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?
Algebra
cube
integer
But try the problem first...
Answer:$58$
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
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
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?
Algebra
cube
integer
But try the problem first...
Answer:$58$
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
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
