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To find the least positive integer ( k ) for which ( 15k ) is a cube, we start with the prime factorization of 15, which is ( 3^1 \times 5^1 ). For ( 15k ) to be a perfect cube, the exponents in its prime factorization must be multiples of 3. Thus, we need to make the exponents of both 3 and 5 in ( 15k ) equal to 3. Therefore, ( k ) must contribute ( 3^2 ) (to make the exponent of 3 equal to 3) and ( 5^2 ) (to make the exponent of 5 equal to 3). Thus, ( k = 3^2 \times 5^2 = 9 \times 25 = 225 ). Therefore, the least positive integer ( k ) is ( 225 ).

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