Problem 663

If Lucas chooses 1 number only, there are 8 possibilities for the sum, namely the 8 numbers themselves: 2, 5, 7, 12, 19, 31, 50, 81.

To count the number of additional sums to be included when Lucas chooses two numbers, we make a table, adding the number on left to the number on top when it is less than the number on top (we don’t need to add the numbers in both directions or a number to itself): $$\begin{array}{c|cccccccc} + & 2 & 5 & 7 & 12 & 19 & 31 & 50 & 81 \\ \hline 2 & & 7 & 9 & 14 & 21 & 33 & 52 & 83 \\ 5 & & & 12 & 17 & 24 & 36 & 55 & 86 \\ 7 & & & & 19 & 26 & 38 & 57 & 88 \\ 12 & & & & & 31 & 43 & 62 & 93 \\ 19 & & & & & & 50 & 69 & 100 \\ 31 & & & & & & & 81 & 112 \\ 50 & & & & & & & & 131 \\ 81 \end{array}$$ We note two things from this table. First, any time we add two consecutive numbers from the original list who sum is not too large, we obtain another number in the list. We do not include this as a new sum as these are already accounted for as sums of 1 number only. Second, the remaining sums are all distinct and there are 20 additional sums that are less than or equal to 100.

Lastly, we consider sums formed by three numbers from the list.

The fact that the sum of any two consecutive numbers from the list equals the next number in the list becomes very important in this case.

If the three numbers chosen are three consecutive numbers in the list and their sum is not too large, then their sum is actually equal to the sum of two numbers from the list. This is because the largest two of the three numbers can be combined into one yet larger number from the list.

For example, $5+7+12 = 5+(7+12) = 5+19$, which is already counted above.

If any two of the three numbers chosen are consecutive in the list, the same thing happens. For example, $12+19+50 = (12+19) + 50 = 31+ 50$ and $2+31+50 = 2+(31+50) = 2+ 81$.

Therefore, any additional sums that are created must come from three numbers, no two of which are consecutive.

We count these cases individually and sequentially, knowing that we are only interested in the sums less than 100 and remembering that we cannot include consecutive numbers from the list:

$2+7+19=28$; $2+7+31=40$; $2+7+50=59$; $2+7+81=90$
$2+12+31=45$; $2+12+50=64$; $2+12+81=95$
$2+19+50=71$
$5+12+31=48$; $5+12+50=67$; $5+12+81=98$
$5+19+50=74$
$7+19+50=76$

Every other combination of 3 integers from the list either includes 2 consecutive numbers (and so has been counted already) or includes both 81 and one of 31 and 19 (and so is too large).

In this case, there are 13 additional sums.

Putting the three cases together, there are $8+20+13=41$ different sums less than or equal to 100.