Competition · AMC preparation · step 4 of 4
AMC 8 · 2004 · #19
Grade 6 number-theoryPick an answer.
AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Four remainder conditions look like four separate problems, but Tool #5 (Find a Pattern) shows they all share the same shape: N - 2 is a multiple of 3, of 4, of 5, and of 6. That reframing collapses the four conditions into one: N - 2 is a common multiple of {3, 4, 5, 6}. Tool #7 (Break into Subproblems) then splits the work cleanly — first find the smallest common multiple (the LCM of the four divisors), then add the +2 offset to recover N. Two short steps replace a messy hunt.
Spot the shared pattern
Shared pattern: a remainder of 2 for every divisor means N - 2 is a multiple of each of 3, 4, 5, and 6.
Grade 4 "multiples" language: a remainder of 2 is just the multiple N - 2 shifted up by 2.
Saying that N leaves a remainder of 2 when divided by each of 3, 4, 5, and 6 is the same as saying that N - 2 is an exact multiple of each of 3, 4, 5, and 6.
▸ Why?
A remainder of 2 on dividing N by a divisor d means you pack N into whole groups of d and exactly 2 are left over, so N is some whole number of d-sized groups plus 2.
▸ Why?
The remainder is precisely what stays after taking out as many full groups of d as possible, so the rest of N is a whole number of equal groups of d — that is, a multiple of d.
▸ Why?
Write the division out as N = q · d + 2, where the leftover 2 is a valid remainder since 0 ≤ 2 ＜ d; removing it gives N - 2 = q · d, which divides into groups of d with remainder 0, and remainder 0 is exactly the condition for N - 2 to be a multiple of d — and this holds identically for each of the four divisors.
Reduce to a least common multiple
Subproblem 1: since N - 2 is a common multiple of {3, 4, 5, 6}, its smallest positive value is their LCM.
Tool #7 splits the work: pin down N - 2 first, then handle the +2. Grade 6 number theory says the smallest shared multiple is the LCM.
6.NS.B.4Identify SubproblemsCompute the LCM
Compute the LCM from primes: the highest power of each is 2², 3, and 5, so lcm(3, 4, 5, 6) = 60.
Highest power of 2 is 2² (from 4); highest power of 3 is 3¹ (from 3 or 6); highest power of 5 is 5¹ (from 5). Multiply: 4 · 3 · 5 = 60.
6.NS.B.4Identify SubproblemsAdd the 2 back
Subproblem 2: undo the shift — N = 60 + 2 = 62, which satisfies N ＞ 2.
62 ＞ 2 holds, and 62 falls inside the interval 60-79 from choice (B).
4.OA.A.3Identify SubproblemsWhen the same remainder shows up for several divisors, subtract that remainder first — what is left is a plain LCM, and adding the remainder back gives the answer.
- Spot the shared pattern
- Reduce to a least common multiple
- Compute the LCM
- Add the 2 back
A parent dashboard for the family lives at sensimlab.com.