AMC 10 · 2013 · #19

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
base-conversiondivisor-countprime-factorization convert-to-algebrasystematic-enumeration ↑ Prerequisites: base-conversion
📏 Medium solution 💡 2 insights
Problem
In base 10 the number 2013 ends in the digit 3, and in base 9 the same number is written (2676)₉ and ends in the digit 6. Now write 2013 in base b, where b is a positive integer. For how many bases b does that base-b representation end in the digit 3?

Pick an answer.

(A)
6
(B)
9
(C)
13
(D)
16
(E)
18

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

The words "ends in the digit 3" hide a clean arithmetic statement: the last digit of a number in base b is just its remainder on division by b. Tool #13 (Convert to Algebra) turns "2013 ends in 3 in base b" into the congruence 2013 ≡ 3 (mod b), which rearranges to "b divides 2010." That converts a base-representation puzzle into a divisor-counting problem. Tool #2 (Make a Systematic List) then counts the divisors of 2010 from its prime factorization, and Tool #3 (Eliminate Possibilities) removes the bases that are too small for the digit 3 to be legal.

1STEP 1

Last digit is a remainder

In base b the units digit is the remainder on division by b, so "ends in 3" says 2013 leaves remainder 3.

2013 ends in 3 (base b) ⇔ 2013 ≡ 3 (mod b)
2STEP 2

Turn it into divisibility

Peel off the leftover 3: 2013 - 3 = 2010 must be a whole multiple of b, so b is exactly a divisor of 2010.

2013 ≡ 3 (mod b) ⇔ b ∣ (2013 - 3) = 2010
3STEP 3

Count the divisors of 2010

2010 = 2 · 3 · 5 · 67: four distinct primes, each an in-or-out switch, so 2 · 2 · 2 · 2 = 16 divisors.

2010 = 2 · 3 · 5 · 67 → (1+1)⁴ = 16 divisors
4STEP 4

Drop bases too small for digit 3

A digit must be smaller than its base, so a digit 3 needs b ＞ 3 — the divisors 1, 2, and 3 are thrown out.

b ＞ 3 → exclude b ∈ {1, 2, 3}
5STEP 5

Final count

Take the 16 divisors, drop the 3 bases that are too small, and 13 bases survive — answer (C).

16 - 3 = 13 → (C) 13
Answer
13
Spot-check the endpoints of the valid list. Base 5: 2013 = 5 · 402 + 3, remainder 3 — ends in 3. Base 2010: 2013 = 2010 · 1 + 3, so (2013)₁₀ = (1 3)₂₀₁₀ — ends in 3. A non-divisor like b = 4 gives 2013 = 4 · 503 + 1, remainder 1, correctly excluded. The three thrown-out divisors 1, 2, 3 are exactly the ones where digit 3 is illegal, so removing 3 from 16 is right. The count 13 matches (C), and it sensibly sits between the raw divisor count 16 and the smaller distractors.
💡Key takeaway

The last digit of a number in base b is just the remainder after dividing by b; "ends in 3" means b divides 2013 - 3 = 2010, and 2010 has 16 divisors, so tossing out the bases 1, 2, 3 (too small to hold a digit 3) leaves (C) 13.

  • Last digit is a remainder
  • Turn it into divisibility
  • Count the divisors of 2010
  • Drop bases too small for digit 3
  • Final count

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