Competition · AMC preparation · step 4 of 4
AMC 8 · 1999 · #15
Grade 5 countingPick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Only a handful of distributions exist for 2 extra letters across 3 sets, so Tool #2 (Make a Systematic List) sweeps them all without missing any: 3 ways to put both letters in one set plus 3 ways to put one letter in each of two sets — exactly 6 cases. Tool #6 (Guess and Check) is the natural companion: compute the new product for each case and pick the largest. The size of the case list is small enough that listing beats algebra, which keeps the solution accessible.
Count the original plates
Multiply the three set sizes: 60 plates is the baseline we subtract at the end.
Three independent choices multiply — the Grade 5 multiplication-as-counting move.
5.NBT.B.5Make A Systematic ListList the ways to add 2 letters
List every way to spread 2 letters: both in one set (3 ways) or one each in two sets (3 ways) — 6 cases.
When the case count is small, writing the full list is the safest way not to miss the winner.
There are exactly 6 different ways to distribute the 2 new letters among the three sets.
▸ Why?
Every way of adding the two letters falls into one of two separate groups — both letters into a single set, or one letter into each of two different sets — and since these groups share nothing and leave nothing out, the total count is the size of the first group plus the size of the second.
▸ Why?
For both letters in a single set, the only decision is which one of the three sets receives them; each set gives exactly one such way, so pairing the ways with the three sets one for one gives 3 ways.
▸ Why?
For one letter into each of two different sets, the only decision is which two of the three sets get a letter; those pairs are exactly (1st, 2nd), (1st, 3rd), and (2nd, 3rd), so pairing the ways with these pairs one for one gives 3 ways.
▸ Why?
Adding the two group sizes is legitimate because every possible distribution belongs to exactly one of the groups — none is skipped and none is counted twice — so the two groups reassemble into all the ways.
Multiply each case out
Multiply the three updated sizes in each case to get its new plate total.
One multiplication per row turns the case list into a column of totals you can scan.
5.NBT.B.5Guess And CheckSubtract to find the increase
The best new total is 100 (from 5×5×4); subtract the original 60 for the ADDITIONAL plates.
"Additional" means the gain over the starting count — subtract the old total from the best new one.
4.OA.A.3Guess And CheckOnly 6 ways exist to spread 2 new letters across 3 sets — list them, multiply, and the best one is 5 × 5 × 4 = 100, which is 40 more plates than the original 60.
- Count the original plates
- List the ways to add 2 letters
- Multiply each case out
- Subtract to find the increase
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