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Question

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:

A
1323
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B
7803
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C
35721
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D
2560
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Solution

The correct option is A 1323
(i)1323
We see that 1323=3×3×3×7×7
So 7 does not occur in triplets,
1323 is not a perfect cube.
So, we multiply 7 to make triplet
So, our number becomes 1323×7=3×3×3×7×7×7=9621
Hence 7 is the smallest number to be multiplied with 1323 to get a perfect cube number 9621
Cube root of 9621=3×7=21
(ii)7803
We see that 7803=3×3×3×17×17
So 17 does not occur in triplets,
7803 is not a perfect cube.
So, we multiply 17 to make triplet
So, our number becomes 7803×17=3×3×3×17×17×17=132,651
Hence 17 is the smallest number to be multiplied with 7803 to get a perfect cube number 132,651
Cube root of 132,651=3×17=51
(iii)35721
We see that 35721=3×3×3×3×3×3×7×7
35721=9×9×9×7×7
So 7 does not occur in triplets,
35721 is not a perfect cube.
So, we multiply 7 to make triplet
So, our number becomes 35721×7=9×9×9×7×7×7=250,047
Hence 7 is the smallest number to be multiplied with 35721 to get a perfect cube number 250,047
Cube root of 250,047=9×7=63
(iv)2560
We see that 2560=5×2×2×2×2×2×2×2×2×2
2560=5×8×8×8
So 5 does not occur in triplets,
2560 is not a perfect cube.
So, we multiply 25 to make triplet
So, our number becomes 2560×25=8×8×8×5×25=64,000
or 2560×25=8×8×8×5×5×5=64,000
Hence 25 is the smallest number to be multiplied with 2560 to get a perfect cube number 64,000
Cube root of 64,000=8×5=40

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