The least number which when divides 37044, gives the result a perfect cube, is:
4
To find the least number which when divides 37044 gives a perfect cube, we need to find the prime factorization of 37044.
The prime factorization of 37044 is \(2^2 \times 3^2 \times 1029\).
Since 1029 = 3 x 7 x 49 = 3 x 7 x 72 = 3 x 73, we can rewrite the prime factorization as:
\(37044 = 2^2 \times 3^3 \times 7^3\).
A perfect cube has exponents that are multiples of 3 for each prime factor in its factorization. In this case, the exponent of 2 is 2, which is not a multiple of 3. To make it a multiple of 3, we need at least one more factor of 2. Therefore, we need to divide 37044 by \(2^2\) to obtain a perfect cube.
Dividing 37044 by 4 (which is \(2^2\)) gives:
\(\frac{37044}{4} = 9261\)
Now let's find the cube root of 9261:
\(\sqrt[3]{9261} = 21\)
Since 21 is an integer, 9261 is a perfect cube. Therefore, the least number that, when dividing 37044, results in a perfect cube is 4.
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