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?
Answer and Explanation
Before we just start whipping out prime numbers, it is a good idea to identify what this week’s question is asking. “As many dowels as is necessary” does not translate to exactly one dowel per side. Of course, you could make a triangle with one dowel per side, but you’d have to make sure that the longest side is less than the sum of the two shortest sides.
Therefore, 2, 3, and 5 do not count.
So let’s move up in numbers, and that will give us the smallest triangle we can construct out of exactly three dowels, without violating the laws of geometry: 3 cm, 5 cm, 7 cm.
But before start just constructing triangles out of three dowels, let’s not forget that we can use as many dowels as is necessary to create a triangle. That means we can use a 2 cm and a 3 cm dowel to construct one side, and, say, a 5 cm dowel and a 7 cm to construct the other two sides, giving us an isosceles triangle with lengths of 5 cm, (2,3) cm and 7 cm.
Notice the notation I used (enclosing the centimeter lengths of the dowels in parentheses when they combine to make one side). Continuing on in this vein, I get the following:
(2,3), 5, 7
(2,5), 3, 7
(3,5), 2, 7
At this point, we have four possible triangles using the four possible dowels. The next step is to move up one prime number, by adding the 11 cm dowel.
(2,3), 7, 11
(2,5), 7, 11
(2,7), 3, 11
(3,5), 7, 11
(2,7), 5, 11
(3,7), 5, 11
(3, 7), 2, 11
(5, 7), 2, 11
(5, 7), 3, 11
That is nine more possible triangles, giving us a total of 13. At this point, it is tempting to assume that by adding the 13 cm dowel, we will be able to easily make it to 24. But that’s the thing: by adding the 11 cm dowel, it gave us more than twice the number of possibilities than when we had just the first four dowels. By adding 13 cm the complexity only increases, so we should definitely go up by more than the 11 triangles needed to get to 24 triangles.
Granted that’s kind of subtle and not a slam-dunk conviction. Nonetheless, a little more experimentation with the five dowels (and a little out of the box thinking) should lead to the insight that we can use two dowels on different sides.
(2,3), (5, 7), 11
(2,5), (3,7), 11
(2,7), (3,5), 11
(3,5), (11,2), 7
(3,7), (5,2), 11
(3,7), (2,11), 5
(2, 11), (7,5), 3
(2,7), (3,5), 11
You probably don’t even need to go that far to realize that you are almost at 24 triangles. Adding the 13 cm would surely lead to vastly more possible combinations of dowels. Therefore, the answer must be (B). For the purpose of thoroughness, however, here are two more, giving us a total of 24:
(2,3,5), 7, 11
(2,3,7), 5, 11
Now this question was slightly more involved than something on the GRE (that’s why it is a brain twister). Yet, the question is still solvable in a relatively short time if you pick up on the subtle twist of multiple dowels per side, and you realize that you don’t need to make all the possible combinations that include the 11 cm dowel to reach 24 triangles, since adding a 13 cm dowel will add far more than 24 triangles.
Addendum: I confess: I actually missed something. It’s a tiny something, given how much information was in the text–yet it makes a big difference. The triangle (2,3,5), 7, 11 is redundant with (2,5), (3,7), 11 (thanks Payal for catching this!).
So Teddy can only make a total of 23 triangles. It seems that this only affected one student, since out of all of the explanations many didn’t notice the multiple dowels per side. Out of those who did, it wasn’t a question of one or two triangles.
For those who didn’t reply but counted 23 and had (C) as the answer, I apologize. So, the question should be rephrased as Teddy being able to make 23 distinct triangles out of his “prime” assortment of dowels.