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Analyse the prime factors of 2750. For 2750/k to be a perfect cube, the exponents of all prime factors in its prime factorisation must be multiples of 3.
To make the exponent of 2 and 11 a multiple of 3, we need to remove 2^1 and 11^1. Since 5^3 is already a multiple of 3, no need to include it in k.
Thus, to make 2750/k a perfect cube, k should be 2 x 11 = 22.
To make the exponent of 2 and 11 a multiple of 3, we need to remove 2^1 and 11^1. Since 5^3 is already a multiple of 3, no need to include it in k.
Thus, to make 2750/k a perfect cube, k should be 2 x 11 = 22.
Date Posted:
2 months ago