Find the smallest number by which 10985 may be divided to get a perfect cube

Complete step-by-step answer:
A) The given value here is $ 1536 $ .
By prime factorization,
 $ 1536 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 $
Now, to obtain a perfect cube, group the factors into triplets i.e., to the power of $ 3 $ .
 $ 1536 = {2^3} \times {2^3} \times {2^3} \times 3 $
Now, by grouping we have one factor $ 3 $ , is left.
Therefore, $ 1536 $ is not a perfect cube.
So, now to make $ 1536 $ as a perfect cube, we have to divide by $ 3 $ .
Hence, the smallest number by which $ 1536 $ must be divided to obtain a perfect cube is $ 3 $ .

B) The given value is $ 10985 $ .
Similarly, by prime factorization,
 $ 10985 = 5 \times 13 \times 13 \times 13 $
Now, to obtain a perfect cube, group the factors into triplets i.e., to the power of $ 3 $ .
 $ 10985 = 5 \times {13^3} $
Now, by grouping we have one factor $ 5 $ , is left.
Therefore, $ 10985 $ is not a perfect cube.
So, now to make $ 10985 $ as a perfect cube, we have to divide by $ 5 $ .
Hence, the smallest number by which $ 10985 $ must be divided to obtain a perfect cube is $ 5 $ .

C) The given value is $ 28672 $ .
Similarly, by prime factorization,
 $ 28672 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 $
Now, to obtain a perfect cube, group the factors into triplets i.e., to the power of $ 3 $ .
 $ 28672 = {2^3} \times {2^3} \times {2^3} \times {2^3} \times 7 $
Now, by grouping we have one factor $ 7 $ , is left.
Therefore, $ 28672 $ is not a perfect cube.
So, now to make $ 28672 $ as a perfect cube, we have to divide by $ 7 $ .
Hence, the smallest number by which $ 28672 $ must be divided to obtain a perfect cube is $ 7 $ .

D) The given value is $ 13718 $ .
Similarly, by prime factorization,
 $ 13718 = 2 \times 19 \times 19 \times 19 $
Now, to obtain a perfect cube, group the factors into triplets i.e., to the power of $ 3 $ .
 $ 13718 = 2 \times {19^3} $
Now, by grouping we have one factor $ 2 $ , is left.
Therefore, $ 13718 $ is not a perfect cube.
So, now to make $ 13718 $ as a perfect cube, we have to divide by $ 2 $ .
Hence, the smallest number by which $ 13718 $ must be divided to obtain a perfect cube is $ 2 $

Note: In this question, it is important to note that a number is a perfect cube only when each factor in the prime factorization is grouped in triples. Be careful during prime factorization. Prime factorization is also known as prime decomposition. And, it is a method of breaking a number down into the set of prime numbers which multiply together to result in the original number.

We have 10985 = 5 × 13 × 13 × 13

= 5 × 13 × 13 × 13

Here we have a triplet of 13 and we are left over with 5.

If we divide 10985 by 5, the new number will be a perfect cube.

∴ The required number is 5.

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