. Find all integers $a, b, c$ satisfying $1<a<b<c$ such that $(a-1)(b-1)(c-1)$ divides $a b c-1$.
. This is another type of exercise, where the fundamental remark is that if $a, b, c$ are "large," then $(a-1)(b-1)(c-1)$ is too close to $a b c-1$ to be a divisor of it.

It is clear that $(a-1)(b-1)(c-1)<(a-1) b c=a b c-b c<a b c-1$. On the other hand, if $(a-1)(b-1)(c-1) \geqslant \frac{a b c}{2}$, then $\frac{a b c-1}{(a-1)(b-1)(c-1)}<2$ cannot be an integer, as there is no integer strictly between 1 and 2. The function $\frac{x-1}{x}=1-\frac{1}{x}$ is increasing and tends to 1 as $x$ grows to infinity. In particular, since $1<a<b<c$, which implies $a \geqslant 2, b \geqslant a+1, c \geqslant b+1 \geqslant a+2, \frac{b-1}{b} \geqslant \frac{a}{a+1}, \frac{c-1}{c} \geqslant \frac{a+1}{a+2}$, so that $\frac{(a-1)(b-1)(c-1)}{a b c} \geqslant \frac{a-1}{a+2}$.

The quotient $q=\frac{a b c-1}{(a-1)(b-1)(c-1)}$ therefore satisfies:

if $a=2,1<q<\frac{a+2}{a-1}=4$, so $q=2$ or $q=3$,

(B) if $a=3,1<q<\frac{a+2}{a-1}=5 / 2$, so $q=2$,

(C) if $a \geqslant 4,1<q<\frac{a+2}{a-1} \leqslant 2$ since the function $\frac{x+2}{x-1}=1+\frac{3}{x-1}$ is decreasing.

The only possible solutions are therefore for $a=2$ or $a=3$.

For $a=2$, we are reduced to seeing when $\frac{2 b c-1}{(b-1)(c-1)}$ is 2 or 3. Since the numerator is odd, the quotient $q$ cannot be 2, and the equation $2 b c-1=3(b-1)(c-1)$ expands to $b c-3 b-3 c+4=0$, or $(b-3)(c-3)=5$. Since $b<c$, the only solutions are $b=4$ and $c=8$. We verify that $3 \times 7=21$ divides $(2 \times 4 \times 8)-1=63$.

For $a=3$, we similarly reduce to $3 b c-1=4(b-1)(c-1)$, since $q=2$ so $q(a-1)=4$, or $b c-4 b-4 c+5=0$, or $(b-4)(c-4)=11$, hence $b=5$ and $c=15$, which provides the last solution $(3,5,15)$. Indeed, $2 \times 4 \times 14=112$ divides $(3 \times 5 \times 15)-1=224$.

The problem therefore has two solutions: $(2,4,8)$ and $(3,5,15)$.

Solution to Exercise 100. Let $k=2007$, and call $a_{1}, a_{2}, \ldots, a_{k}$ $k$-th powers of distinct prime numbers, hence pairwise coprime.

The Chinese Remainder Theorem allows us to assert that there are infinitely many integers $n$ such that:

$$
\begin{cases}n \equiv 1 & \left(\bmod a_{1}\right) \\ n \equiv 2 & \left(\bmod a_{2}\right) \\ & \vdots \\ n \equiv k & \left(\bmod a_{k}\right)\end{cases}
$$

which precisely means that $n-1$ is divisible by $a_{1}, n-2$ by $a_{2}, \ldots, n-k$ by $a_{k} ; n-1, n-2, \ldots, n-k$ are indeed $k$ consecutive integers satisfying the conditions of the statement, hence the result.