The Cryptic Cube Conundrum

Question

Consider a perfect cube $N^3$ where $N$ is a positive integer. You are given the following cryptic clues related to $N$:

1. The sum of the digits of $N$ is a perfect square.
2. $N$ is divisible by the sum of its digits.
3. The prime factorization of $N$ includes exactly three distinct primes.

Determine the smallest possible value of $N$ that satisfies all these conditions.

$\textbf{My Work:}$

Let $N = abc$ be the three-digit number representing the cube. Without loss of generality, assume $a$, $b$, and $c$ are the digits of $N$.

Despite my efforts in analyzing the conditions and attempting various approaches, I couldn't determine the smallest possible value of $N$ that satisfies all these conditions. Any guidance, insights, or a solution from the community would be greatly appreciated.

Answer 1

$N$ can’t be one-digit. Let us try to find two-digit $N$. The sum of its digits is a square, so $1$, $4$, $9$ or $16$. There is only $10$ with sum $1$, it doesn’t fit.

If the sum if digits of $N$ is $4$, then $N$ is $13$, $22=2\cdot11$, $31$, $40=2^3\cdot5$. Doesn’t work again.

If the sum of digits is $9$, then we have to try $18=2\cdot3^2$, $27=3^3$, $36=2^2\cdot3^2$, $45=5\cdot3^2$, $54=2\cdot3^3$, $63=3^2\cdot7$, $72=2^3\cdot3^2$, $81=3^4$, $90=2\cdot3^2\cdot5$.

If the sum of digits is $16$, then we look at $79$, $88=2^3\cdot11$, $97$.

The sum can’t be greater than $18$.

$90$ is divisible by $9+0=9$, so this is the answer.

Answer 2

So it’s not necessarily true that N is three digits.

Let N be denoted by $$\sum_{k = 0}^{n - 1} 10^k a_k$$ where n is the number of digits.

Also $$\sum_{k = 0}^{n - 1} a_k = m^2$$

And we know that $$N \equiv 0 (\mod m^2)$$

And that N = p_1^{j_1}p_2^{j_2}p_3^{j_3}

And we know because $N$ is divisible by $m^2$ at least one of the $j$’s is even.

Suppose $p_1 = 2, p_2 = 3, p_3 = 5, j_1 = 2, j_2 = 1, j_3 = 1$

Then $N = 60$ and N is the smallest value which is divisible by a perfect square and that has 3 distinct factors, but 6 is not a perfect square.

If we find adjust the values to make the next 2 smallest numbers with those properties we see that for $j_1 = 3$, we get $N = 80$, but 8 is not a perfect square, then changing $j_1 = 1, j_2 = 2$ which is the third smallest we get $N = 90$ then 9 is a perfect square and $N$ is divisible by 9. Therefore $N = 90$ is the smallest.

However, this doesn’t use $N^3$ anywhere, so if we were to interpret the question as what is $N^3$ then we see that $N^3 = 729000$.

However if we are to interpret it to be such that $N$ itself IS a perfect cube, then we need to start over and look at $N$ such that $j_1, j_2, j_3$ all are divisible by 3, so we get the smallest $N = 27000$ which the sum of the digits is 9 which is a perfect square, and 27000 is divisible by 9, so if that is the interpretation then $N = 27000$.