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Find the least positive integer k such that 24 × 35 × 7 × 112 × k is a perfect cube (that is, it equals an integer raised to the third power). a) 2 b) 3 c) 4 d) 5

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To make the expression 24 × 35 × 7 × 112 × k a perfect cube, we need to find the least positive integer value of k that will make it a perfect cube.

First, let's break down the given expression into its prime factors:

24 = 2^3 × 3
35 = 5 × 7
112 = 2^4 × 7

Now, let's multiply these prime factors together:

(2^3 × 3) × (5 × 7) × (2^4 × 7) × k = 2^(3+4) × 3 × 5 × 7 × k = 2^7 × 3 × 5 × 7 × k

For the expression to be a perfect cube, each prime factor must have an exponent that is a multiple of 3.

We already have 2^7, which has an exponent of 7, so we need to adjust the other prime factors:

3 × 5 × 7 × k

To make this a perfect cube, we need to have exponents of 3 for each prime factor.

3 = 3^1 (already has an exponent of 1)
5 = 5^1 (already has an exponent of 1)
7 = 7^1 (already has an exponent of 1)

For k, we need to find the least positive integer value such that k = 3^2, which will give it an exponent of 2.

So, the least positive integer k is 3^2 = 9.

Therefore, the answer is (b) 3.
User David Momenso
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