AMC 10 · 2009 · #19
Grade 5 countingPick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Reframe: correct means no digit is 1
The clock spoils only the digit 1, so the display is right exactly when not a single digit in it is a 1.
If the machine only ever mangles the digit 1, then avoiding every 1 guarantees a perfect display.
4.NBT.A.2Change Focus Count The ComplementCount the correct hours
Hours 1, 10, 11, 12 contain a 1, so only 2 through 9 work: 8 good hours out of 12, a fraction of .
Just cross out every hour that has a 1 somewhere and count what survives.
4.NF.A.1Make A Systematic ListCount the correct minutes
Tens digit 0,2,3,4,5 (5 ways) times ones digit 0,2,…,9 (9 ways) gives 45 good minutes out of 60, a fraction of .
Every allowed tens digit can sit next to every allowed ones digit, so multiply the choices.
Every allowed tens digit can sit next to every allowed ones digit, so the choices multiply.
▸ Why?
The two places are chosen without regard to each other, so every combination occurs exactly once.
▸ Why?
A number is its digits sitting in fixed places, so the places can be counted separately.
Combine the two subproblems
Hour and minute must both be good, and the same sits inside every hour, so — choice (A).
Correct hour and correct minute must both happen, so the fractions multiply.
5.NF.B.4Identify SubproblemsSplit 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