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

Remember to check in on Thursday for the answer and explanation!

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Answer should be ‘A’.

Here we have to find the longest dowel length (least value), take first 3 prime number

i.e, 2,3,5 (but we can’t make any triangle with this), so next value would be 7

here we can make triangle by 2+3,5,7 or 3,2+5,7.

Hence the longest dowel length would be 7, which is least as well.

It seems that you can only make two triangles that way :). The answer and explanation are up now.

If I understand correctly, there are N numbers of total dowels and each side can contain N number of dowels, couldn’t you make any length side by combining any number of 3cm and 2cm dowels? Why would you need a dowel longer than 3cm?

Okay answer is 11 . The number of distinct triangles are :

P.S. the twist is that to make a side of length 13 , we will use 11+2 and not 13 itself.

Here are triangles.

2,8,7

3,5,7

5,5,7

3,7,7

5,7,11

7,7,11

5,9,11

8,7,11

5,10,11

7,10,11

8,9,11

9,3,11

2,10,11

2,12,11

5,12,11

5,5,11

14,5,11

14,7,11

3,12,11

3,12,13

8,7,13

14,5,13

8,11,13

5,10,13

Now if anyone is wondering that how come sides are not primes , the sides have been made by adding primes . And no prime is used twice .

If you want I can provide with complete solution.

Answer is therefore 11. What say Chris.

B

With 1 2 3 and 7 I could manage only 19 distinct triangles (considering the fact that sum of any two sides should be greater than the third side)

…obviously, the answer is B (11). Can’t anyone see that?!!!

If Teddy is allowed to form a triangle without the constraint that the ends of the dowel

should form the vertices of the triangle being formed, then:

The answer is A.

This is because once the conditions to form a triangle is met, Teddy can form several

distinct triangles using 3,5,7. If the base is 5 and one of the sides is 3, the length of the

third side can be anything between (3 and 7). Therefore using only 3 distinct dowels(as

many as the sides of a triangle), Teddy can form 24 distinct triangles.

This may be a little out of the box thinking. May be this time I am correct ๐

I could count only 14 triangles (not equilateral, not isosceles) with prime numbers up to 19. So the answer has to be 23. This took me too long, so even if the answer happens to be correct, the approach is probably wrong.

Also, i could not decipher the significance of saying “as many dowels as is necessary…”

There is definitely some profound chicanery, so to say, that I am overlooking. ๐

There is definitely some “profound chicanery” afoot :). And you are right that the clue is “as many dowels as is necessary for each triangle”. Hint: 2 + 3 = 5.

Ahha! So a dowel with length 5 can be replaced with 3+2.

17=11+5+1 and so on. These replacements are of no use since all dowels have a distinct length. So this will only further reduce the number of valid triangles.

Though 23 is not enough to form the required number of triangles, we can use the length 29(=19+7+3) too.

{29, 23,11}, {29,17,11}. {29,13,11} are the new triangles that can be formed

So the answer is 23?

Hi Aditya,

Just put up the explanation today. I hope that helps illuminate things ๐

7 is the answer according to me.

The sides of triangle must follow the law that the third side, say c, must lie betwwen the sum and differences of the third side i.e.

a-b <c< a+b

Given that sides are prime , 3 , 5 , and 7 only satisfies the rule.

Hence 7.

What about the higher primes? You’ll need to be able to make 24 distinct triangles :).

4 distinct dowels are enough to form 24 distinct triangles.

First 4 prime numbers are 2,3,5,7.

However, 2 3 5 does not meet the conditions for forming a triangle. Lets take the next prime number, 11.

3 7 11 too does not meet the conditions for forming a triangle. Lets include 13, next prime number.

Now our set is {3,5,7,11,13}

3,5,7; 5,7,11; 7,11,13; forms 18 distinct triangles.

3,7,13; 5,7,13 too does not meet the conditions for forming a triangle.

We need next prime number 17, in the set as well.

Our new set is {3,5,7,11,13,17}

3,5,7; 5,7,11; 7,11,13; 11,13,17. will form 24 distinct triangles.

Therefore the answer is D.

There is a subtle twist in there :). Hint: “as many dowels as is necessary for each triangle”.

Ans C 13

Close :).

Assuming triangle with sides 11, 7, 5 is identical to 11, 7, 2+3. Option C – 13.

Close. You are the only one to get the twist. But not quite right ๐

I took 23 as test case as it is the maximum :

prime numbers till 23 are 2,3,5,7,11,13,17,19,23

the sets of triangles that can be made are :

3,5,7

3,11,13

3,17,19

5,7,11

5,11,13

5,13,17

5,17,19

5,19,23

7,11,13

7,13,17

7,17,19

7,17,23

11,13,17

11,13,19

11,17,19

11,19,23

11,17,23

13,17,19

13,17,23

13,19,23

17,19,23

these are only 21, with the maximum option the possible triangles are coming out to be 21.

what did I miss?

There is a twist :). The wording “as many dowels as is necessary for each triangle” is the key.

Is the answer D?

Nope, there is a little twist that nobody has really picked up on yet ๐