Sum it up!

Notice that $T = \{5+3k : k \in \{0, 1 \dots 15\}\}$ so we might as well consider $T^* = \{0 \dots 15\}$ for simplicity. The minimum sum is then $0+1+2=3$ and the maximum is $13+14+15 = 42$ . All that remains is to show that every number in between can be reached too. Let $S$ be some sum. If $3 \leq S \leq 16$ then $S = 0 + 1 + (S-1)$ . If $16 < S \leq 29$ then $S = 0 + 15 + (S-15)$ . If $29 < S \leq 42$ then $S = 15 + 14 + (S-29)$ . It is trivial to verify that all terms are within the right bounds and distinct. Finally, there are $40$ integers in $\{3 \dots 42\}$ and we are done.

First realize that each number in $T$ is of the form $2+3n$ where n ranges from 1 to 16. Thus this problem can be simplified to "how many unique sums can be made with three different numbers from the set $\{1,2,3,...15,16\}$ "

The minimum sum is $1+2+3 = 6$ and the maximum sum is $14+15+16 = 45$ . Every sum in between is possible (since we can always change one of the three numbers in the sum by one) and thus by taking the difference $45 - 6 = 39$ and realizing that when counting this difference is off by one, we get that the number of unique sums is $39 + 1 = 40$ which is the answer.

We can rewrite $T= \{ 5+3 i|i=0,1 \ldots 15 \}$ .

Sums of three different numbers from $T$ are equal with $5+3( i+j+k)$ where $0 \leq i,j,k \leq 15$ are different. So we have to evaluate all possible values for $i+j+k$ .

Due to fact that summation is commutative we can consider only sums $i+j+k$ with $i<j<k$ .

If $i$ and $j$ are fixed then possible values for $k$ are $j+1,j+2, \ldots ,15$ that is all possible values for $i+j+k$ are $\{ i+2j+1, \ldots ,i+j+15 \}$

Now if we fix $i$ the possible values for $j$ are $\{ i+1, \ldots ,14 \}$

So for fixed $i$ we have to make the following union

$\{ i+2(i+1)+1, \ldots ,i+(i+1)+15 \} \cup$

$\{ i+2(i+2)+1, \ldots ,i+(i+2)+15 \} \cup$

$\ldots \cup$

$\{ i+2(14)+1, \ldots ,i+(14)+15 \}$

which results in $\{ 3i+3, \ldots ,i+2 \times 15-1 \}$

$i$ takes values from the set $0,1, \ldots ,15-2 \}$

Then the final result is the union from below

$\{ 3 \times 0+3, \ldots ,0+2 \times 15-1 \} \cup$

$\{ 3 \times 1+3, \ldots ,1+2 \times 15-1 \} \cup$

$\ldots$

$\{ 3 \times (15-2)+3, \ldots ,(15-2)+2 \times 15-1 \} \cup$

which is equal with $\{ 3,4, \ldots ,3 \times 15 -3 \}$ that means $40$ distinct values.

The problem can be generalized for any arithmetic progression of $n$ therms with $n \geq 6$ . The proof is as previous by substituting $15$ with $n$ .

By counting, one would find that there are 16 elements in T. By assigning an index 1-16 to each one and using index sets (a,b,c), we see that any sets of three different indexes whose sums are equal (e.g. (1,2,5) and (1,3,4)) would not be distinct. Starting with the set (1,2,n), we see there are 14 possible values for n (3-16).

From here on out, we have to worry about non-distinct sets. It's easy to see that a set (a,b,c) would have the same sum as (a,b+1,c-1), (a,b-1,c+1), etc.

Since (1,2,16) would be equal to (1,3,15), we can't let c go below 16 (except when a is 1 and b is 2, which we have already accounted for). This leaves 13 values for b (3-15). On the same premise, we can't let b go below 15, leaving another 13 values for a (2-14). With a total of 14+13+13=40 sets, we have our answer.

Because we only need a count of the numbers, and not the sums themselves, we can modify the numbers as long as we modify them all equally. First subtract five from each number, then divide them all by three, giving the set {0,1,2,...15}. From this, note that 3 is the smallest sum that can be made, 0+1+2, and 42 is the biggest sum that can be made, 13+14+15. By increasing or decreasing one of the numbers in the triplets by 1, the sum can be increased or decreased by 1, as long as it doesn't exceed these bounds. This means any sum between 3 and 42, inclusive, is possible, and nothing outside is possible. There are 40 numbers between 3 and 42.

Every element of T could be written as 5 + 3(k-1) for some positive integers k. Thus, S= [ 5 + 3(k 1 - 1)] + [ 5 + 3(k 2 - 1)] + [ 5 + 3(k 3 -1)]= 6 + 3(k 1 + k 2 + k 3), where k 1 \neq k 2 \neq k_3.

24 \leq S= 6 + 3(k 1 + k 2 + k 3) \leq 141. 6 \leq X= k 1 + k 2 + k 3 \leq 45

X has 45 - 6 +1 = 40 possible values. Thus, S also has 40 distinct possible values.

All numbers are congruent to $2$ mod $3$ , $3|S$ . The minimum for $S$ is $5+8+11=24$ , while the maximum is $44+47+50=141$ . We claim all multiples of 3 between these two numbers can be obtained, which is the maximum possible answer. Let $S=11+8+x$ , where $x$ is any other term gives all multiples of 3 from $24$ to $69$ . Let $S=y+47+44$ , where $y$ is any other term gives all multiples of 3 from $96$ to $141$ . Let $S=8+50+z$ , where $z$ is $\{14, 17, ... , 35\}$ gives us multiples of 3 from $72$ to $93$ . This justifies our claim. So the answer is $\frac{141-24}{3}+1=40$ .

first we find the minimum sum of 3 numbers in T. next we find the maximum sum of 3 numbers in T. if we sort them from minimum to maximum we'll see that the difference between 2 consecutive numbers in this sorted set is 3. with having these three information above we can find the number of different sums. 24,27,.............,141 we should only find the number of the above sequence.

We notice that each number in $T$ is congruent to $2 \pmod{3}$ . Thus, the sum of three distinct integers will be congruent to $0 \pmod{3}$ . Thus, the smallest possible sum is $S_{\mbox{min}} = 5 + 8 + 11 = 24 = 3(8)$ and the largest possible sum is $S_{\mbox{max}} = 44 + 47 + 50 = 141 = 3(47)$ .

To show that each of the multiples of $3$ from $3 \times 8$ to $3 \times 47$ can be achieved, we start with $\{5, 8, 11\}$ , and add $3$ to $\{11\}$ each time till we hit $50$ . Then we add $3$ to $\{8\}$ each time till we hit $47$ . Finally, we add $3$ to $\{ 5\}$ each time till we hit $44$ . This ensures that we still have $3$ pairwise distinct elements each time.

Thus, there are $47 - 8 + 1 = 40$ distinct values of $S$ .

Lowest value = 5+8+11 = 24 ( sum of the least 3 values ) Highest value = 50+47+44 = 141 ( sum of the highest 3 values ) Smallest step = 3 So , the number of possible distinct values of S = 1 + ((141-24)/3) = 40

Triplets in this set can yield sums ranging from 1+2+3 = 6 to 14+15+16 = 45. Thus, there are 45-6+1 = 40 distinct values for S.

All the numbers are 1 (mod 3), so the sum of three distinct numbers must be a multiple of 3. The smallest possible sum is 24, or 8 times 3, and the largest is 141, or 47 times 3. All multiples of three between the high and low are all of them. So the answer is 47-8+1=40.

Simply pick 3 numbers, and the sum will be the multiple of 3. Therefore, the possible values are between the sum of the 3 smallest numbers (5+8+11) and the sum of the 3 biggest numbers (44+47+50). So, the distinct possible values of S is 24, 27, 30, 33,......., 138, 141. So, by the arithmetic theorem, there are 40 values of S.

let 5 be the U1, 8 the U2 and so on.. the formula for UN : is 2+3n U1: 2+3 1 U2 = 2+3 2 U3= 2+3 3 sum from U1 until U3 are the smallest possible value for S because it is the sum of smallest Un and U14 = 2+3 14 U15 = 2+3 15 U16 = 2+3 16 sum from U14 until U16 are the biggest possible value for S because it is the sum of biggest Un

I count the sum of the distinct possible value from T and there is (14+15+16) - (1+2+3) +1 possible value of T

+1 because the total doesn't included the last possible values of T from the U14-U16

for example from you thumb to your smallest finger there are 3 others finger but the total of a space between your fingers is 3+1 = 4

note the sum of U1, U2 and U6 is the same as U2, U3 and U4 because 2+3n + 2+6n +2+ 18n = 6+ 27n and 2+6n + 2+9n +2+ 12n = 6+ 27n both is the same and it give the same number of its sum

40 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141

This is the code I used in Wolfram Mathematica

t = Table[5 + 3 n, {n, 0, 15}]; a = Subsets[t, {3}]; b = Map[Total, a]; c = DeleteDuplicates[b] Length[c]