Quickly write a twelve-digit number that is not a perfect cube.
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147. It is sufficient for the last three digits not to be a perfect cube of a single-digit number; it is also sufficient for the number to end in a number of zeros not divisible by 3.
Source: NuminaMath-1.5,
licensed Apache-2.0.
Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.