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Given:

CU and BBUE are concatenated Cubes with all prime digits B, C, U, E.

Which are these two Unique Cubes?

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  • $\begingroup$ Err, there are only two 2-digit cubes, and only one of those has all-prime-digits. [Not even a spoiler, IMO]. Really we only need to identify BBUE $\endgroup$ – smci May 21 at 9:35
  • $\begingroup$ Then, we don’t get to see C U B E in the answer. $\endgroup$ – Uvc May 21 at 11:07
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Solution

$$C=2\,\,\,\,\,\,\, U=7\,\,\,\,,\,\,\,\, B=3\,\,\,\,,\,\,\,\, E=5$$

Reasoning

The only $2$-digit cubes are $27$ and $64$ which gives us $C$ and $U$ straight away. Then, $BB7E$ must be the cube of a $2$-digit number $x$ (since it is greater than $1000$). Let $x = 10a+b$. Then, we have $$(10a+b)^3 = 1000a^3 + 300a^2 b + 30ab^2 + b^3 = BB7E$$. This means that $b$ is either $7$ or $5$ (to make $E$ prime and different to $U$). If $b$ is $7$, then we have $$1470a + 343 \equiv 70a + 43 \equiv 73 \mod 100.$$ The smallest solution to this equation is $a=9$ which is too large to cube to a $4$-digit number.
Otherwise, if $b=5$, we have $$250a + 125 \equiv 75 \mod 100$$ which means $a$ is odd. Since, $35^3 > 10000$, the only possibility is $a=1$.
Checking this, we have $$ 15^3 = 15 \times 225 = 3375 $$

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