# These Two Cubes are The Only Ones That Are All Pure Prime..name them

Given:

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

Which are these two Unique Cubes?

• 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 – smci May 21 '19 at 9:35
• Then, we don’t get to see C U B E in the answer. – Uvc May 21 '19 at 11:07

$$C=2\,\,\,\,\,\,\, U=7\,\,\,\,,\,\,\,\, B=3\,\,\,\,,\,\,\,\, E=5$$
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 $$15^3 = 15 \times 225 = 3375$$