The integers 2, 2, 5, 5, 8, and 9 are written on six cards, as shown. Any number of the six cards is chosen, and the sum of the integers on these cards is determined. Note that the integers 1 and 30 cannot be obtained as sums in this way. How many of the integers from 1 to 31 cannot be obtained as sums?
(A) 4
(D) 10
(B) 22
(E) 6
(C) 8

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-093.jpg?height=331&width=359&top_left_y=1518&top_left_x=1379)
The integers 2, 2, 5, 5, 8, and 9 are written on six cards, as shown. Any number of the six cards is chosen, and the sum of the integers on these cards is determined. Note that the integers 1 and 30 cannot be obtained as sums in this way. How many of the integers from 1 to 31 cannot be obtained as sums?
(A) 4
(B) 22
(C) 8
(D) 10
(E) 6

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-415.jpg?height=328&width=353&top_left_y=720&top_left_x=1401)

## Solution

First, we observe that the sum of the digits on all 6 cards is 31.

Next, we see that if we cannot get a sum of $S$, then we cannot get a sum of $31-S$. This is an important point. If we could get $31-S$, then we could take the cards not used and their digits would add to $S$. Simply stated, our inability to get $S$ means that we would be unable to get $31-S$.

So we need to check which sums from 1 to 15 cannot be obtained, and then double this total number. Checking possibilities, we see that we can get 2, 4, 5, 7, 8, 9, 10,11,12, 13, 14, 15 but cannot get 1 , 3 , 6. From above, we thus cannot get $31-1=30,31-3=28$ or $31-6=25$. So there are 6 sums that cannot be obtained.

ANSWER: (E)