What is the smallest number by which 8192 must be divided so that quotient is A?

Solution : By factorization,
`8192=2xx2xx2xx2xx2xx2xx2xx2xx2xx2xx2`
`8192=2^3xx2^3xx2^3xx2`
Since, one triples remain incomplete, hence `8192` is not a perfect cube.
So, we divide `8192` by `2` to make its quotient a perfect cube.
Since, `8192/2=4096`
`4096=2xx2xx2xx2xx2xx2xx2xx2xx2xx2xx2xx2`
`4096 =2^3times2^3xx2^3xx2^3`
Cube root of `4096`,
`root(3)(4096)=root(3)(2^3xx2^3xx2^3xx2^3)`
`root(3)(4096)=2xx2xx2xx2=16`

Given:

210125

To do:

We have to find the smallest number by which 8192 must be divided so that quotient is a perfect cube and find the cube root of the product.

Solution: 

Prime factorisation of 8192 is,

$8192=2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2$

$=2^3\times2^3\times2^3\times2^3\times2$

Grouping the factors in triplets of equal factors, we see that $2$ is left.

In order to make 8192 a perfect cube, we have to divide it by $2$.

$8192\div2=2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\div2$

$=2^3\times2^3\times2^3\times2^3\times2\div2$

$\sqrt[3]{4096}=\sqrt[3]{2^3\times2^3\times2^3\times2^3}$

$=2\times2\times2\times2$

$=16$

The smallest number by which 8192 must be divided so that the quotient is a perfect cube is 2 and the cube root of the quotient is 16.

Here,

8192 = 2×2×2×2×2×2×2×2×2×2×2 

= 23 ×23 ×23×2

Since, one triples remain incomplete, hence 8192 is not a perfect cube.

So, we divide 8192 by 2 to make its quotient a perfect cube.

Since,

8192/2 = 4096

4096 = 2×2×2×2×2×2×2×2×2×2×2×2 

= 23 ×23 ×23 ×23

Cube root of 4096, 

∛4096 = ∛(23×23×23×23) 

= 2×2×2×2 

= 16

Given:

210125

To do:

We have to find the smallest number by which 8192 must be divided so that quotient is a perfect cube and find the cube root of the product.

Solution: 

Prime factorisation of 8192 is,

$8192=2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2$

$=2^3\times2^3\times2^3\times2^3\times2$

Grouping the factors in triplets of equal factors, we see that $2$ is left.

In order to make 8192 a perfect cube, we have to divide it by $2$.

$8192\div2=2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\div2$

$=2^3\times2^3\times2^3\times2^3\times2\div2$

$\sqrt[3]{4096}=\sqrt[3]{2^3\times2^3\times2^3\times2^3}$

$=2\times2\times2\times2$

$=16$

The smallest number by which 8192 must be divided so that the quotient is a perfect cube is 2 and the cube root of the quotient is 16.

Updated on 10-Oct-2022 12:46:54

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On factorising 8192 into prime factors, we get:

\[8192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

On grouping the factors in triples of equal factors, we get:

\[8192 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times 2\]

It is evident that the prime factors of 8192 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8192 is not a perfect cube. However, if the number is divided by 2, the factors can be grouped into triples of equal factors such that no factor is left over.
Hence, the number 8192 should be divided by 2 to make it a perfect cube.
Also, the quotient is given as:

\[\frac{8192}{2} = \frac{\left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times 2}{2}\]
\[ \Rightarrow 4096 = \left\{ 2 \times 2 \times 2 \right\}\times\left\{ 2 \times 2 \times 2 \right\}\times\left\{ 2 \times 2 \times 2 \right\}\times\left\{ 2 \times 2 \times 2 \right\}\]

To get the cube root of the quotient 4096, take one factor from each triple. We get: 
Cube root =  \[2 \times 2 \times 2 \times 2 = 16\]

Hence, the required numbers are 2 and 16.

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