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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