AMC 10 Step by Step

2006 AMC 10A Problem 4

P1-10~2 minPrint

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

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Key insight

Make the hour digits and the minute digits as big as possible separately. The hour 9 beats 10, 11 and 12, and the minutes can be 59.

Solution

The hour and the minutes do not affect each other, so make each one as big as possible on its own.

  1. The hour. Hours go from 11 to 1212. The one-digit hours give digit sums 11 to 99. The two-digit hours give only 1+0=11 + 0 = 1, 1+1=21 + 1 = 2 and 1+2=31 + 2 = 3. So the best hour is 99, with digit sum 99.
  2. The minutes. Minutes go from 0000 to 5959. The first digit can be at most 55 and the second at most 99, and 5959 has both. Its digit sum is 5+9=145 + 9 = 14.
  3. Put them together. The display 9:599{:}59 is a real time, and its digits add to 9+14=239 + 14 = 23. No display can beat both parts at once, so 2323 is the largest.

The answer is (E) 23\boxed{\textbf{(E)}\ 23}.

Why this works

When a total is made of separate pieces that do not affect each other, make each piece as large as it can be, then check the best pieces can happen together. Be careful that "biggest number" is not the same as "biggest digit sum": 1212 is a later hour than 99, but its digits add to only 33.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The hour part and the minute part are independent, so maximize each on its own.

Hours run from 11 to 1212. A one-digit hour contributes its own value, at most 99; the two-digit hours 1010, 1111, 1212 contribute only 11, 22, 33. So the best hour is 99, contributing 99.

Minutes run from 0000 to 5959. The tens digit is at most 55 and the ones digit at most 99, and 5959 achieves both, contributing 5+9=145 + 9 = 14.

The display 9:599{:}59 therefore has digit sum 9+14=239 + 14 = 23, and nothing larger is possible.

The answer is (E) 23\boxed{\textbf{(E)}\ 23}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2006 12A #4.