AMC 10 · 2009 · #10

Grade 5 counting
complementary-countingdigit-countingfraction-arithmetic identify-subproblems ↑ Prerequisites: complementary-counting
📏 Medium solution 💡 2 insights
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Problem
A clock shows one digit wrongly wherever it should appear. Find the fraction of a day on which the display happens to be correct.

Pick an answer.

(A)
$\frac 12$
(B)
$\frac 58$
(C)
$\frac 34$
(D)
$\frac 56$
(E)
$\frac {9}{10}$
How to solve
Strategy Identify Subproblems

The display has two independent parts: the hour and the minute. A time is correct only when the hour is correct AND the minute is correct, and the set of correct minutes is the same in every hour. So split the problem: find the fraction of hours with no 1, find the fraction of minutes with no 1, then multiply the two fractions.

1STEP 1

Reframe: correct means no digit is 1

Correct means that digit never appears.

display correct ⇔ no digit equals 1
2STEP 2

Count the correct hours

Counting the hours gives 2/3.

good hours = {2,3,4,5,6,7,8,9} → 8/12 = 2/3
3STEP 3

Count the correct minutes

Counting the minutes gives 3/4.

5 × 9 = 45 good minutes → 45/60 = 3/4
4STEP 4

Combine the two subproblems

The two are independent, so multiplying gives 1/2, choice (E).

2/3 × 3/4 = 6/12 = 1/2
Answer
1/2
Sanity-check with raw counts: there are 12 × 60 = 720 possible times, and 8 × 45 = 360 of them have no 1. That is exactly 360/720=1/2, matching the multiplied fractions. Half is also believable: a 1 is a fairly common digit, so it is reasonable that it spoils about half of all readings.
💡Key takeaway

Split a two-part display into its parts, find the fraction right in each part, then multiply the fractions to get the fraction right overall.

  • Reframe: correct means no digit is 1
  • Count the correct hours
  • Count the correct minutes
  • Combine the two subproblems