All buzzkill for just one number-Part 2

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

• If $m,n$ are positive integers and coprime to each other, if the number of integer solution(s) to the equation $m+n=a$ is 36 then the 5th smallest such $a$ will give you my first two digits.

• The sum of all the quadratic residues of 472 is 177 times $k$ , here $k$ gives the next three digit of my number.

• The 7th smallest positive integer $p$ which is 2 less than a prime number and whose sum of quadratic residue is 1 less than a prime number, then reversing digits of $p$ gives you the next 2 digits of my number.

• Let $1<p,q<1000$ , such that $p,q$ are both prime numbers with $\phi(p)$ is the sum of all quadratic residues of $q$ and if for largest $p$ , $q=10A+B$ where $A,B$ are positive integers. $A$ is the third last digit of my number and $B$ is the last digit of my number.

• The second last digit of my number is 1.

Details and assumptions :

• 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 $0+1+4+5+6+9=25$

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

• Euler totient function : $\phi$

• Quadratic Residues

You may try an easier version of the problem by Aditya Raut here