AMC 10 · 2024 · #11
Grade 8 number-theoryPick an answer.
The question stacks four jobs on top of each other: (i) translate 2024_b into base ten, (ii) turn "divisible by 16" into a clean congruence on b, (iii) count the bases in [5, 2024] that pass that congruence, (iv) take a digit sum. That stacking is the trigger for Tool #7 (Identify Subproblems) — solve each piece on its own, then stitch. Inside step (ii) we use Tool #9 (Solve an Easier Related Problem): every coefficient in 2b³ + 2b + 4 is even, so divide the whole congruence by 2 to drop the modulus from 16 to 8. Working mod 8 is much friendlier because for any odd b, b² ≡ 1 (mod 8) — that single fact collapses the odd case to a linear congruence.
Expand the numeral
The numeral is two b cubed plus two b plus four.
Place value in base b is the Grade 5 rule "each place is b times the next", extended one variable at a time.
5.NBT.A.1Identify SubproblemsPull out the 2
The condition lightens to mod 8.
Pulling a common factor out of every term is the Grade 6 distributive-property trick, applied to a congruence instead of a sum.
6.NS.B.4Solve An Easier Related ProblemHandle odd bases
An odd cube leaves the same remainder as b.
"Odd squares are 1 mod 8" is one of the most useful integer-exponent facts in number theory — it turns the cubic into a linear congruence.
8.EE.A.1Identify SubproblemsThe surviving odd remainders
Among odds, 3 and 7 pass.
Going from "step by 4" to "step by 8" splits each residue class in two — the Grade 4 "continue a pattern" rule applied to residues.
4.OA.C.5Identify SubproblemsHandle even bases
Among evens only 6 passes.
Substituting b = 2k kills the cubic term because (2k)³ is already a multiple of 8 — a clean cube-power fact.
8.EE.A.1Identify SubproblemsRead the density
Three out of every eight survive.
Residues mod 8 repeat every 8 steps — once you know which residues qualify, counting is just "how many full blocks fit, plus leftovers".
Remainders repeat on a fixed period, so counting inside a window is counting full blocks plus leftovers.
▸ Why?
Once the remainders come back around, everything that followed them the first time follows again.
▸ Why?
Any window splits into whole periods plus one short leftover, and that split is unique.
Count inside the range
Trimming the ends gives K equals 758.
The "count in a window" trick is to count an easy super-window then subtract the part you do not want.
4.OA.C.5Identify SubproblemsAdd the digits
Seven plus five plus eight is 20.
Digit sum is base-10 place value at its most direct — add the digits and stop.
5.NBT.A.1Identify SubproblemsWhen divisibility hides inside a polynomial in b, split the work: translate the base-b number, shrink the modulus by pulling out common factors, then handle odd b and even b separately. Once you know which residues mod 8 qualify, counting bases in a long range is just 3/8 of the window, adjusted for the edges.