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Find $Ain mathbb{N}$ s.t. $A=5cdot d_1^3 + 5 cdot d_2^3 + d_3^3 + d_4^3$ and $d_1, d_2, d_3, d_4$ are the smallest divisors of $A$ but not in this order.

It's easy to see that $1$ and $2$ are among them.

I have no idea how to continue.

I'll show how one might stumble across a few solutions and then indicate how to systematically find them all.

First, we verify that $A$ must be even. If $A$ were odd, each of $d_1,d_2,d_3,d_4$ would be odd, so $5 d_1^3 + 5 d_2^3 + d_3^3 + d_4^3$ would be even. But that sum is supposed to be $A$; hence, $A$ is even and its two smallest divisors are $1$ and $2$.

Next, we try to judge the scale of the problem. How big must $A$ be? The smallest possible values for the divisors are $1,2,3,4$, and using those values $5 d_1^3 + 5 d_2^3 + d_3^3 + d_4^3$ will be smallest if $d_1$ and $d_2$ are $1$ and $2$. Thus $A$ is at least $5cdot1^3 + 5cdot2^3 + 3^3 + 4^3 = 136$.

But look: that almost solves the problem. $136$ is divisible by $1,2$ and $4$; it's only missing the factor of $3$. Can we fix up this near miss?

If we switch which divisors have the coefficients of $5$ or $1$, that won't change divisibility by $4$. There aren't many possibilities to check, but we can figure out what the change $bmod 3$ will be in advance: if $5a^3 + b^3$ is changed to $5b^3 + a^3$, the difference is $4b^3 - 4a^3 equiv b^3 - a^3 equiv b - a pmod 3$. Since $136 equiv 1 pmod 3$, an increase of $2 bmod 3$ is needed. This can be done by either swapping $1$ with $3$, yielding
$$
240 = 5cdot 3^3 + 5cdot 2^3 + 1^3 + 4^3,
$$
or swapping $2$ with $4$:
$$
360 = 5cdot 1^3 + 5cdot 4^3 + 3^3 + 2^3.
$$

Now we try to be more thorough. We consider two cases: either $A$ is divisible by $4$ or not.

If $A$ is divisible by $4$, three of its four smallest divisors are necessarily $1,2$ and $4$. The remaining one must be odd so that $5 d_1^3 + 5 d_2^3 + d_3^3 + d_4^3$ will be even; hence it is the smallest odd divisor of $A$ apart from $1$. It must be a prime, so call it $p$. Since $p$ divides both $A$ and the term on the right-hand side in which it appears, it must also divide the sum of the remaining three terms. Depending on where factors of $5$ appear, that sum is one of the following:
$$
begin{align}
5cdot 1^3 + 5cdot 2^3 + 4^3 &= 109\
5cdot 1^3 + 2^3 + 5cdot 4^3 &= 333 = 3^2cdot 37\
1^3 + 5cdot 2^3 + 5cdot 4^3 &= 361 = 19^2\
5cdot 1^3 + 2^3 + 4^3 &= 77 = 7cdot 11\
1^3 + 5cdot 2^3 + 4^3 &= 105 = 3cdot 5cdot 7\
1^3 + 2^3 + 5cdot 4^3 &= 329 = 7cdot 47
end{align}
$$
There are thus $11$ cases, where $p$ is one of the prime factors of these sums. Adding in the remaining term of $p^3$ or $5p^3$ may not produce a multiple of $4$ (e.g., $109 + 109^3$ doesn't work) or may introduce a new divisor smaller than $p$ (e.g., $361 + 19^3$ is divisible by $5$), but in four cases everything works out. Two were found above; the other two are
$$
begin{align}
1792 &= 5cdot 1^3 + 5cdot 7^3 + 2^3 + 4^3\
2044 &= 5cdot 4^3 + 5cdot 7^3 + 1^3 + 2^3
end{align}
$$

If $A$ is not divisible by $4$, then its third smallest divisor is an odd prime $p$. The fourth must be an even divisor; since it is not divisible by $4$, it is twice an odd divisor. Thus the smallest it can be is $2p$. Now $p$ divides $A$ and two terms on the right-hand side, so it divides the sum of the remaining two terms. That sum is one of the following:
$$
begin{align}
5cdot 1^3 + 5cdot 2^3 &= 45 = 3^2cdot 5\
5cdot 1^3 + 2^3 &= 13\
1^3 + 5cdot 2^3 &= 41\
1^3 + 2^3 &= 9 = 3^2
end{align}
$$
There are five prime factors but seven cases, since for $p = 13$ the remaining terms could be $13^3 + 5cdot 26^3$ or $5cdot 13^3 + 26^3$ and likewise for $p = 41$. However, in each case the resulting total is either a multiple of $4$ or has an odd prime factor smaller than $p$, so there are no further solutions.

So there are exactly four possible values of $A$: $240, 360, 1792$ and $2044$.