Somewhat Big Numbers

The first pair is $(x_0$ ; $y_0$ ) and I can choose $x_0$ =A and $y_0$ =B.

Calculating the next pairs we notice that $x_n$ contains the factors $A * B * (A+B) * (2A+B) * (3A+B) ... ((n-1)A+B)$

The most problematic factors of $100!$ are prime numbers: if A and B are the prime numbers we need there is no problem. If they aren't then the $x_p$ pair (where p is a prime number) can be divided by p.

We can demonstrate it using remainder classes: A can have remainder from 1 to p-1 in the division by p and when we multiplicate A by every n drom 1 to p-1 we obtain every reaminder fro 1 to p-1; the same happend to B that summed to n*A gives 0 (at least once).

In conclusion, if we want $x_n$ divided by 97 (the last prime number less than 100) n must be exactly 97.

$\Rightarrow$ N=97

I'm sorry for my bad english but I often make so big errors!

The recursive formulas defining the pairs also hold modulo every prime $p$ . Note that if $x_i$ is not zero modulo $p$ , then $\frac{y_{i+1}}{x_{i+1}}=\frac{y_i}{x_i}+1.$ Therefore, after $p-1$ or fewer steps $\frac{y_i}{x_i}$ will be zero modulo $p$ . This means that $y_i$ is zero modulo $p$ , so after one more step $x_i$ will be zero modulo $p$ . Once $x_k\equiv 0\pmod p$ , the same is true for all $x_n$ with $n\geq k$ . Therefore, for any prime $p$ , $x_k\equiv 0 \pmod p$ for all $k\geq p$ .

Consider $x_{97}$ , with $97$ being the largest prime less than $100.$ By the above, $x_{97}$ is divisible by all primes from $53$ to $97$ , that appear once in the product $100!$ . For the primes less than $50$ , notice that once $p$ divides $x_{50}$ and $y_{50}$ . From this, the order of $p$ in $x_{51}$ and $y_{51}$ is at least $3$ , the order of $p$ in $x_{52}$ and $y_{52}$ is at least $3^2$ , and so on. The order of $p$ in $x_{97}$ is at least $3^{47}$ , which is much bigger than the order of $p$ in $100!$ (the largest such order, the order of $2$ , is given by the sum $\sum \limits_{j=1}^{\infty} \lfloor \frac{100}{2^j} \rfloor$ , which is at most $\sum \limits_{j=1}^{\infty} \frac{100}{2^j} = 100$ ).

On the other hand, $x_{96}$ is not divisible by $97$ if $x_0=y_0=1$ . So the answer is $97$ .

We will first show for $n \geq 2$ that $x_n = x_0^(3^n+1)$ and $y_n =$

Then $x_{n+1}=$ and $y_{n+1}$

It is well known that $n!| \displaystyle\prod_{i=1}^{n}(a+(i-1)*d)$ for $n, a, d \in Z^+$ .

By substituting x0= y0 = 1, we find out that y0 = x0, y1 = 2x1, y2=3x2 and so on. By prime factorising xk, we find out that he factorisation of x2 contain 2, factorisation of x3 contain 3 and so on, so the factorisation of x97 contain 97. Since the largest prime number obtained by prime factorising 100! is 97, x97 must be the smallest value that is divisible by 100!.