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How many perfect cubes are between $2^8+1$ to $2^{18}+1$ ( inclusively)
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How many perfect cubes are between $2^8+1$ to $2^{18}+1$ ( inclusively)
$2^9,2^{12},2^{15},2^{18}$ are all perfect cube.there are many other. I try to use modulo 2 .but it won't work, and no other methods i tried get me nowhere
Any ideas?
elementary-number-theory
edited Jan 8 at 16:34
ajotatxe
53.8k23890
asked Jan 8 at 16:31
Cloud JRCloud JR
883517
$endgroup$
add a comment |
$begingroup$
How many perfect cubes are between $2^8+1$ to $2^{18}+1$ ( inclusively)
$2^9,2^{12},2^{15},2^{18}$ are all perfect cube.there are many other. I try to use modulo 2 .but it won't work, and no other methods i tried get me nowhere
Any ideas?
elementary-number-theory
edited Jan 8 at 16:34
ajotatxe
53.8k23890
asked Jan 8 at 16:31
Cloud JRCloud JR
883517
$endgroup$
3
$begingroup$
Hint : Which is the smallest and which the largest number , such that its cube is in the given range ?
$endgroup$
– Peter
Jan 8 at 16:33
add a comment |
$begingroup$
How many perfect cubes are between $2^8+1$ to $2^{18}+1$ ( inclusively)
$2^9,2^{12},2^{15},2^{18}$ are all perfect cube.there are many other. I try to use modulo 2 .but it won't work, and no other methods i tried get me nowhere
Any ideas?
elementary-number-theory
edited Jan 8 at 16:34
ajotatxe
53.8k23890
asked Jan 8 at 16:31
Cloud JRCloud JR
883517
$endgroup$
How many perfect cubes are between $2^8+1$ to $2^{18}+1$ ( inclusively)
$2^9,2^{12},2^{15},2^{18}$ are all perfect cube.there are many other. I try to use modulo 2 .but it won't work, and no other methods i tried get me nowhere
Any ideas?
elementary-number-theory
elementary-number-theory
edited Jan 8 at 16:34
ajotatxe
53.8k23890
asked Jan 8 at 16:31
Cloud JRCloud JR
883517
edited Jan 8 at 16:34
ajotatxe
53.8k23890
asked Jan 8 at 16:31
Cloud JRCloud JR
883517
edited Jan 8 at 16:34
ajotatxe
53.8k23890
edited Jan 8 at 16:34
ajotatxe
53.8k23890
edited Jan 8 at 16:34
ajotatxe
53.8k23890
53.8k23890
asked Jan 8 at 16:31
Cloud JRCloud JR
883517
asked Jan 8 at 16:31
Cloud JRCloud JR
883517
asked Jan 8 at 16:31
Cloud JRCloud JR
883517
883517
3
$begingroup$
Hint : Which is the smallest and which the largest number , such that its cube is in the given range ?
$endgroup$
– Peter
Jan 8 at 16:33
add a comment |
3
$begingroup$
Hint : Which is the smallest and which the largest number , such that its cube is in the given range ?
$endgroup$
– Peter
Jan 8 at 16:33
3
3
$begingroup$
Hint : Which is the smallest and which the largest number , such that its cube is in the given range ?
$endgroup$
– Peter
Jan 8 at 16:33
$begingroup$
Hint : Which is the smallest and which the largest number , such that its cube is in the given range ?
$endgroup$
– Peter
Jan 8 at 16:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
For every positive integer $x$:
$$2^8+1le x^3le2^{18}+1iffsqrt[3]{2^8+1}le xle sqrt[3]{2^{18}+1}iff 7le xle 2^6$$
answered Jan 8 at 16:33
ajotatxeajotatxe
53.8k23890
$endgroup$
$begingroup$
Thanks i got it
$endgroup$
– Cloud JR
Jan 8 at 16:41
add a comment |
$begingroup$
One way to get around your very limited human sense of numbers is to take advantage of logarithms and roots whenever you can.
For problems like the one you're dealing with today, what you need is the computation of cubic roots. For example, $10$ is not a perfect cube, but we see that $root 3 of {10} approx 2.15$, and then we check that $2^3 < 10 < 3^3$. What's more, this "$approx 2.15$" should tell you that $10$ is much closer to the next lower perfect cube than it is to the next higher cube.
So for your particular problem, you need to find that $root 3 of {2^8 + 1} approx 6.35$ and $root 3 of {2^{18} + 1} approx 64.00008$.
answered Jan 8 at 23:58
The Short OneThe Short One
5941624
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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votes
$begingroup$
Hint:
For every positive integer $x$:
$$2^8+1le x^3le2^{18}+1iffsqrt[3]{2^8+1}le xle sqrt[3]{2^{18}+1}iff 7le xle 2^6$$
answered Jan 8 at 16:33
ajotatxeajotatxe
53.8k23890
$endgroup$
$begingroup$
Thanks i got it
$endgroup$
– Cloud JR
Jan 8 at 16:41
add a comment |
$begingroup$
Hint:
For every positive integer $x$:
$$2^8+1le x^3le2^{18}+1iffsqrt[3]{2^8+1}le xle sqrt[3]{2^{18}+1}iff 7le xle 2^6$$
answered Jan 8 at 16:33
ajotatxeajotatxe
53.8k23890
$endgroup$
$begingroup$
Thanks i got it
$endgroup$
– Cloud JR
Jan 8 at 16:41
add a comment |
$begingroup$
Hint:
For every positive integer $x$:
$$2^8+1le x^3le2^{18}+1iffsqrt[3]{2^8+1}le xle sqrt[3]{2^{18}+1}iff 7le xle 2^6$$
answered Jan 8 at 16:33
ajotatxeajotatxe
53.8k23890
$endgroup$
Hint:
For every positive integer $x$:
$$2^8+1le x^3le2^{18}+1iffsqrt[3]{2^8+1}le xle sqrt[3]{2^{18}+1}iff 7le xle 2^6$$
answered Jan 8 at 16:33
ajotatxeajotatxe
53.8k23890
answered Jan 8 at 16:33
ajotatxeajotatxe
53.8k23890
answered Jan 8 at 16:33
ajotatxeajotatxe
53.8k23890
answered Jan 8 at 16:33
ajotatxeajotatxe
53.8k23890
53.8k23890
$begingroup$
Thanks i got it
$endgroup$
– Cloud JR
Jan 8 at 16:41
add a comment |
$begingroup$
Thanks i got it
$endgroup$
– Cloud JR
Jan 8 at 16:41
$begingroup$
Thanks i got it
$endgroup$
– Cloud JR
Jan 8 at 16:41
$begingroup$
Thanks i got it
$endgroup$
– Cloud JR
Jan 8 at 16:41
add a comment |
$begingroup$
One way to get around your very limited human sense of numbers is to take advantage of logarithms and roots whenever you can.
For problems like the one you're dealing with today, what you need is the computation of cubic roots. For example, $10$ is not a perfect cube, but we see that $root 3 of {10} approx 2.15$, and then we check that $2^3 < 10 < 3^3$. What's more, this "$approx 2.15$" should tell you that $10$ is much closer to the next lower perfect cube than it is to the next higher cube.
So for your particular problem, you need to find that $root 3 of {2^8 + 1} approx 6.35$ and $root 3 of {2^{18} + 1} approx 64.00008$.
answered Jan 8 at 23:58
The Short OneThe Short One
5941624
$endgroup$
add a comment |
$begingroup$
One way to get around your very limited human sense of numbers is to take advantage of logarithms and roots whenever you can.
For problems like the one you're dealing with today, what you need is the computation of cubic roots. For example, $10$ is not a perfect cube, but we see that $root 3 of {10} approx 2.15$, and then we check that $2^3 < 10 < 3^3$. What's more, this "$approx 2.15$" should tell you that $10$ is much closer to the next lower perfect cube than it is to the next higher cube.
So for your particular problem, you need to find that $root 3 of {2^8 + 1} approx 6.35$ and $root 3 of {2^{18} + 1} approx 64.00008$.
answered Jan 8 at 23:58
The Short OneThe Short One
5941624
$endgroup$
add a comment |
$begingroup$
One way to get around your very limited human sense of numbers is to take advantage of logarithms and roots whenever you can.
For problems like the one you're dealing with today, what you need is the computation of cubic roots. For example, $10$ is not a perfect cube, but we see that $root 3 of {10} approx 2.15$, and then we check that $2^3 < 10 < 3^3$. What's more, this "$approx 2.15$" should tell you that $10$ is much closer to the next lower perfect cube than it is to the next higher cube.
So for your particular problem, you need to find that $root 3 of {2^8 + 1} approx 6.35$ and $root 3 of {2^{18} + 1} approx 64.00008$.
answered Jan 8 at 23:58
The Short OneThe Short One
5941624
$endgroup$
One way to get around your very limited human sense of numbers is to take advantage of logarithms and roots whenever you can.
For problems like the one you're dealing with today, what you need is the computation of cubic roots. For example, $10$ is not a perfect cube, but we see that $root 3 of {10} approx 2.15$, and then we check that $2^3 < 10 < 3^3$. What's more, this "$approx 2.15$" should tell you that $10$ is much closer to the next lower perfect cube than it is to the next higher cube.
So for your particular problem, you need to find that $root 3 of {2^8 + 1} approx 6.35$ and $root 3 of {2^{18} + 1} approx 64.00008$.
answered Jan 8 at 23:58
The Short OneThe Short One
5941624
answered Jan 8 at 23:58
The Short OneThe Short One
5941624
answered Jan 8 at 23:58
The Short OneThe Short One
5941624
answered Jan 8 at 23:58
The Short OneThe Short One
5941624
5941624
add a comment |
add a comment |
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3
$begingroup$
Hint : Which is the smallest and which the largest number , such that its cube is in the given range ?
$endgroup$
– Peter
Jan 8 at 16:33