Competition · AMC preparation · step 4 of 4
AMC 8 · 2006 · #23
Grade 6 number-theoryPick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Each remainder condition pins N to a clear arithmetic sequence — "start at 4, step by 6" for the first, "start at 3, step by 5" for the second — so Tool #2 (Systematic List) writes both sequences cleanly and the first shared value is the smallest N. Tool #5 (Look for a Pattern) sharpens the search: the gap to the next multiple of 6 is 2, and the gap to the next multiple of 5 is also 2, so N + 2 must be a common multiple of 5 and 6. That pattern means we only need the smallest common multiple of 5 and 6, which is 30, and then N = 30 - 2 = 28. Both tools land on the same N, after which the final remainder by 7 is one division.
List the first condition
The numbers leaving remainder 4 when divided by 6 start at 4 and grow by 6.
Grade 4 "generate a number pattern from a rule" — the rule here is +6, starting at 4.
4.OA.C.5Make A Systematic ListList the second condition
The numbers leaving remainder 3 when divided by 5 start at 3 and grow by 5.
Same Grade 4 pattern move with a different rule: +5, starting at 3.
4.OA.C.5Make A Systematic ListFind the smallest shared value
Scanning both lists in order, the first shared value is 28, so that is the smallest N.
Intersecting two short lists is the cleanest Grade 4 way to pin down the smallest number that fits two multiple-style rules at once.
The smallest number of coins that leaves 4 left over when shared among six and 3 left over when shared among five is 28.
▸ Why?
28 satisfies both sharing conditions at once, and it is the first number that does, since the numbers that fit each condition arrive at fixed, evenly spaced steps that first line up at 28.
▸ Why?
Sharing 28 among six leaves 4: each of the six takes 4 coins, which uses 24, and the remaining 28 - 24 = 4 coins are too few to give everyone one more.
▸ Why?
Six equal shares of 4 coins each account for 6 × 4 = 24 coins.
▸ Why?
The 24 handed out and the 4 held back together rebuild all 28, so exactly 4 are left over.
▸ Why?
Sharing 28 among five leaves 3: each of the five takes 5 coins, which uses 25, and the remaining 28 - 25 = 3 coins are too few to give everyone one more.
▸ Why?
Five equal shares of 5 coins each account for 5 × 5 = 25 coins.
▸ Why?
The 25 handed out and the 3 held back together rebuild all 28, so exactly 3 are left over.
▸ Why?
No number below 28 fits both, because the numbers leaving 4 under six-way sharing recur every 6 and those leaving 3 under five-way sharing recur every 5, so writing each list in order up to 28 covers every earlier candidate and turns up no earlier shared value.
▸ Why?
The numbers that leave remainder 4 when shared six ways make the steady pattern 4, 10, 16, 22, 28, each one 6 past the last, so this one pattern already holds every candidate for the first condition.
▸ Why?
The numbers that leave remainder 3 when shared five ways make the steady pattern 3, 8, 13, 18, 23, 28, each one 5 past the last, so this one pattern already holds every candidate for the second condition.
Divide 28 by 7
Dividing 28 by 7 gives exactly 4 with remainder 0.
Grade 4 division-with-remainder closes the problem: 7 divides 28 evenly, so nothing is left over.
4.NBT.B.6Make A Systematic ListWhen a count leaves two awkward remainders, two short arithmetic lists usually meet within a few terms — and spotting that both shortfalls are the same number turns the search into a quick LCM.
- List the first condition
- List the second condition
- Find the smallest shared value
- Divide 28 by 7
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