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Narcissistic cube asks who are we?

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Narcissistic cube asks who are we?

Cutting from a cube (visualization test)Who likes Actress E?Nine identical spheres fit exactly into a cubeMaximise Steve's step counter reset processMake numbers 1 - 32 using the digits 2, 0, 1, 7Escape the cube puzzleEscape the cube 2 (Theseus and Minotaur variant)How many liars are in the room?Make numbers 1 - 30 using the digits 2, 0, 1, 8Make numbers 93 using the digits 2, 0, 1, 8

5

$begingroup$

I am a five digit narcissistic cube who likes to display my root right upfront. I also have hidden it in the back of my four digit cube brother. Who are we?

mathematics logical-deduction no-computers

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asked 4 hours ago

UvcUvc

3686

$endgroup$

add a comment | 

5

$begingroup$

I am a five digit narcissistic cube who likes to display my root right upfront. I also have hidden it in the back of my four digit cube brother. Who are we?

mathematics logical-deduction no-computers

share|improve this question

asked 4 hours ago

UvcUvc

3686

$endgroup$

add a comment | 

$begingroup$

I am a five digit narcissistic cube who likes to display my root right upfront. I also have hidden it in the back of my four digit cube brother. Who are we?

mathematics logical-deduction no-computers

share|improve this question

asked 4 hours ago

UvcUvc

3686

$endgroup$

share|improve this question

asked 4 hours ago

UvcUvc

3686

share|improve this question

asked 4 hours ago

UvcUvc

3686

asked 4 hours ago

UvcUvc

3686

asked 4 hours ago

UvcUvc

3686

1 Answer
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oldest

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7

$begingroup$

$32768 = 32^3$ and $5832 = 18^3$

Process of finding them:

The cube root of a 5 digit number would have to be 2 digits. So say, $AB^3 = ABCDE$. Then, divide by $AB$ to get an approximation $AB^2 approx 1000$. $sqrt1000 approx 31.6$, so it makes sense to try slightly larger (since our approximation is a clear underestimate). We can try $32^3 = 2^15 = 32768$ and sure enough it satisfies our condition. ($33^3 = 1089 cdot 33 > 1100 cdot 32 = 35200$ can be tested to not work, so this is our five digit cube).

To get the four-digit number, we know that $32$ appears at the end of the cube. The cubes have residues $0, 1, 8, 7, 4, 5, 6, 3, 2, 9 bmod 10$, so we know that the cube root must be $8 bmod 10$. Since our four-digit number must have a two-digit cube root, $18^3$ makes the most sense to try ($28^3 > 784 cdot 20 > 15000$ would be way too large). So therefore we have our answer.

Afternote

Narcissistic numbers have an alternative meaning, so this was a little misleading for awhile.

Perhaps I should justify $1089 cdot 33 > 1100 cdot 32$. Well, $1100$ is roughly 1% larger than 1089 (it's 11 more), whereas $32$ is roughly 3% smaller than $33$.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

phenomistphenomist

9,1423555

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You just beat me to the writeup. :)
  $endgroup$
  – Rubio♦
  4 hours ago
  

  
  
  
  

add a comment | 

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7

$begingroup$

$32768 = 32^3$ and $5832 = 18^3$

Process of finding them:

The cube root of a 5 digit number would have to be 2 digits. So say, $AB^3 = ABCDE$. Then, divide by $AB$ to get an approximation $AB^2 approx 1000$. $sqrt1000 approx 31.6$, so it makes sense to try slightly larger (since our approximation is a clear underestimate). We can try $32^3 = 2^15 = 32768$ and sure enough it satisfies our condition. ($33^3 = 1089 cdot 33 > 1100 cdot 32 = 35200$ can be tested to not work, so this is our five digit cube).

To get the four-digit number, we know that $32$ appears at the end of the cube. The cubes have residues $0, 1, 8, 7, 4, 5, 6, 3, 2, 9 bmod 10$, so we know that the cube root must be $8 bmod 10$. Since our four-digit number must have a two-digit cube root, $18^3$ makes the most sense to try ($28^3 > 784 cdot 20 > 15000$ would be way too large). So therefore we have our answer.

Afternote

Narcissistic numbers have an alternative meaning, so this was a little misleading for awhile.

Perhaps I should justify $1089 cdot 33 > 1100 cdot 32$. Well, $1100$ is roughly 1% larger than 1089 (it's 11 more), whereas $32$ is roughly 3% smaller than $33$.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

phenomistphenomist

9,1423555

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You just beat me to the writeup. :)
  $endgroup$
  – Rubio♦
  4 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

$32768 = 32^3$ and $5832 = 18^3$

Process of finding them:

The cube root of a 5 digit number would have to be 2 digits. So say, $AB^3 = ABCDE$. Then, divide by $AB$ to get an approximation $AB^2 approx 1000$. $sqrt1000 approx 31.6$, so it makes sense to try slightly larger (since our approximation is a clear underestimate). We can try $32^3 = 2^15 = 32768$ and sure enough it satisfies our condition. ($33^3 = 1089 cdot 33 > 1100 cdot 32 = 35200$ can be tested to not work, so this is our five digit cube).

To get the four-digit number, we know that $32$ appears at the end of the cube. The cubes have residues $0, 1, 8, 7, 4, 5, 6, 3, 2, 9 bmod 10$, so we know that the cube root must be $8 bmod 10$. Since our four-digit number must have a two-digit cube root, $18^3$ makes the most sense to try ($28^3 > 784 cdot 20 > 15000$ would be way too large). So therefore we have our answer.

Afternote

Narcissistic numbers have an alternative meaning, so this was a little misleading for awhile.

Perhaps I should justify $1089 cdot 33 > 1100 cdot 32$. Well, $1100$ is roughly 1% larger than 1089 (it's 11 more), whereas $32$ is roughly 3% smaller than $33$.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

phenomistphenomist

9,1423555

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You just beat me to the writeup. :)
  $endgroup$
  – Rubio♦
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

$32768 = 32^3$ and $5832 = 18^3$

Process of finding them:

The cube root of a 5 digit number would have to be 2 digits. So say, $AB^3 = ABCDE$. Then, divide by $AB$ to get an approximation $AB^2 approx 1000$. $sqrt1000 approx 31.6$, so it makes sense to try slightly larger (since our approximation is a clear underestimate). We can try $32^3 = 2^15 = 32768$ and sure enough it satisfies our condition. ($33^3 = 1089 cdot 33 > 1100 cdot 32 = 35200$ can be tested to not work, so this is our five digit cube).

To get the four-digit number, we know that $32$ appears at the end of the cube. The cubes have residues $0, 1, 8, 7, 4, 5, 6, 3, 2, 9 bmod 10$, so we know that the cube root must be $8 bmod 10$. Since our four-digit number must have a two-digit cube root, $18^3$ makes the most sense to try ($28^3 > 784 cdot 20 > 15000$ would be way too large). So therefore we have our answer.

Afternote

Narcissistic numbers have an alternative meaning, so this was a little misleading for awhile.

Perhaps I should justify $1089 cdot 33 > 1100 cdot 32$. Well, $1100$ is roughly 1% larger than 1089 (it's 11 more), whereas $32$ is roughly 3% smaller than $33$.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

phenomistphenomist

9,1423555

$endgroup$

share|improve this answer

edited 4 hours ago

answered 4 hours ago

phenomistphenomist

9,1423555

answered 4 hours ago

phenomistphenomist

9,1423555

answered 4 hours ago

phenomistphenomist

9,1423555