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Bunuel
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12 + 4 16 is possible, not sure if we can take multiples of each pen type or just 1 of each type

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We can’t ignore that he has a total of 4 pens!

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14 = 4+4+4+2
16 = 4+4+4+4
17=7+4+4+2
23= 12+7+2+2
21 should be the answer.
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Steve has 4 pens. The values of the pens are $2, $4, $7, and $12. Which of the following could NOT be the total value of any collection of these pens?

A. $14 -> 2+4+4+4
B. $16 ->4+4+4+4
C. $17 ->2+4+4+7
D. $21 -> not possible with 4 pens
E. $23 ->2+2+7+12

So, I think D. :)
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Possibly list all of the ways in which you can get a sum for the prices given. So, $12 and $2 makes $14. You can eliminate that answer and other answers that have a sums as answer choices. Leaving you with C. $17 as the only balance that could not be a total of the collection.



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TarunKumar1234
Steve has 4 pens. The values of the pens are $2, $4, $7, and $12. Which of the following could NOT be the total value of any collection of these pens?

A. $14 -> 2+4+4+4
B. $16 ->4+4+4+4
C. $17 ->2+4+4+7
D. $21 -> not possible with 4 pens
E. $23 ->2+2+7+12

So, I think D. :)



Same calculation performed! Moreover, 21 requires at least one 7, since it's the only odd number, and that means that the other 3 should sum to 14, clearly not possible. So I'm pretty sure the right answer is D :)
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21 is not possible
17=7+4+4+2
Answer should be D
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