, 4 numbers.
▸ Why?
The three complete cycles drop each remainder 0 through 6 into its bin exactly three times.
▸ Why?
The remainders keep repeating in the same fixed block of seven, so every group of seven even numbers is one identical copy of that cycle.
▸ Why?
That block (2, 4, 6, 1, 3, 5, 0) shows all seven remainders once each, so the seven numbers pair one-to-one with the seven bins.
▸ Why?
Running a block that holds one of each remainder three times means three equal groups of one, which is three.
▸ Why?
The four numbers left over after the full cycles add one more to the bins for remainders 2, 4, 6, and 1.
▸ Why?
The leftovers pick up right where the last full cycle ended, and after a full period the pattern restarts from its first entries 2, 4, 6, 1.
▸ Why?
Those four leftover numbers land in four different bins, so each of those bins gains exactly one.
Write the bar heights
In remainder order 0–6 the bar heights are 3, 4, 4, 3, 4, 3, 4, summing to 25 — matching the even-number count.
Reading bar heights off a frequency tally is the Grade 3 scaled-bar-graph skill.
3.MD.B.3Eliminate PossibilitiesMatch the right histogram
Only histogram (A) shows [3, 4, 4, 3, 4, 3, 4]; (B), (C), (D), (E) each differ in a bar, so the answer is (A).
Comparing our computed frequencies to each histogram is a Grade 3 bar-graph interpretation task.
3.MD.B.3Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 division-with-remainder and pattern-spotting you already know!
- Count the even numbers
- List the remainders
- Confirm the cycle repeats
- Count the complete cycles
- Tally each remainder
- Write the bar heights
- Match the right histogram
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