If n is a positive integer such that 72*n is a perfect cube, what is the smallest possible value of n?

Answer and explanation

Correct answer: B. 3

  1. Step 1: Find the prime factorization of 72: 72 = 8 9 = 2^3 3^2.
  2. Step 2: For 72*n to be a perfect cube, every prime exponent in its prime factorization must be a multiple of 3.
  3. Step 3: The exponent of 2 is already 3 (a multiple of 3).
  4. Step 4: The exponent of 3 is 2. To reach the next multiple of 3 (which is 3), we need one more factor of 3.
  5. Step 5: Thus the smallest positive integer is n = 3^1 = 3.
Answer
3.
Why not the others: 9 makes 72n = 2^3 3^4 (exponent 4 is not divisible by 3); 6 introduces an extra factor of 2 giving 2^4; 12 introduces 2^5.

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