Bunuel wrote:
Teddy has a set of dowels, each with a distinct length in centimeters, represented by a prime number. If Teddy can create twenty-three distinct triangles using as many dowels as is necessary for each triangle, what is the least possible value of the longest dowel?
(A) 7
(B) 11
(C) 13
(D) 17
(E) 23
If 7 is the longest dowel, then Teddy can have at most dowels of length 2, 3, 5 and 7. The only triangles that can be formed using dowels of length 2, 3, 5 and 7 are: {3, 5, 7}, {2+3, 5, 7}, {3, 5 + 2, 7} and {2, 7, 5 + 3}. If 7 is the longest dowel, there’s no way we can construct 23 distinct triangles.
If we add 11 to the list as the longest dowel, then we can choose any 3 numbers among 3, 5, 7, and 11 to construct a triangle, for a total of 4C3 = 4 triangles. Notice that we cannot add 2 to this list because any combination of 2 and other two numbers in the set will not form a triangle. For example, {2, 3, 5} is not a triangle since 2 + 3 = 5, and {2, 7, 11} is not a triangle since 2 + 7 = 9 (recall that the sum of the lengths of the shortest sides of a triangle has to be greater than the length of the longest side).
In addition to the four triangles formed by only three dowels, there are also triangles formed using more than three dowels; such as: {2 + 3, 5, 7}, {3, 5 + 2, 7}, which are formed using four dowels; or {2 + 5, 3 + 7, 11}, {2 + 7, 3 + 5, 11}, which are formed using five dowels. It will be tricky to list all triangles and show that the number of possible triangles is exactly 23; therefore we will instead show that if we add the next prime to the list, there will be too many triangles.
If Teddy has dowels of lengths 2, 3, 5, 7, 11 and 13, then we can choose three dowels from any one of the dowels excluding 2 (by the same reason discussed above), and there are 5C3 = 10 triangles that can be formed using just three dowels. Here are 14 more triangles (among the many more that are possible) that can be formed using the provided dowels:
{2 + 3, 5, 7}, {3, 2 + 5, 7}, {2, 3 + 5, 7}, {2 + 3, 7, 11}, {7, 2 + 5, 11}, {7, 2 + 5, 13}, {3, 2 + 7, 11}, {7, 3 + 5, 11}, {7, 3 + 5, 13}, {5, 2 + 7, 11}, {5, 2 + 7, 13}, {5, 3 + 7, 11}, {5, 3 + 7, 13}, {2, 11, 5 + 7}
As we have shown that at least 24 triangles can be constructed using the numbers 2, 3, 5, 7, 11 and 13; we know that the answer is B.
Answer: B
If he is creating first triangle with (3,5,7) then he can create next triangle with (6,10,14) and so on. So the minimum dowels he will need is 3, out of which 7 is the longest.