IMO 1968 Problem 2

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m57s

Problem

Find all natural numbers $x$ such that the product of their digits (in decimal notation) is equal to $x^2 - 10x - 22$.

Exploration

The expression on the right side is quadratic in $x$, while the left side is a product of decimal digits, hence grows at most exponentially in the number of digits but remains extremely constrained digitwise. For large $x$, the quadratic term dominates, while the digit product is bounded above by a power of $9$ depending on the number of digits.

A necessary condition is $x^2 - 10x - 22 \ge 0$, which forces $x$ to lie outside the interval between the roots of $t^2 - 10t - 22$. The discriminant equals $188$, giving roots $5 \pm \sqrt{47}$, so admissible natural numbers satisfy $x \ge 12$.

For numbers with at least three digits, say $x \ge 100$, the right-hand side exceeds $100^2 - 1000 - 22 = 8978$, while any three-digit digit product is at most $9^3 = 729$, so no solution can have three or more digits. Hence only two-digit candidates remain.

Writing $x = 10a + b$ reduces the equation to a finite system over digits $a \in {1,\dots,9}$ and $b \in {0,\dots,9}$. Each fixed $a$ leads to a quadratic condition in $b$, which can be checked exhaustively.

The most restrictive case appears to be small $a$, since the quadratic term in $x$ grows quickly compared to the linear digit product $ab$.

Problem Understanding

The problem asks for all natural numbers whose decimal digits multiply to a value equal to a quadratic expression in the number itself. This is a classification problem of Type A, since every solution must be identified and no extraneous candidates may remain.

The main difficulty lies in balancing a global expression $x^2 - 10x - 22$ with a highly local digit-dependent quantity. The quadratic grows rapidly with digit length, while the digit product remains small, forcing strong structural restrictions on $x$. The task reduces to ruling out all but finitely many digit configurations and then checking them exactly.

The only plausible candidate region consists of small two-digit numbers, and a direct digit analysis should determine the full solution set.

The expected answer is $x = 12$, since it already satisfies the equation upon substitution and matches the necessary bounds.

Proof Architecture

The first lemma establishes that any solution must satisfy $x \ge 12$ by solving the quadratic inequality $x^2 - 10x - 22 \ge 0$.

The second lemma proves that any solution has at most two digits by comparing $9^k$ with the quadratic lower bound for $k \ge 3$.

The third lemma reduces the problem to two-digit numbers and derives the digit equation in variables $a$ and $b$ for $x = 10a + b$.

The fourth lemma performs an exhaustive check over $a = 1$ through $9$, showing that only $a = 1$ admits a valid digit $b$, namely $b = 2$.

The hardest part is the elimination of all cases except $a = 1$, since each requires careful comparison of a quadratic in $b$ with a linear term.

Solution

Lemma 1

Any solution $x$ satisfies $x \ge 12$.

The expression $x^2 - 10x - 22$ must be nonnegative, since it equals a product of digits. The roots of $t^2 - 10t - 22 = 0$ are $5 \pm \sqrt{47}$. The smaller root lies between $-1$ and $-2$, and the larger root lies between $11$ and $12$. The quadratic is nonnegative for $t \ge 5 + \sqrt{47}$, hence for all natural numbers this implies $x \ge 12$.

This establishes that only integers at least $12$ can satisfy the condition, since smaller values force a negative right-hand side incompatible with digit products.

Lemma 2

Any solution $x$ has at most two decimal digits.

If $x$ has $k \ge 3$ digits, then $x \ge 100$. The digit product is at most $9^k$. For $k = 3$, this gives an upper bound $729$.

For $x \ge 100$, the quadratic expression satisfies $x^2 - 10x - 22 \ge 10000 - 1000 - 22 = 8978$. Since $8978 > 729$, no three-digit number can satisfy the equation. Larger values only increase the gap between the two sides.

This shows that every solution lies in the range $12 \le x \le 99$.

Lemma 3

Every solution can be written as $x = 10a + b$ with digits $a \in {1,\dots,9}$ and $b \in {0,\dots,9}$, and satisfies

$ab = (10a + b)^2 - 10(10a + b) - 22.$

Expanding gives

$ab = 100a^2 + 20ab + b^2 - 100a - 10b - 22.$

Rearranging yields

$0 = 100a^2 + 19ab + b^2 - 100a - 10b - 22.$

This reformulation converts the problem into a finite set of algebraic constraints over digit pairs $(a,b)$, which fully captures the original condition in two variables.

This step ensures that solving the original equation is equivalent to solving this digit polynomial identity.

Lemma 4

The only digit pair $(a,b)$ satisfying the equation is $(1,2)$.

For $a = 1$, the equation becomes

$b = (10 + b)^2 - 10(10 + b) - 22.$

Expanding gives

$b = 100 + 20b + b^2 - 100 - 10b - 22 = b^2 + 10b - 22.$

Rearranging yields

$0 = b^2 + 9b - 22.$

The discriminant is $81 + 88 = 169$, hence $b = \frac{-9 \pm 13}{2}$. The only digit solution is $b = 2$.

For $a \ge 2$, the left-hand side $ab$ grows linearly in $b$, while the right-hand side becomes a quadratic expression in $b$ with a large positive constant term. Substituting $x = 10a + b$ into the equation gives

$0 = 100a^2 + 19ab + b^2 - 100a - 10b - 22.$

For each fixed $a \ge 2$, the left-hand side is a quadratic in $b$ with discriminant

$\Delta_a = (19a - 10)^2 - 4(100a^2 - 100a - 22).$

Direct computation yields $\Delta_2 = 72$, while for $a \ge 3$ the discriminant is negative since the quadratic term dominates. For $a = 2$, the candidate values of $b$ are non-integral or outside digit range, so no solution arises.

This completes the elimination of all cases except $a = 1$.

Combining with the valid pair $(1,2)$ gives $x = 12$.

Final conclusion

The only natural number satisfying the condition is

$\boxed{12}.$

Verification of Key Steps

The reduction to two-digit numbers relies on comparing $x^2 - 10x - 22$ with the maximal digit product $9^k$. Recomputing for $k = 3$ confirms that even the smallest three-digit value $x = 100$ already produces a right-hand side far exceeding $9^3$, so no boundary case is missed.

The case $a = 1$ is verified directly by substitution, and the algebraic reduction to $b^2 + 9b - 22 = 0$ is checked by expanding all terms independently, ensuring no sign error occurs in the rearrangement.

For $a \ge 2$, recomputing the discriminant shows it becomes increasingly negative as $a$ increases, confirming that no hidden integer solutions can appear for larger digits.

Alternative Approaches

A different strategy uses modular constraints, comparing the equation modulo $9$ or $11$ to restrict possible digit sums and digit products simultaneously. Another approach uses bounding arguments based on the growth rate of digit products versus quadratic polynomials, leading quickly to the restriction $x \le 99$ and then an exhaustive search. The direct digit expansion method remains the most transparent since it converts the problem into finitely many explicit quadratic equations.