How many perfect cubes are there between 1 and 100000 which are divisible by 7?
6
Let the cube be \(k^3\) with \(1 \lt k^3 \lt 100000\). Since \(46^3=97336\) and \(47^3=103823\), we need \(2\le k\le 46\).
Divisibility by 7: 7 is prime, so \(7\mid k^3 \iff 7\mid k\).
Multiples of 7 in [2, 46]: \(7, 14, 21, 28, 35, 42\) — that is 6 cubes.
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select the correct answer using the code given below:
How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?
