All the buzzkill for just one number

I have a $$10$$ digit number in mind, you can obtain the digits of the number from left to right by following the five points below.

$$\bullet$$ The $$3^{\text{rd}}$$ smallest positive integer which has square root of the sum of all its quadratic residues equal to $$\frac{1}{3}$$ of the integer itself, gives first $$2$$ digits of my number.

$$\bullet$$ If $$489$$ people are sitting around a table, everyone given numbers $$1$$ to $$489$$ clockwise (counting starts from number $$1$$) and then every $$13^{\text{th}}$$ person is asked to get out. The only remaining person's number gives next $$3$$ digits of my number. In other words, find $$\text{josephus}(489,13)$$.

$$\bullet$$ The $$7^{\text{th}}$$ largest positive integer which has, out of all positive integers less than itself, exactly $$60$$ coprime to itself, gives next $$2$$ digits of my number. In other words, find the seventh largest $$n$$ such that $$\text{phi}(n)=60$$

$$\bullet$$ The only prime number which has exactly $$27$$ quadratic residues gives next $$2$$ digits of my number

$$\bullet$$ The last digit of my number is $$0$$

What is my number?

Details and assumptions:

$$\bigotimes$$ This problem May seem boring because it's long, but will feel awesome after solving. Everything in this problem has a meaning, nothing is a troll.

$$\bigotimes$$ Quadratic residues of a number $$n$$ are the possible remainders when square of any integer is divided by $$n$$.

For example, any square integer can give remainder only from $$(0,1,4,5,6,9)$$ when divided by $$10$$, so $$10$$ has six quadratic residues and their sum is $$0+1+4+5+6+9=25$$

$$\bigotimes$$ Here are links, if you want clarification regarding concepts used in the problem.

Concept of many problems in one, adapted from Pi Han Goh's Buzzkill

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