AMC 10 · 2017 · #18
Grade 6 number-theoryPick an answer.
We can't see n, so we work with the one thing we control: what adding 1 does to a digit sum. Tool #5 (Look for a Pattern) on small numbers reveals a clean rule — adding 1 raises the digit sum by 1, unless trailing 9s roll over to 0. Tool #4 (Introduce a Variable) names k, the count of trailing 9s, and turns the rule into a formula for S(n+1). Then, because this is multiple choice with a finite list, Tool #3 (Eliminate Possibilities) tests each choice against the formula and keeps only the one that gives a whole-number k.
See what adding 1 does
Adding one turns trailing nines into zeros.
A 9 at the end can't go higher, so adding 1 flips it to 0 and pushes the carry left.
A nine at the end cannot go higher, so adding one flips it to zero and pushes the carry left.
▸ Why?
A number is its digits weighted by their places, and each place holds only up to nine.
▸ Why?
Reaching ten in a place means one full bundle moves up, leaving the remainder zero behind.
Write the rule as a formula
So the new sum is the old plus one minus nines.
Counting the trailing nines is the only thing that changes the answer, so name that count and the rest is fixed.
6.EE.A.2Introduce A VariableTest each choice for a whole-number k
Most choices fail that divisibility.
If 1275-c isn't a clean stack of 9s, no count of trailing nines can produce that drop.
4.OA.B.4Eliminate PossibilitiesConfirm the survivor
Only 1239 survives, choice (D).
A whole-number k corresponds to a real number n, so the formula's solution is genuinely achievable.
6.EE.B.5Eliminate PossibilitiesAdding 1 raises a digit sum by 1, but every trailing 9 that rolls over to 0 knocks it down by 9.
- See what adding 1 does
- Write the rule as a formula
- Test each choice for a whole-number k
- Confirm the survivor