Cube sum of digits

Find all integers that is equal to the cube of sum of its digits.

Solution

Clearly 0 satisfies the problem statement, and negative numbers cannot satisfy. So now we may assume that the integer is positive.

Suppose $N = a_1 \dots a_d$ where each $a_i$ is a single digit, and $a_1 \neq 0$. Because $N \geq {10}^{d-1}$ and $N = (a_1 + \dots + a_d)^3 \leq (9d)^3$ so we must have: $$10^{d-1} \leq 9d^3 $$

For large $d$ this inequality fails to hold. In fact, it's only true for $d \leq 4$. So $N$ is a 4-digit number, which is also a cube. Therefore $N \leq 21^3$ because $22^3 > 10^5$.

Now consider modulo 9: we know that $N \equiv (a_1 + \dots + a_d) \mod 9$ so: $$ N = (a_1 + \dots + a_d)^3 \equiv N^3 \mod 9$$ Among all the residues modulo 9, only 0, 1, 8 satisfy $x \equiv x^3 \mod 9$. Therefore $N \equiv 0, 1, 8 \mod 9$.

So now our candidates are: $N = 1^3, 8^3, 10^3, 17^3, 18^3$. By inspection, only $N = 1^3 = 1, 8^3 = 512, 17^3 = 4913, 18^3 = 5832$ satisfy.

a,b,c fraction integer

Given positive integers $a,b,c$ such that
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$$
is also an integer, show that
$$\frac{a^2b^2}{c},\frac{b^2c^2}{a}, \frac{c^2a^2}{b}$$
are all cubic numbers.

a,b,c integers and cubic number

Suppose $a,b,c$ are positive integers such that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer. Prove that $abc$ is a cubic number.

Solution

We have $ab^2 + bc^2 + ca^2 = kabc$ for some k.

Let $d = \gcd(a,b,c)$. We can replace $a,b,c$ by $a/d, b/d, c/d$ respectively and the problem does not change. Thus, without loss of generality, we may assume that $d = 1$.

Let $p$ be a prime that divides $abc$, which means $p$ divides at least one of $a,b,c$. We also know that $p$ cannot divide all three, since $d = 1$.

If $p$ divides exactly one of $a,b,c$, for example $a$, then $ab^2, ca^2, kabc$ are all divisible by $p$, but not $bc^2$. Impossible. Thus, $p$ must divide exactly two of $a,b,c$.

Suppose $p$ divides $a$ and $b$. Furthermore, let $x$ be the largest integer such that $p^x$ divides $a$. Likewise, let $y$ be the largest integer such that $p^y$ divides $b$.

Since $ab^2, bc^2, kabc$ are all divisible by $b$, then so is $ca^2$. Thus $y \leq 2x$.

Since $ab^2, ca^2, kabc$ are all divisible by $a$, then so is $bc^2$, Thus $y \geq x$, which means $x \leq y \leq 2x$.

Now, since $ab^2, ca^2, kabc$ are all divisible by $p^{2x}$, then so is $bc^2$, which means $y \geq 2x$.

Therefore, $y = 2x$, which means that the degree of $p$ in the factorization of $abc$ is $x+y = 3x$.

For each prime $p$ that divides $abc$, it must occur as a cubic number in its prime factorization. Thus $abc$ is a cubic number.

A non-trivial example is $a=1,b=2,c=4$.