AMC 10 · 2013 · #18

Grade 6 counting
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
digit-sumplace-valuesystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: digit-sumplace-value
📏 Medium solution 💡 2 insights
Problem
A four-digit number has the special property that its units digit equals the sum of its other three digits (for example 2013, since 2+0+1=3). The numbers under consideration are the whole numbers strictly between 1000 and 2013. How many of them have this property?

Pick an answer.

(A)
33
(B)
34
(C)
45
(D)
46
(E)
58

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

How to solve
Strategy Make a Systematic List

Name the four digits so the property becomes one clean equation, then note the units digit can be at most 9 to turn it into a size limit on the other digits. The thousands digit is only 1 or 2, so I split into two subproblems and count each with an organized list: for thousands digit 1 the job is counting digit pairs whose sum is small enough, and for thousands digit 2 only the numbers 2000-2012 are in range, so I just check them.

1STEP 1

Name the four digits

Call the digits a, b, c, d, so the rule reads d = a + b + c — and since we stay in 1001 to 2012, a is 1 or 2.

abcd, d = a + b + c, a ∈ {1, 2}
2STEP 2

The units digit caps the sum

A single digit tops out at 9, so d = a + b + c pins the other three digits to a sum of at most 9.

d ≤ 9 → a + b + c ≤ 9
3STEP 3

Count the case a = 1

With a = 1, b + c is at most 8; grouping the pairs by their sum gives 1 + 2 + ... + 9 = 45 numbers.

Σ_s=0⁸(s+1) = 1+2+…+9 = 45
4STEP 4

Count the case a = 2

With a = 2 the range is only 2000 to 2012, where d = 2 + c; checking them leaves exactly one hit, 2002.

2000-2012: d = 2 + c → 2002 only
5STEP 5

Add the two cases

The two cases never overlap, so 45 from the 1000s plus 1 from the 2000s gives 46, choice (D).

45 + 1 = 46
Answer
46
The 1000s case alone gives 45, and the 2000s add just one more (2002), landing at 46, which matches choice (D). Choice (C) 45 is the trap for anyone who forgets the lone 2002; note that 2013 itself is excluded because we need numbers strictly below 2013, so it never gets counted. The count 45 = 1+2+...+9 also makes sense: as the hundreds digit drops from 8 down to 0, each step allows one more valid tens digit.
💡Key takeaway

Since the units digit maxes out at 9, the other three digits must add to 9 or less; count those digit pairs for the 1000s (45 of them) and add the single 2000s number 2002 to get 46.

  • Name the four digits
  • The units digit caps the sum
  • Count the case a = 1
  • Count the case a = 2
  • Add the two cases

More like this

Same archetype — closest grade level first.