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2020 AMC 10A Problem 6

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2020 AMC 10A Problem 6AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Counting · difficulty 2 of 5 · about 2 min · position P1-10

How many 44 -digit positive integers (that is, integers between 10001000 and 99999999 , inclusive) having only even digits are divisible by 5?5?

  1. A)

    8080

  2. B)

    100100

  3. C)

    125125

  4. D)

    200200

  5. E)

    500500

Answer

B

Key insight

Divisible by 5 with an even last digit forces the units digit to be 0; the thousands digit has 4 even choices and each middle digit 5.

Solution

A number is divisible by 55 exactly when its last digit is 00 or 55. Since every digit must be even, the last digit is forced to be 00: one choice.

The thousands digit must be even and nonzero: 2,4,6,82, 4, 6, 8, four choices.

The hundreds and tens digits may be any even digit 0,2,4,6,80, 2, 4, 6, 8: five choices each.

By the multiplication principle the count is
4⋅5⋅5⋅1=100.4 \cdot 5 \cdot 5 \cdot 1 = 100.

The answer is (B) 100\boxed{\textbf{(B)}\ 100}.

Why this works

Digit-counting problems are products of independent choices once each position's restriction is identified. Handle the most restrictive positions first (the last digit here is completely forced), and remember that the leading digit of a 44-digit number is never 00.

The trap

Allowing 0 as the leading digit (5 choices, giving 125) or allowing 5 as the last digit (200).

Common mistakes

  • Allowing 0 as the leading digit (5 choices, giving 125) or allowing 5 as the last digit (200).
  • Using 44 choices for every digit (excluding 00 everywhere), which gives 6464 or 8080-type answers.

Techniques

Set up the equation/formula and compute; no special trick needed