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Perfect cubes | Algebra | AMC 8, 2018 | Problem 25

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

Perfect Cubes | AMC-8, 2018| Problem 22


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

  • 55
  • 60
  • 58

Key Concepts


Algebra

cube

integer

Check the Answer


Answer:$58$

AMC-8, 2018 problem 25

Challenges and Thrills in Pre College Mathematics

Try with Hints


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

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

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

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

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

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

Perfect Cubes | AMC-8, 2018| Problem 22


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

  • 55
  • 60
  • 58

Key Concepts


Algebra

cube

integer

Check the Answer


Answer:$58$

AMC-8, 2018 problem 25

Challenges and Thrills in Pre College Mathematics

Try with Hints


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

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

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

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

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

Subscribe to Cheenta at Youtube


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