[superprismatic]

Three different numbers are chosen

at random from the integers 1 to N,

inclusive. The probability that

these could be the lengths of the

sides of a (nondegenerate) triangle

is 75/161. Find N.

[Quantum.Mechanic]

N=23 (825/1771)

[EventHorizon]

N=23 (825/1771)

because my first guess was 23 due to the prime factors of 161.

[Guest]

i have a whole in my logic somewhere and help would be appreciated by whomever

so take this from the perspective of the largest number taken A

there are summation (0 to A-1)=A(A-1)/2 ways that this number can be the biggest number and no triangle be made

for example for 5

4 with 1

3 with (1,2)

2 with (1,2,3)

1 with (1,2,3,4)

this counts the points being chosen each way second and last

notice if there are two of the biggest number there are no bad triangles so this is all the bad triangles

now you multiply by three because the largest number can be chosen in any position (but you have already accounted for the other two being chosen at different times)

now integrate for all A

N^3/2-3N^2/4

so by my logic it should be all - bad

N^3-N^3/2-3N^2/4

divide by n^3 and this should be the probability but it doesnt work

anything?

[superprismatic]

i have a whole in my logic somewhere and help would be appreciated by whomever

so take this from the perspective of the largest number taken A

there are summation (0 to A-1)=A(A-1)/2 ways that this number can be the biggest number and no triangle be made

for example for 5

4 with 1

3 with (1,2)

2 with (1,2,3)

1 with (1,2,3,4)

this counts the points being chosen each way second and last

notice if there are two of the biggest number there are no bad triangles so this is all the bad triangles

now you multiply by three because the largest number can be chosen in any position (but you have already accounted for the other two being chosen at different times)

now integrate for all A

N^3/2-3N^2/4

so by my logic it should be all - bad

N^3-N^3/2-3N^2/4

divide by n^3 and this should be the probability but it doesnt work

anything?

Although I can't follow your argument completely, it is clear that you allow 2 of the three numbers to be the same. The problem says "Three different numbers...." Perhaps that's the flaw in your logic.

[Guest]

As the first responder noted, 825/1771 which is 75/161.

We know that the total number of ways to pick the 3 integers out of N is C(N,3). Just check different values of N (via a spreadsheet) to see the first value of C(N,3) that is exactly divisible by 161. This turns out to be 23. Then apply some brute force. For N=23, just count the number of possible triplets that form triangles. Start with the largest of the three being 23, then 22 and so on. You get the sum

110+100+90+81+72+64+56+49+42+36+30+25+20+16+12+9+6+4+2+1 = 825.