arora007 wrote:This is a card game problem from Cambridge.
The even integers 2, 4, 6, and 8 can be added to get any even integer from 2 to 20 (because 2 is the smallest possible even value, and 2+4+6+8 = 20 is the largest possible even sum).
Between 2 and 20 inclusive, the number of even integers = (biggest - smallest)/2 + 1 = (20-2)/2 + 1 = 10 possible even point totals.
If we add the 1 card to each of these even point totals, we get an additional 10 possible point totals.
The 1 card by itself gives us the smallest possible point total, which is 1.
Thus, 10 + 10 + 1 = 21 point totals are possible.
The correct answer is
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