# 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.

×