AMC 10 · 2023 · #16
Grade 6 number-theoryPick an answer.
The famous two-coin formula ab - a - b is not available: it needs exactly two coprime values, and here every pair shares a factor (gcd(6,10) = 2, gcd(6,15) = 3, gcd(10,15) = 5). With three coins there is no formula at all, so the answer has to be built. Tool #15 (Organize Information in More Ways) is the whole solve: instead of listing prices 1, 2, 3, … in size order, sort every price by its remainder on division by 6, giving six lanes. Adding one 6-cent coin moves a price up its own lane and never leaves it, so each lane is unpayable up to some threshold and payable forever after. Tool #13 (Convert to Algebra) writes payability as N = 6a + 10b + 15c; tool #2 (Make a Systematic List) finds the cheapest payable price in each of the six lanes; tool #14 (Extreme Principle) picks the largest of the six per-lane gaps; tool #6 (Guess and Check) confirms 29 itself fails while 30 through 35 all succeed; tool #3 (Eliminate Possibilities) reads the digit sum against the five choices.
Write payability as an equation
Write payability as an equation.
Paying exactly means landing on N on the nose with a pile of coins — no change back means nothing can ever be subtracted.
6.EE.B.6Convert To AlgebraThe six-cent coin is a one-way step
Once payable, it stays payable.
Adding a 6-cent coin keeps you in the same remainder lane and moves you one step up it, so every lane is a ray: blocked below a threshold, open from the threshold on.
A six-cent coin keeps you in the same remainder lane and moves you one step up it.
▸ Why?
Adding a multiple of six leaves the remainder after dividing by six untouched.
▸ Why?
Every step up a lane is the same fixed size, so each lane is an evenly spaced list with a first member.
Which coins change lanes
The remainder splits values into lanes.
A 6-coin is invisible to the remainder, so a lane can only be entered by the 10s and 15s — and the first entry found is the cheapest one.
6.NS.B.2Organize Information In More WaysCheapest entry per lane
Find each lane's cheapest entry.
Lane 5 is the stubborn one: a price with remainder 5 is odd, an odd total needs an odd number of 15s, and one 15 still needs two 10s to land on remainder 5 — that is 15 + 10 + 10 = 35.
4.OA.B.4Make A Systematic ListEach lane's last gap
The last gap sits six below the entry.
The biggest hole overall must live in the lane that opens last, and it sits exactly one 6-step below that lane's door.
6.EE.B.8Extreme PrincipleCheck the boundary
Confirm everything above works.
Six consecutive payable prices cover all six remainder lanes at once, so from 30 on nothing can ever be missed.
4.OA.A.3Guess And CheckAdd the digits
Adding the digits gives 11.
The question asks about the digits of x, not x itself, so the last move is place value, not more number theory.
4.NBT.A.2Eliminate PossibilitiesBecause one coin is worth 6, sort every price into six lanes by its remainder after dividing by 6. A lane opens at its cheapest payable price and stays open forever, so the very last unpayable price sits 6 below the door of the lane that opens last: 35 - 6 = 29, and 2 + 9 = 11.
- Write payability as an equation
- A 6-coin is a one-way step
- Only 10s and 15s change lanes
- Cheapest entry in each lane
- Each lane's last gap is 6 below
- Check 29 fails and 30 up works
- Add the digits