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 \(0+1+4+5+6+9=25\)
Here are links, if you want clarification regarding concepts used in the problem.