A nine digit number given as “89x84620y” is divisible by 8 as well as 9. Find the value of √2xy. [x > 0 and y > 0]

To solve this problem, we need to understand the divisibility rules of 8 and 9.

1. A number is divisible by 8 if the last three digits of the number form a number that is divisible by 8.
2. A number is divisible by 9 if the sum of its digits is divisible by 9.

Let's apply these rules to the given number 89x84620y.

1. For the number to be divisible by 8, the last three digits 20y must be divisible by 8. The possible values of y that make 20y divisible by 8 are 0, 4. But y > 0, so y = 4.

2. For the number to be divisible by 9, the sum of its digits must be divisible by 9. So, 8 + 9 + x + 8 + 4 + 6 + 2 + 0 + 4 = 41 + x must be divisible by 9. The smallest number greater than 41 that is divisible by 9 is 45, so x = 45 - 41 = 4.

So, the values of x and y are 4 and 4 respectively.

Now, we need to find the value of √2xy. Substituting the values of x and y, we get √244 = √64 = 8.