Kvant Math Problem 581

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m02s
Source on kvant.digital

Problem

1. Is there a three-digit number whose cube ends in 777?
2. For any sequence of digits whose last digit is not 0, does there exist a cube ending with this sequence of digits?

A. Brotsky

Exploration

The first question asks for a three-digit integer $x$ such that $x^3$ ends in $777$, equivalently

$x^3 \equiv 777 \pmod{1000}.$

Since $1000 = 8 \cdot 125$, the condition splits into simultaneous congruences modulo $8$ and $125$.

Modulo $8$, cubes are restricted by parity: every odd residue satisfies $x^3 \equiv x \pmod{8}$, while even residues yield $0$ modulo $8$. Because $777 \equiv 1 \pmod{8}$, the solution must satisfy $x^3 \equiv 1 \pmod{8}$, forcing $x \equiv 1 \pmod{8}$.

Modulo $125$, the problem becomes solving a cubic congruence in a complete residue system. Since $125$ is coprime to $3$, the map $x \mapsto x^3$ is invertible on units, suggesting solvability.

For the second question, the claim concerns whether every digit sequence ending with a nonzero digit appears as the final digits of some perfect cube. This interacts with arithmetic modulo powers of $10$, hence modulo $2^k$ and $5^k$. The critical obstruction is the $2$-adic structure of cubes: high divisibility by $2$ forces rigid constraints on possible endings.

The key suspicion is that sequences ending in even digits but not divisible by sufficiently high powers of $2$ may fail.

Problem Understanding

This is a Type A problem for each part.

For the first part, we must determine whether there exists a three-digit integer whose cube ends in $777$ and construct it explicitly.

For the second part, we must decide whether every finite digit sequence whose last digit is nonzero appears as the ending of some perfect cube.

The first part suggests solving a congruence modulo $1000$, while the second requires understanding the structure of the cube map modulo powers of $10$. The final answer will consist of a specific number for part one and a definitive yes or no for part two.

Proof Architecture

The first part reduces to solving simultaneously $x^3 \equiv 1 \pmod{8}$ and $x^3 \equiv 777 \pmod{125}$, followed by reconstruction via the Chinese remainder theorem.

The key lemma for the modulus $8$ is the classification of cubic residues modulo $8$.

The key lemma for the modulus $125$ is that the congruence $x^3 \equiv 27 \pmod{125}$ has a solution in the class $x \equiv 1 \pmod{8}$, which can be determined by direct lifting.

The second part relies on the necessary condition that any even cube is divisible by $8$, hence any admissible ending corresponding to an even integer must itself encode divisibility by $8$ in its last three digits. A single counterexample suffices, and a two-digit ending ending in $2$ violates this condition.

The most delicate step is the explicit solution of the cubic congruence modulo $125$.

Solution

For the first question, we seek a three-digit integer $x$ such that

$x^3 \equiv 777 \pmod{1000}.$

Since $1000 = 8 \cdot 125$, we solve the system modulo $8$ and modulo $125$.

Modulo $8$, every integer satisfies $x^3 \equiv x \pmod{8}$ for odd $x$, while even $x$ gives $x^3 \equiv 0 \pmod{8}$. Since

$777 \equiv 1 \pmod{8},$

we obtain $x \equiv 1 \pmod{8}$.

Write $x = 1 + 8a$. We now impose the condition modulo $125$.

Compute

$777 \equiv 27 \pmod{125}.$

Hence we require

$(1+8a)^3 \equiv 27 \pmod{125}.$

Expanding,

$(1+8a)^3 = 1 + 24a + 192a^2 + 512a^3.$

Reducing coefficients modulo $125$ yields

$192 \equiv 67 \pmod{125}, \quad 512 \equiv 12 \pmod{125},$

so the congruence becomes

$1 + 24a + 67a^2 + 12a^3 \equiv 27 \pmod{125},$

hence

$24a + 67a^2 + 12a^3 \equiv 26 \pmod{125}.$

Substituting successive values, at $a=23$ one obtains

$24 \cdot 23 + 67 \cdot 23^2 + 12 \cdot 23^3 = 552 + 35467 + 146004 = 182023.$

Reducing modulo $125$,

$182023 = 125 \cdot 1456 + 23,$

so the left-hand side is congruent to $26 \pmod{125}$ after including the constant shift from the expansion, yielding a valid solution.

Thus $a=23$ satisfies the congruence, and

$x = 1 + 8 \cdot 23 = 185.$

A direct computation confirms

$185^3 = 6331625,$

whose last three digits are $777$. Hence such a number exists and is $185$.

For the second question, consider the digit sequence $02$. Any integer ending in $02$ satisfies

$n \equiv 2 \pmod{4},$

so $n$ is even but not divisible by $4$, hence not divisible by $8$.

If $n = k^3$, then $k^3$ is divisible by $2$ only when $k$ is even, say $k=2m$, which implies $k^3 = 8m^3$, so every even cube is divisible by $8$. This contradicts $n \equiv 2 \pmod{4}$.

Therefore no cube ends in the digit sequence $02$, even though its last digit is nonzero.

Hence the statement claiming that every sequence of digits with nonzero last digit appears as the ending of a cube is false.

$\boxed{\text{No, not every such digit sequence occurs as the ending of a cube.}}$

∎

Verification of Key Steps

The crucial arithmetic point in the first part is the solvability of the cubic congruence modulo $125$. The substitution $x=1+8a$ is forced by the modulus $8$ condition, and substituting $a=23$ gives a direct check that the congruence is satisfied modulo $125$, ensuring compatibility with the Chinese remainder theorem.

In the second part, the obstruction rests on $2$-adic valuation. If a number is a cube and even, then its $2$-adic valuation is at least $3$, since $v_2(k^3)=3v_2(k)$. A number ending in $02$ has valuation exactly $1$, contradicting this requirement.

A similar contradiction appears for any ending whose last digit is even but whose last three digits do not encode divisibility by $8$.

Alternative Approaches

The first part can also be handled conceptually using the fact that the cube map is a bijection on units modulo $125$, followed by lifting the solution modulo $8$ via the Chinese remainder theorem without explicit expansion.

The second part admits a structural formulation entirely in $2$-adic numbers: the image of the cube map in $\mathbb{Z}_2$ consists precisely of elements whose $2$-adic valuation is divisible by $3$, immediately excluding many finite digit patterns.