3
$\begingroup$

Let $S$ be the set of all positive integers such that each element $n$ in $S$ satisfies the following conditions:

  1. The sum of the digits of $n$ is a prime number.
  2. The sum of the digits of $n^2$ is a perfect square.

Prove or disprove the existence of an infinite subset $T$ of $S$ such that for every $n$ in $T$, the sum of the digits of $n^3$ is a perfect cube.

Attempt

To prove the existence of an infinite subset $T$ of $S$ such that for every $n$ in $T$, the sum of the digits of $n^3$ is a perfect cube, we need to show that there are infinitely many positive integers $n$ satisfying the given conditions.

I analyzed the conditions provided:

  1. The sum of the digits of $n$ is a prime number.
  2. The sum of the digits of $n^2$ is a perfect square.

I tried to construct such an infinite subset $T$ by considering the properties of the conditions.

First, i started with the condition that the sum of the digits of $n$ is a prime number. This condition restricts the possible values of $n$ to those whose digit sums are primes. Since there are infinitely many prime numbers, there are infinitely many such $n$.

Second, i considered the condition that the sum of the digits of $n^2$ is a perfect square. For any given prime digit sum $p$, there are infinitely many numbers whose digit sum is $p$. Let's denote one such number as $n_1$. Now, if $n_1$ satisfies the condition that the sum of the digits of $n_1^2$ is a perfect square, we can include it in our subset $T$.

Now if I continue this process to construct infinitely many numbers in $T$, each satisfying the given conditions. Since there are infinitely many prime numbers and infinitely many numbers for each prime digit sum, we can conclude that there exists an infinite subset$T$ of $S$ satisfying the condition that the sum of the digits of $n^3$ is a perfect cube for every $n$ in $T$.

Therefore, can we say that the existence of such infinite subset $T$ is proven?

I am unsure of my answer. Is this correct?

$\endgroup$
8
  • 1
    $\begingroup$ Note: I don't understand your argument. You appear to simply assert that you can find natural numbers with some desired properties, but this isn't obvious. $\endgroup$
    – lulu
    Commented Mar 18 at 14:22
  • 1
    $\begingroup$ The set $T = \{2 \times 10^n\}$ indeed gets the job done, as $2$ is a prime, $4$ is a perfect square, and $8$ is a perfect cube. $\endgroup$
    – D S
    Commented Mar 18 at 14:27
  • $\begingroup$ @DS Not sure what I was thinking there. Thanks! Of course $\{2\times 10^n\}$ works, as you point out. $\endgroup$
    – lulu
    Commented Mar 18 at 14:28
  • 1
    $\begingroup$ To the OP: Your attempt is flawed. If $A$ is infinite, $B$ is infinite, then $A \cap B$ is not necessarily infinite. $\endgroup$
    – D S
    Commented Mar 18 at 14:29
  • $\begingroup$ @DS It's easier than that because $11$ works as well. In fact, I believe you can construct numbers with two ones separated by zeroes so that the square has non-zero digits $1,2,1$ and the digits of the cube are $1,3,3,1$. That would give the infinite set. $\endgroup$
    – John Douma
    Commented Mar 18 at 14:52

1 Answer 1

2
$\begingroup$

Numbers of the form $10^n+1$ work for all positive $n$.

The digits of $10^n+1$ sum to $2$.

$(10^n+1)^2=10^{2n}+2\times 10^n+1$ so the digits sum to $4$.

$(10^n+1)^3=10^{3n}+3\times 10^{2n}+3\times 10^n+1$ and the digits sum to $8$.

$\endgroup$
4
  • $\begingroup$ Also, $(10^n+1)10^k$ work too. $\endgroup$
    – D S
    Commented Mar 18 at 17:59
  • $\begingroup$ I still don't get the point you were trying to make in the comments. Doesn't the set $\{2,20,200,2000\ldots \}$ work too? $\endgroup$
    – D S
    Commented Mar 18 at 18:00
  • $\begingroup$ @DS Yes, it does which shows that there are more than one. $\endgroup$
    – John Douma
    Commented Mar 18 at 18:55
  • $\begingroup$ No, $p=5$doesn't work, neither does p = 7. $\endgroup$
    – D S
    Commented Mar 18 at 21:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .