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Try this beautiful problem from Algebra based on Perfect cubes from AMC-8, 2018, Problem -25.

How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?

- 55
- 60
- 58

Algebra

cube

integer

But try the problem first...

Answer:$58$

Source

Suggested Reading

AMC-8, 2018 problem 25

Challenges and Thrills in Pre College Mathematics

First hint

Find the vale 0f \(2^8+1\) and \(2^{18} +1\)

Can you now finish the problem ..........

Second Hint

Find the lest and largest cubes betwwen \(2^8+1\) and \(2^{18} +1\)

can you finish the problem........

Final Step

The value of \(2^8+1=257\)

Now less than 257 the perfect cube is \(6^3\)=216 and \(7^3=343\),which is greater than 257

Now \(2^{18}=(2^6)^3=(64)^3\) which will be the largrst cube less than \(2^{18 }+1\)

Hece the required number of cubes is 64-7+1=58

- https://www.cheenta.com/problem-based-on-integer-prmo-2018-problem-6/
- https://www.youtube.com/watch?v=ilbU2K2wS1I

Contents

[hide]

Try this beautiful problem from Algebra based on Perfect cubes from AMC-8, 2018, Problem -25.

How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?

- 55
- 60
- 58

Algebra

cube

integer

But try the problem first...

Answer:$58$

Source

Suggested Reading

AMC-8, 2018 problem 25

Challenges and Thrills in Pre College Mathematics

First hint

Find the vale 0f \(2^8+1\) and \(2^{18} +1\)

Can you now finish the problem ..........

Second Hint

Find the lest and largest cubes betwwen \(2^8+1\) and \(2^{18} +1\)

can you finish the problem........

Final Step

The value of \(2^8+1=257\)

Now less than 257 the perfect cube is \(6^3\)=216 and \(7^3=343\),which is greater than 257

Now \(2^{18}=(2^6)^3=(64)^3\) which will be the largrst cube less than \(2^{18 }+1\)

Hece the required number of cubes is 64-7+1=58

- https://www.cheenta.com/problem-based-on-integer-prmo-2018-problem-6/
- https://www.youtube.com/watch?v=ilbU2K2wS1I

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