Question

How ;many perfect cubes are there between 1 and 100000 which are divisible by 7?

Answer 1

Answer:

Step-by-step explanation:

If we are finding the numbers which are divided by 7 and the number should be a perfect cube.

So, there will be two-component of number, one is divided by 7, and the second is multiple. But here as it is asked a perfect cube so we have to cube both the numbers.

The format of the number will be like this :

$= 7^3 * A^3$

So the numbers are :

• 349 * 1
• 349 *8
• 349*27
• 349*64
• 349*125
• 349*216

The next will be 349*349 which will be more than 100000.

So, there will be 6 Numbers which will follow this criterion.

Answer 2

$(7 \times n)^{3} \:is \: always \: divisible \:by \:7$

$\blue { ( Where \: n \: is \: a \: natural \: number)}$

1. If n = 1 then (7× 1)³ = 7³ = 343
2. If n = 2 then (7×2)³ = 14³= 2744
3. If n = 3 then (7×3)³ = 21³= 9261
4. If n = 4 then (7×4)³ = 28³= 21952
5. If n = 5 then (7×5)³ = 35³= 42875
6. If n = 6 then (7×6)³ = 42³= 74088
7. If n = 7 then (7×7)³ = 49³= 117649 > 100000

$perfect \:cubes \: between \: 1 \:and \\100000\:which \:are \: divisible \:by \:7 \\= \{ 343,2744,9261,21952,42875,74088\}$

Therefore.,

$\red{ Number \:of \: perfect \:cubes \: between }\\\red{ 1 \:and \:100000 \:which \:are }\\\red{divisible \:by \:7}\green {= 6 }$

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