Question

what is the smallest number by which 13122 must be divided to make it a perfect cube?
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Answer 1

Answer:

1536=2×2×2×2×2×2×2×2×2×3

After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.

1536=(2×2×2)×(2×2×2)×(2×2×2)×3

So, in order to make it a perfect cube, it must be divided by 3.

Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.1536=2×2×2×2×2×2×2×2×2×3

After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.

1536=(2×2×2)×(2×2×2)×(2×2×2)×3

So, in order to make it a perfect cube, it must be divided by 3.

Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.1536=2×2×2×2×2×2×2×2×2×3

After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.

1536=(2×2×2)×(2×2×2)×(2×2×2)×3

So, in order to make it a perfect cube, it must be divided by 3.

Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.1536=2×2×2×2×2×2×2×2×2×3

After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.

1536=(2×2×2)×(2×2×2)×(2×2×2)×3

So, in order to make it a perfect cube, it must be divided by 3.

Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.1536=2×2×2×2×2×2×2×2×2×3

After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.

1536=(2×2×2)×(2×2×2)×(2×2×2)×3

So, in order to make it a perfect cube, it must be divided by 3.

Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.1536=2×2×2×2×2×2×2×2×2×3

After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.

1536=(2×2×2)×(2×2×2)×(2×2×2)×3

So, in order to make it a perfect cube, it must be divided by 3.

Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.