The number of ways there are to arrange 6 distinct rooks on a 10x10 board such that none of them can attack each other is n. Find the product of the non-zero digits of n.
To find the number of ways to arrange 6 distinct rooks on a 10x10 board such that none of them can attack each other, we can use combinatorics.
First, let's consider the placement of the first rook. Since there are 100 squares on the board, there are 100 choices for the first rook.
After placing the first rook, we need to place the second rook. Since the second rook cannot be in the same row or column as the first rook, there are 90 available squares left (10 rows and 10 columns excluding the ones occupied by the first rook). Thus, there are 90 choices for the second rook.
Similarly, for the third rook, there are 80 available squares left (10 rows and 10 columns excluding the ones occupied by the first two rooks). Thus, there are 80 choices for the third rook.
Continuing this pattern, the number of choices for the fourth rook is 70 (10 rows and 10 columns excluding the ones occupied by the first three rooks), for the fifth rook is 60, and for the sixth rook is 50.
To find the total number of arrangements, we multiply the number of choices for each rook together:
Total number of arrangements = 100 × 90 × 80 × 70 × 60 × 50 = 15,120,000,000
Now, we need to find the product of the non-zero digits of this number. To do this, we can first find the prime factorization of the number:
15,120,000,000 = 2^9 × 3 × 5^7
From the prime factorization, we see that the non-zero digits are 2, 3, and 5. So the product of the non-zero digits is:
Product of non-zero digits = 2 × 3 × 5 = 30
Therefore, the product of the non-zero digits of the number of arrangements, n, is 30.