Teddy has a set of dowels, each with a distinct length in centimeters, : Problem Solving (PS)

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01 Oct 2017, 03:04

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i tried to follow the pattern

1,3,5,7--- 4c3 out of 4 only one is valid triangle with sides 3,5,7
1,3,5,7,11 5c3 out of 10 only 2 are valid triangles 3,5,7 5,7,11
1,3,5,7,11,13 out of 6c3 only 4 valid tangles 3,5,7 5,7,11 3,11,13 5,11,13

if we follow this pattren

with 4 prime nos we get one valid trangle

4---1
5---2
6---4
7---8
8---16 here at when one no is 23 we get 16 valid trangle
now at 9---(means 29) we get 32 valid trangles , which should be the ans and not an option..

BUNUEL please suggest..

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09 Oct 2017, 12:14

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Bunuel

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

Say, we take 2,3,5 and 7.
Then we can create 4C3 + 4*3 (two times 2 and other 3 numbers+two times 3 and other 3 numbers ...) = 4+12 =16

If we take 2,3,5,7,11.
Then we can create 5C3 + 5*4 = 10+20= 30. Which is sufficient.

Thus answer is (B) 11.

Correct me if I am wrong.

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10 Nov 2017, 09:39

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VeritasPrepKarishma please share an approach for this.

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11 Nov 2017, 20:37

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HI

4C3 + 4*3 , why hav u included 4*3 here...
even I m getting 4C3, but what is this 4*3?

anirudsuhag.25

Bunuel

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

Say, we take 2,3,5 and 7.
Then we can create 4C3 + 4*3 (two times 2 and other 3 numbers+two times 3 and other 3 numbers ...) = 4+12 =16

If we take 2,3,5,7,11.
Then we can create 5C3 + 5*4 = 10+20= 30. Which is sufficient.

Thus answer is (B) 11.

Correct me if I am wrong.

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26 Jun 2019, 08:40

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chetan2u mikemcgarry

I have no idea where to begin! Please explain..

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08 Jul 2019, 20:29

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@Bunnel : if you may please help here with the solution

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10 Jul 2019, 10:54

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Take the nos 2,3,5 & 7.

No of equilateral triangles= 4
No of isosceles triangles* = 11
No of triangles having 3 different sides = 4

Total = 19 triangles.

So next higher number will be the right answer ie 11.

* Hint for isosceles triangles - sum of two sides to be greater than third side . therefore 2 will make one triangle with 3 as third side
3 will make two triangles with 2 and 5 ( 7 not possible for the same reason)
5 will make 4 triangles and 7 will make 4 triangles

Total =11 triangles

Took me 15 mins to come at a logical solution. If some one has something better and simple please advise.

Regards

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04 Jan 2021, 20:03

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ScottTargetTestPrep

Bunuel

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

Shouldn't the answer be 7?
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.
Please let me know if this is a correct reasoning.

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10 Jan 2021, 17:39

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RahulSha

ScottTargetTestPrep

Bunuel

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

Shouldn't the answer be 7?
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.
Please let me know if this is a correct reasoning.

Solution:

If he has dowels only of lengths 3, 5, and 7, then he can form only one triangle, namely the triangle with sides 3, 5, and 7. I think you are misinterpreting the question; we are not told that the prime factors of the side lengths must contain 3, 5, and 7. Notice that we are told that Teddy can create exactly 23 triangles; if he were allowed to create the triangles with side lengths of 6,10, 14, and so on, he would be able to create infinitely many triangles, not just 23.

The question is instead saying this: Teddy has a number of wooden dowels, each with a different length that is a prime number. He can join two or more dowels to use as a side of the triangle (because we are told that “he can use as many dowels as is necessary for each triangle”). We are also told that the number of distinct triangles that he can form is 23. Under these restrictions, we need to determine how small the largest prime can be.

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18 Apr 2022, 06:39

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ScottTargetTestPrep

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).

We can only create 2 triangles by choosing 3 numbers among 3, 5, 7, and 11 because {3, 5, 11} and {3, 7, 11} are invalid for reasons mentioned above.

ScottTargetTestPrep

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.

This statement is again flawed for the above-mentioned reason.

Teddy has a set of dowels, each with a distinct length in centimeters,

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