A Not-So-Simple Graph (Part I)

Consider a graph with vertices t 1 , t 2 , t 3 , . . . , t 15 t_{1},t_{2},t_{3},...,t_{15} . For every integer pair ( m , n ) (m,n) where 15 m > n 1 15≥m>n≥1 , there are n n distinct edges connecting t m t_{m} to t n t_{n} . No edges intersect. An ant moves along the edges of the graph. Given that it can only move from t a t_{a} to t b t_{b} if a > b a>b , the number of distinct ways it can move from t 15 t_{15} to t 1 t_{1} is A A . Find the sum of digits of A A .

By the way, check out Part II of the problem! It is a lot more challenging... But more challenging = more fun, right?


The answer is 45.

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1 solution

Sam Zhou
May 29, 2019

Let x n x_{n} denote the number of ways the ant can move from t n t_{n} to t 1 t_{1} . Clearly, x 1 = x 2 = 1 x_{1}=x_{2}=1 .

We can define x n x_{n} recursively. For every k < n k<n , the number of ways to move from t n t_{n} to t k t_{k} is k k . Therefore, we can deduce that

x n + 1 = i = 1 n i x i x_{n+1}=\sum_{i=1}^{n}ix_{i}

We also know that

x n = i = 1 n 1 i x i = x n + 1 n x n x_{n}=\sum_{i=1}^{n-1}ix_{i}=x_{n+1}-nx_{n}

So x n + 1 = ( n + 1 ) x n x_{n+1}=(n+1)x_{n}

We can see that x n = ( n ) ( n 1 ) ( n 2 ) . . . ( 3 ) x 2 = n ! 2 x_{n}=(n)(n-1)(n-2)...(3)x_{2}=\frac{n!}{2}

A = x 15 = 15 ! 2 = 653837184000 A=x_{15}=\frac{15!}{2}=653837184000 . The sum of digits of A A is therefore 45 \boxed{45} .

Very nice!

Chris Lewis - 2 years ago

Oddly enough, if you actually did 16 ! 2 = 10461394944000 \frac{16!}{2} = 10461394944000 , you also get a digit sum of 45. ;(

Samuraiwarm Tsunayoshi - 1 year, 6 months ago

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