AMC 8 Daily Practice Round 9

Complete problem set with solutions and individual problem pages

Problem 25 Medium

How many four-digit numbers \overline{abcd} can be formed using the digits 1, 2, 3, 4 (digits may be repeated) such that they satisfy the condition:   a + c = b + d?

  • A.

    36

  • B.

    40

  • C.

    44

  • D.

    48

  • E.

    52

Answer:C

We classify the four-digit numbers into four cases:

① All four digits are the same. Possible numbers: 1111, 2222, 3333, 4444

Total: 4 numbers.

② Two different digits appear twice each. For example, numbers like 1221, 1122, 2112, 2211.

We choose two different digits from \{1,2,3,4\}, which can be done in \binom{4}{2} = 6 ways.

Each choice allows 4 different digit arrangements, so this case contributes:      6 \times 4 = 24 \text{ numbers.}

③ Three different digits appear, with one digit repeated. Examples include 1232, 3212, 2123, 2321.

The valid digit sum conditions are 1+3 = 2+2 and 2+4 = 3+3, giving 2 valid digit sets.

Each set allows 4 different digit arrangements, so this case contributes: 2 \times 4 = 8 \text{ numbers.}

④ All four digits are different. Valid numbers include 1243, 1342, 4213, 4312, 2134, 2431, 3124, 3421, totaling:      8 \text{ numbers.}

Adding all cases together:   4 + 24 + 8 + 8 = 44.

The answer is \text{C}.