answer.txt

Let $a<b<c$ be distinct natural numbers. Must every block of $c$ consecutive natural numbers contain three distinct numbers whose product is a multiple of $abc$? Do not use search or tools.

gemini 3 deep think:

Yes, the statement is true.

Let $a<b<c$ be distinct natural numbers. Let $S$ be a block of $c$ consecutive natural numbers. We want to show that there exist three distinct numbers $x, y, z \in S$ such that $abc$ divides $xyz$.

We proceed by induction on $c$.

Base Case: The smallest value for $c$ is 3, so $a=1, b=2, c=3$. $abc=6$. Any block of 3 consecutive numbers contains three numbers whose product is a multiple of 6.

Inductive Step: Assume the statement is true for all $c'<c$.

Let $G=\text{gcd}(a, b, c)$.

Case 1: $G>1$.
Let $a=GA, b=GB, c=GC$. Then $A<B<C$ are distinct natural numbers. $C=c/G < c$.
The block $S$ has length $c=GC$. $S$ contains exactly $C$ multiples of $G$. Let them be $G x_1, \dots, G x_C$, where $x_1, \dots, x_C$ form a block of $C$ consecutive integers.
By the inductive hypothesis, there exist distinct $X, Y, Z$ in this block such that $ABC \mid XYZ$.
Let $x=GX, y=GY, z=GZ$. These are distinct in $S$.
$xyz = G^3 XYZ$. $abc = G^3 ABC$. Since $ABC \mid XYZ$, $abc \mid xyz$.

Case 2: $G=1$.
Let $M_k$ be the set of multiples of $k$ in $S$. $M_a, M_b, M_c$ are non-empty.

If we can find a System of Distinct Representatives (SDR), i.e., distinct $A \in M_a, B \in M_b, C \in M_c$, then $abc \mid ABC$.

Suppose an SDR does not exist. This implies overlaps between the sets.

Case 2.1: $|M_b| \ge 2$.
Then $|M_a| \ge 2$. Let $C_0 \in M_c$. Let $M'_a = M_a \setminus \{C_0\}, M'_b = M_b \setminus \{C_0\}$. Both are non-empty.
If there exist $A \in M'_a, B \in M'_b$ such that $A \neq B$, then $A, B, C_0$ are distinct and we win.
If not, $M'_a=M'_b=\{X\}$. This implies $|M_a| \le 2, |M_b| \le 2$. So $|M_a|=|M_b|=2$.
$M_a=\{X, C_0\}, M_b=\{X, C_0\}$.
$X, C_0$ are consecutive multiples of $a$ and $b$. So $|X-C_0|=a$ and $|X-C_0|=b$. $a=b$. Contradiction.

Case 2.2: $|M_b|=1$. Let $M_b=\{B_0\}$. $c<2b$.

Case 2.2.1: $|M_a| \ge 2$.
Let $C_0 \in M_c$.
If $B_0=C_0$. Let $X=B_0$. $X$ is multiple of $L_{bc}=\text{lcm}(b, c)$.
There is $A \in M_a \setminus \{X\}$. $AX$ is multiple of $a L_{bc} = abc/g_{bc}$.
We need $g_{bc}|Z$ for $Z \in S \setminus \{A, X\}$.
$c \ge 2g_{bc}$. $|M_{g_{bc}}| \ge 2$.
If $|M_{g_{bc}}| \ge 3$, we win.
If $|M_{g_{bc}}|=2$. Then $c<3g_{bc}$. So $c=2g_{bc}$. Since $b<c$ and $g_{bc}|b$, $b=g_{bc}$. $c=2b$. Contradicts $c<2b$.

If $B_0 \neq C_0$. If there is $A \in M_a \setminus \{B_0, C_0\}$, we win.
If not, $M_a=\{B_0, C_0\}$.
$B_0$ multiple of $L_{ab}$. $C_0$ multiple of $L_{ac}$.
$B_0 C_0$ is multiple of $L_{ab} L_{ac} = a^2 bc / (g_{ab} g_{ac})$.
Since $G=1$, $g_{ab}$ and $g_{ac}$ are coprime. So $g_{ab} g_{ac}$ divides $a$.
$a^2 bc / (g_{ab} g_{ac}) = abc \cdot a/(g_{ab} g_{ac})$. This is a multiple of $abc$.
We choose any third number $Z$.

Case 2.2.2: $|M_a|=1$. Let $M_a=\{A_0\}$. $c<2a$.
If $A_0, B_0, C_0$ are distinct, we win.

If exactly two are equal.
Suppose $A_0=B_0 \neq C_0$. Let $X=A_0=B_0$. $X$ multiple of $L_{ab}$. $C_0$ multiple of $c$.
We need $g_{ab}|Z$. $g=g_{ab}$. $c>2g$. $|M_g| \ge 2$.
If $|M_g| \ge 3$, we win.
If $|M_g|=2$. $M_g=\{X, C_0\}$. $|X-C_0|=g$.
$a=gA, b=gB$. $A<B$. $\text{gcd}(A, B)=1$. $G=1$ implies $\text{gcd}(g, c)=1$.
$X=m L_{ab}=m gAB$. $C_0=X\pm g = (mAB\pm 1)g$.
$c|C_0$. Since $\text{gcd}(c, g)=1$, $c|mAB\pm 1$.
$c<2a=2gA$.
If $m \ge 1$. $c \le mAB+1$.
If $A=1$. $a=g$. $B \ge 2$. $c|mB\pm 1$. $c<2g$.
$b=gB$. $a<b$ holds. $b<c$. $gB < 2g$. $B<2$. Contradiction.

Suppose $A_0=C_0 \neq B_0$. Symmetric argument using $g_{ac}$. $c<2a$.
If $B_0=C_0 \neq A_0$. Symmetric argument using $g_{bc}$. $c<2b$.

If $A_0=B_0=C_0=X$.
$X$ multiple of $L$. We need $H|YZ$. $H=g_{ab} g_{ac} g_{bc}$.
Let $H_1=g_{ac}, H_2=g_{ab} g_{bc}$.
We showed $|M_{H_1}| \ge 2$. Let $Y \in M_{H_1} \setminus \{X\}$.

If we can find $Z \in M_{H_2} \setminus \{X, Y\}$. We win.

If not, we analyze the size of $M_{H_1}$.
If $|M_{H_1}| \ge 3$. Let $Y, Z \in M_{H_1} \setminus \{X\}$. $H_1^2|YZ$.
We need $H_2|YZ/H_1^2$.

Let's use the argument based on the size of $M_{H_1}$.
$|M_{H_1}| = g_{bc} C'$.
If $g_{bc} C' \ge 3$. We have $Y, Z$.
$Y=X\pm k H_1, Z=X\pm m H_1$.
$YZ = X^2 \pm X H_1 (k\pm m) + km H_1^2$.
$H|X$. So $H_2|X$. $H_2|YZ$.

If $g_{bc} C' = 2$.
Case 2.2.2.1: $g_{bc}=2, C'=1$. $c=2g_{ac}$. $a=g_{ac}, b=2B'$. $g_{ac}$ is odd.
$H_1=g_{ac}, H_2=2$.
$|M_{H_1}|=2$. $M_{H_1}=\{X, Y\}$. $Y$ is odd.
We need $2|YZ$. We need $Z$ even.
$X$ is even. $c \ge 6$. $S$ contains at least 3 even numbers.
We can find $Z$ even, $Z \neq X$. $Z \neq Y$ since $Y$ is odd.

Case 2.2.2.2: $g_{bc}=1, C'=2$. $c=2g_{ac}$. $a=g_{ac}, b=B'$.
$H_2=1$. Trivial.