All about 2310

Say $a+b=2310=2\cdot3\cdot5\cdot7\cdot11$

Consider any one of the prime factors $p$ of $2310$ .

$p$ divides $a$ if and only if it divides $b=2310-a$ . So for $2310$ to divide the product $ab$ , every prime factor of $2310$ must also divide $a$ . But the smallest such positive number is $2310$ itself! So no such pair $a,b$ exists.

Note that the "positive" condition is needed, otherwise the pair $0,2310$ would work.

Let $a,b$ be positive integers such that $a+b=2310$ and $ab=2310k$ for some positive integer $k$ . Also keep in mind that $2310= 2\times 3\times 5\times 7\times 11.$

Combining the two equations, we get $a+\frac{2310k}{a}=2310\implies a^2-2310a+2310k=0$ .

Note that this is a quadratic with variable $a$ . Thus the discriminant $\Delta =2310^2-4(2310k)$ is a perfect square as $a$ is a positive integer.

Since $2310^2-4(2310k)=2^2\times 3\times 5\times 7\times 11\times (1155-2k),$ it follows that $1155-2k=3\times 5\times 7\times 11\times m^2=1155m^2$ for some positive integer $m.$

But $1155m^2\geq 1155>1155-2k$ which is a contradiction. So $ab$ is not divisible by 2310.

Let $x,y \in \mathbb{Z} ^{+}$ we have $x + y= 2310$ We notice that $2310$ is divisible by 2, since it's last digit is divisible by 2. Thus we know that both $x \text{ and } y$ are even or they are both odd. If they both odd their product is odd (we know $2310$ is even!), so then they must both be even, but then there exists some $j, k \in \mathbb{Z}^{+}$ such that $x= 2j$ and $y=2k$ Thus the following holds $x y = 4jk$ And so then we require $xy$ to be divisible by 4, but we note half of $2310$ is a number that ends with 5 and let's call that number $p$ , which clearly is not divisble by 2 (it is odd). We have $\frac{xy}{2310} = \frac{4jk}{2p}= \frac{2jk}{p}$

Then this fraction is an even number divided by an odd, and gives a remainder. Thus we conclude that $2310$ does not divide $xy$

Thus such an $x,y$ must not exist. $\blacksquare$

A generic square equation has the form:

(x-a)(x-b)=0

that multipied gives

x^2 - (a+b) x + ab =0

if a+b and ab are both 2310 they would both solve

x^2 -2310 x + 2310 =0

This ordinary second grade equation has only two solutions, and they are:

x = (3310 + Sqrt[3310^2 - 4 * 2310]) / 2 and x = (3310 - Sqrt[3310^2 - 4 * 2310]) / 2

x = 1.00043327564459609367424630743, x = 2308.99956672435540390632575369

And they are not integers. No other solutions are possible.

Unit place digit of the sum of the numbers a and b is 0. So unit place digits of (a, b) can be (1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1). None of the unit place digits of ab is thus 0. So ab is never divisible by 2310

Call the two positive integers M and N. Since the 2310 is even, both M and N must be odd, meaning their product must be odd. Odd numbers cannot be divided be even numbers.