AMC 10 Step by Step

2008 AMC 10B Problem 1

P1-10~1 minPrint

A basketball player made 55 baskets during a game. Each basket was worth either 22 or 33 points. How many different numbers could represent the total points scored by the player?

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Key insight

Each time a 2-point basket becomes a 3-point basket, the total goes up by exactly 1. Six choices for the number of 3-pointers give six totals.

Solution

The only thing that changes the total is how many of the 55 baskets are 33-pointers. So list every possibility.

  1. The number of 33-pointers can be 0,1,2,3,40, 1, 2, 3, 4 or 55. The rest are 22-pointers.
  2. Work out each total:
33-pointers22-pointersTotal
00550+10=100 + 10 = 10
11443+8=113 + 8 = 11
22336+6=126 + 6 = 12
33229+4=139 + 4 = 13
441112+2=1412 + 2 = 14
550015+0=1515 + 0 = 15
  1. The six totals 10,11,12,13,14,1510, 11, 12, 13, 14, 15 are all different.

So there are 66 possible totals.

The answer is (E) 6\boxed{\textbf{(E)}\ 6}.

Why this works

Trading one 22-point basket for a 33-point basket adds exactly 11 point. So the totals climb by one at a time, from 1010 (all twos) to 1515 (all threes), and no total repeats. When a question asks "how many different totals", pick one count to vary and list the cases; remember to include both ends, zero and all.

Concise solution

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

Let kk be the number of 33-point baskets, so the other 5−k5 - k baskets are worth 22 points each. The total is
3k+2(5−k)=10+k.3k + 2(5 - k) = 10 + k.
Since kk can be 0,1,2,3,40, 1, 2, 3, 4 or 55, the possible totals are 10,11,12,13,14,1510, 11, 12, 13, 14, 15: six different values, each achieved by exactly one kk.

The answer is (E) 6\boxed{\textbf{(E)}\ 6}.

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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.