## $ABC$ is an integer sided triangle with incenter $I$ and perimeter $p$.

The segments $IA$, $IB$ and $IC$ have integral length as well.

Let $L = p + |IA| + |IB| + |IC|$.

Let $S(P) = \sum L$ for all such triangles where $p \le P$. For example, $S(10^3) = 3619$.

Find $S(10^7)$.

### To find $M(137)$, we need to calculate the harmonic numbers and check the divisibility of their denominators by $137$ until we find the largest value of $n$ where the denominator is not divisible by $137$.

We can solve this problem by using Python. Below is the implementation:

“`python

from fractions import Fraction

def gcd(a, b):

# Function to calculate the greatest common divisor

while b:

a, b = b, a % b

return a

def M(p):

n = 1 # Start with the first harmonic number

while True:

# Calculate the denominator of the current harmonic number

denominator = 1

for k in range(1, n+1):

denominator *= k

# Check if the denominator is divisible by p

if denominator % p != 0:

prev_denominator = denominator

else:

return n-1 if prev_denominator is None else n-2

n += 1

# Find M(137)

result = M(137)

print(result)

“`

The code begins by defining a function `gcd` to calculate the greatest common divisor. This will be used later to find the reduced fraction.

The `M` function takes an input `p` and iterates through the harmonic numbers until it finds a denominator that is not divisible by `p`. It initializes `n` to `1` and enters an infinite loop (`while True`).

Inside the loop, the code calculates the denominator of the current harmonic number by multiplying all the positive integers up to `n`. It then checks if the denominator of the current harmonic number is divisible by `p`. If it is not, it stores the previous denominator in a variable `prev_denominator` and continues to the next iteration. If it is divisible by `p`, it returns `n-1` if `prev_denominator` is `None` (indicating that this is the first harmonic number), or `n-2` (indicating the largest value of `n` where the denominator is not divisible by `p`).

Finally, the code calls the `M` function with `137` as the input and prints the result.

Executing this code will provide the value of `M(137)`, which is the largest value of `n` such that the denominator of the harmonic number is not divisible by `137`.

##### More Answers:

Roots on the RiseThe Last Question

Chef Showdown