Question

find the smallest number by which 1800 should be divided so that the quotient is a perfect cube.​

Answer 1

Answer:

The possibilities are endless. The required numbers include 225, 1/15, 1800, 25/3, 9/5, etc. So, let me show you how to find those out.

First, you need to break 1800 into its prime factors.

1800=2³×3²×5²

To make 1800 a perfect cube, you must make sure that it is a product of perfect cubes. For that you'll need to multiply (which is inverse of division) and/or divide it with certain numbers. The ways to do that could be

Multiplying it by 3×5=15

1800×15=2³×3³×5³=(2×3×5)³=30³

Multiplying by 15 can also be written as dividing by 1/15

Multiplying it by 3 and dividing by 25

1800×3÷25=2³×3³=(2×3)³=6³

It can also be written as dividing by 25/3

Multiplying it by 5 and dividing by 9

1800×5÷9=2³×5³=(2×5)³=10³

It can also be written as dividing by 9/5

Dividing it by 25×9=225

1800÷225=2³

As I said, dividing by 225. XD

Now, you can multiply or divide the above results with any perfect cube, the answer will always be a perfect cube. For example:

Dividing 1800/225 by 2³

2³÷2³=1=1³

That is basically dividing 1800 by 1800

This one is my favourite

Dividing 1800/(25/3) by 2³

6³÷2³=3³

That is basically dividing 1800 by 200/3

So as you can see, these random possibilities go on and on forever from now on.

And, if you're anything like me, you'll surely like to question dividing 1800 by ∞

1800÷∞=0=0³

And vice versa too, of course

But, are these true?

Sadly, no, they aren't. As they are not defined :(

Hope you got your answer now.

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