If ‘x’ is both cube and square of an integer and ‘x’ is between 1 and 200. What is the value of ‘x’ ?
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8
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16
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64
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125
A number that is both a perfect square and cube must be a perfect sixth power (since LCM of 2 and 3 is 6). Check sixth powers: 1⁶ = 1 (not in range 1-200, exclusive), 2⁶ = 64, 3⁶ = 729 (too large). So x = 64. Verify: 8² = 64 and 4³ = 64. Options A (8), B (16), and D (125) are not both perfect squares and cubes.
A number that is both a perfect square and a perfect cube must be a perfect sixth power (since it needs to be expressible as both a²and b³, the exponents in its prime factorization must be divisible by both 2 and 3, i.e., by 6). Checking sixth powers: 1⁶=1 (excluded, boundary), 2⁶=64, 3⁶=729 (too large, exceeds 200). So 64 is the only value between 1 and 200 satisfying both conditions — it's 8² (square) and 4³ (cube). The other options (8, 16, 125) are perfect cubes or squares individually but not both simultaneously.