If n is an integer larger than 1, then what is the least value of the integer n + n^{1/2} + n^{1/3}?

Correct — C, 76. Read the exponents carefully: they are one-half and one-third, so the expression is n plus the square root of n plus the cube root of n. The stem says the whole thing is an integer and n is an integer, so the square root and the cube root have to be whole numbers too — the square root of a whole number is either a whole number or irrational, and the same is true of the cube root, and an irrational part cannot cancel out here. A number that is both a perfect square and a perfect cube is a perfect sixth power. The sixth powers are 1, 64, 729 and so on, and the stem rules out 1 by asking for n larger than 1. Take n = 64: its square root is 8, its cube root is 4, and 64 + 8 + 4 = 76.

• (a)1 — Would need n = 0 or a fractional n, and in any case the stem sets n larger than 1. Even at n = 1 the expression is 1 + 1 + 1 = 3.
• (b)3 — This is the value at n = 1, the case the question deliberately excludes. The next admissible n jumps straight to 64, because there is no sixth power in between.
• (d)785 — The value at the next sixth power, n = 729: 729 + 27 + 9 = 765, and 785 is not even that. Either way it is not the least value, since 64 comes first.

The square root of a positive integer is rational only when the integer is a perfect square, and then it is a whole number; the same holds for cube roots and perfect cubes. So an expression that adds an integer, a square root and a cube root can only be an integer when both roots are exact. Being a perfect square and a perfect cube at once means every prime in the number appears to a power divisible by 2 and by 3, that is by 6 — the number is a perfect sixth power.

The question rewards knowing which n are even allowed before any arithmetic starts. The admissible values are 1, 64, 729, 4096 and so on, and once 1 is excluded the answer is decided by a single substitution. Two habits protect you here: read fractional exponents as roots rather than skimming past them, and remember that the least value of a function over a restricted set is found by testing the smallest admissible member of that set, not the smallest number in sight.

• The square root of an integer is either a whole number or irrational; the same is true of the cube root.
• A number that is both a perfect square and a perfect cube is a perfect sixth power.
• The first few sixth powers are 1, 64, 729, 4096 and 15625.
• At n = 64 the expression is 64 + 8 + 4 = 76; at n = 729 it is 729 + 27 + 9 = 765.
• n raised to the one-half is the square root and n raised to the one-third is the cube root.

• Reading the exponents as ordinary powers and computing n + n² + n³.
• Testing n = 2, 3, 4 and finding no integer value, then guessing.
• Forgetting that the stem excludes n = 1, which is the source of option (b).

A least-value item that is really a test of which n make the expression an integer at all.

No directly related past PYQ was found.

• practice — not a real PYQ

  What is the smallest positive integer greater than 1 that is both a perfect square and a perfect cube?

  • (a)16
  • (b)36
  • (c)64
  • (d)216

  Answer(c) 64 — it is 8 squared and 4 cubed, that is 2 to the sixth power.

• practice — not a real PYQ

  If n = 729, the value of n + n^(1/2) + n^(1/3) is

  • (a)756
  • (b)765
  • (c)785
  • (d)793

  Answer(b) 765 — the square root is 27 and the cube root is 9, so 729 + 27 + 9 = 765.