Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #12
Grade 4 number-theoryPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Three simultaneous remainder conditions feel intimidating, but Tool #9 transforms the problem into something much easier: "N has remainder 1 when divided by d" is the same as saying "N - 1 is a multiple of d." So instead of hunting for N directly, we hunt for N - 1, which must be a common multiple of 4, 5, and 6 — pure 4th-grade multiples thinking. Tool #2 (Systematic List) then sweeps multiples of the largest divisor (6) in order and checks each against the other two conditions, which is faster than computing a formal LCM. Tool #3 (Eliminate) closes the problem by matching N against the five labeled ranges.
Rewrite as one divisibility
Remainder 1 when divided by d means N - 1 is divisible by d, so we seek the smallest N - 1 that is a common multiple of 4, 5, and 6.
Remainder 1 means "one past a multiple," so subtracting 1 snaps the number back onto a multiple — a Grade 4 remainder idea.
Requiring N to leave a remainder of 1 when divided by 4, 5, and 6 is exactly the same as requiring N-1 to be a common multiple of 4, 5, and 6.
▸ Why?
A remainder of 1 when dividing N by any one of the divisors d means N is built from a whole number of equal groups of size d with a single unit left over, so N is a multiple of d with 1 sitting on top.
▸ Why?
Stacking a whole number of equal groups of size d is exactly what makes a multiple of d, so the grouped part of N really is a multiple of d.
▸ Why?
Taking that single left-over unit away turns N into N-1, and this lands N-1 right on the grouped part alone — a multiple of d with nothing left over, so d divides N-1 evenly for each of d=4, 5, and 6.
▸ Why?
Subtracting 1 is the exact reverse of the +1 that put the leftover there, so removing it leaves precisely the multiple of d that was underneath.
Hunt for the common multiple
Sweep multiples of 6 — 6, 12, 18, ..., 60 — the first also divisible by 4 and 5 is 60, the smallest common multiple of 4, 5, 6.
Sweeping multiples of one number and checking divisibility by the others is the Grade 4 "factors and multiples" skill in action.
4.OA.B.4Make A Systematic ListAdd 1 to get N
Add 1 to undo the shift: N = 60 + 1 = 61. The next common multiple, 120, gives N = 121 — far bigger — so 61 is the smallest.
Undoing the "-1" shift to translate between the easier problem and the original is exactly the Grade 4 multi-step word-problem move.
4.OA.A.3Solve An Easier Related ProblemMatch against the ranges
Test 61 against the ranges: it misses [2,19], [20,39], [40,59], and is below 80, so it lands in [60, 79] — choice (D).
Comparing 61 to the range endpoints uses Grade 4 multi-digit number comparison.
4.NBT.A.2Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 multiples and remainders you already know — "remainder 1" just means "one more than a multiple"!
- Rewrite as one divisibility
- Hunt for the common multiple
- Add 1 to get N
- Match against the ranges
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