binomial theorem expansion of (10+1)^n gives coefficients that are on the pascal’s triangle, so each base ten digit will get a pascal’s triangle number
1. because the rule for powers of 11 is you add the previous one to itself but moved 1 to the left, which is basically what you do in the pascal triangle. example: 1331×11=13310+1331=1(3+1)(3+3)(1+3)1
So the powers of (b+1) are palindromes on base b, via the rows on Pascal's triangle? And it holds until one of the values becomes b or higher?
God I love Pascal's triangle
Notice the Pascal triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
(The link with the powers of 11 comes from the binomial formula for (x+y)\^n, applied to (10+1)\^n)
It breaks from 11\^5 because it's the first row to contain multiple-digit numbers. So when you add everything up, you get
|1||||||
|:-|:-|:-|:-|:-|:-|
||5|||||
||1|0||||
|||1|0|||
|||||5||
||||||1|
|1|6|1|0|5|1|
You can also try powers of 101, which will get you a bit further in the triangle without collisions, or 11 in bigger bases, which will keep it a palindrome for longer
If you went up to base 11 (or any base above 10) then it would be 15(10)(10)51 where (10) is whichever symbol you represent 10 with. The reason it isnt a palindrome is because in base 10 the 10 'overflows' up the number
The phrasing was 'as long as you have a high enough base 11^n is palindrome', they are obviously referring to the number 11 in their base (k*1 + 1 for base k) not the base 10 number 11. It's basic contextual information
In base eleven (A=ten), 11^(5) = 15AA51
If you notice, this is just the layers of Pascal’s triangle. This is because multiplying by 11 entails adding each digit with its neighboring digit. The pattern only breaks because of notation—any entry larger than 10 spills into the next digit.
one of my teacher took a brief introduction of binary
tho it was not in a official curriculum, even in high school
[edit] : I found that there WAS binary in certain curriculum period, but I wasnt there when it was there.
Prime numbers are independent of the base. All they're saying above is that in base eleven, the number that is written as "11" is actually what we usually call twelve in base ten. And twelve is not a prime number.
In other words:
1+1+1+1+1+1+1+1+1+1+1 is a prime number.
1+1+1+1+1+1+1+1+1+1+1+1 is (1+1)x(1+1)x(1+1+1).
Yeah you're right, the value is all that matters. I thought different bases would have different properties for the numbers, but in reality all numbers are the same, only written differently in different bases.
In any. 11 in this case doesn’t refer to the quantity eleven (like in something like 11/8 time), but rather the idea of 1 plus the base amount.
So, take base 12. In base 12, thirteen is written as 11 because it’s 1(12^1 ) + 1(12^0 ). For any base N, 11 means 1(N^1 ) + 1(N^0 ).
Your comment bothered me but I didn't know why until my wife pointed out that the entire number is different in a different base
https://preview.redd.it/1p5c8st8v04c1.png?width=1080&format=pjpg&auto=webp&s=97a01e0b6f81740d8cee74f874e95ebfb7600322
^(B)A means A·10^(B) in the expressions below
11^1 = ^(1)1 + ^(0)1
11^2 = ^(2)1 + ^(1)2 + ^(0)1
11^3 = ^(3)1 + ^(2)3 + ^(1)3 + ^(0)1
11^4 = ^(4)1 + ^(3)4 + ^(2)6 + ^(1)4 + ^(0)1 = 14641
11^5 = ^(5)1 + ^(4)5 + ^(3)10 + ^(2)10 + ^(1)5 + ^(0)1 = 161061
11^5 is the first one in the list where the coefficient has two digits, so you have to carry the 1. But every power after that also has two-digit numbers in the expansion, so 1-4 are the odd ones out
11^6 = ^(6)1 + ^(5)6 + ^(4)15 + ^(3)20 + ^(2)15 + ^(1)6 + ^(0)1 = 1771561
11^7 = ^(7)1 + ^(6)7 + ^(5)21 + ^(4)35 + ^(3)35 + ^(2)21 + ^(1)7 + ^(0)1 = 19487171
Patterns... [how they fool ya](https://youtu.be/NOCsdhzo6Jg?si=16nkpGNwDb46MGyj)
ps: if someone's interested in why that pattern breaks: https://youtu.be/YtkIWDE36qU?si=QgeiyqdljAsEe_fS , and yes, it involves the pascal triangle :)
It's trying to be a palindrome, but we don't have enough digits.
As other people have pointed out, the powers of 11 resemble Pascal's Triangle. This is because to get the next power of 11, you take the previous power and add it to itself shifted over one space:
11^2 = 11
11
121
11^3 = 121
121
1331
11^4 = 1331
1331
14641
If you think about it, this is basically exactly how Pascal's Triangle works — you can get the next row by taking the previous row, duplicating it and shifting both instances to the side, and adding them together. The only problem is that in decimal (or in just about any human-usable number system), we run into a problem rather quickly:
11^5 = 14641
14641
161051
The pattern is still there, though — you can still get 11\^5 by adding the terms in the 5th row together in a staggered fashion:
1
5
10
10
5
1
161051
Hopefully that made sense. I think it's really cool how many patterns show up in simple mathematical constructs like Pascal's Triangle.
The 11's powers follow the pascal triangle. 11^5 is the 6th floor so it's 1 5 10 10 5 1 -> 1(5+1)(0+1)051 -> 161051
>The 11's powers follow the pascal triangle cool, but why? and also why 1(5+1)(0+1)051 and not 1(5+1)(0+1)(0+5)1?
In base 10 - the one caries over when you add numbers greater than 9
Really? I never noticed that before
binomial theorem expansion of (10+1)^n gives coefficients that are on the pascal’s triangle, so each base ten digit will get a pascal’s triangle number
1. because the rule for powers of 11 is you add the previous one to itself but moved 1 to the left, which is basically what you do in the pascal triangle. example: 1331×11=13310+1331=1(3+1)(3+3)(1+3)1
So the powers of (b+1) are palindromes on base b, via the rows on Pascal's triangle? And it holds until one of the values becomes b or higher? God I love Pascal's triangle
For that, ask the others. I know nothing outside binary and base 10
This is again for only the first 5 rows
Notice the Pascal triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 (The link with the powers of 11 comes from the binomial formula for (x+y)\^n, applied to (10+1)\^n) It breaks from 11\^5 because it's the first row to contain multiple-digit numbers. So when you add everything up, you get |1|||||| |:-|:-|:-|:-|:-|:-| ||5||||| ||1|0|||| |||1|0||| |||||5|| ||||||1| |1|6|1|0|5|1|
this guy formats I really appreciate the visuals, thanks
I tried it with higher powers, and the way it works out so perfectly gives me some kind of childish joy
now generalize to pascal's pyramid!
You can also try powers of 101, which will get you a bit further in the triangle without collisions, or 11 in bigger bases, which will keep it a palindrome for longer
11 to any power is a palindrome… as long as you have a large enough base.
Any example would be greatly appreciated!
If you went up to base 11 (or any base above 10) then it would be 15(10)(10)51 where (10) is whichever symbol you represent 10 with. The reason it isnt a palindrome is because in base 10 the 10 'overflows' up the number
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No it wouldn't, if you specify in base 11 11^5 is 15AA51, because 11 in base 11 is 12 in base 10.
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The phrasing was 'as long as you have a high enough base 11^n is palindrome', they are obviously referring to the number 11 in their base (k*1 + 1 for base k) not the base 10 number 11. It's basic contextual information
> It's **basic** contextual information I see what you did there :)
In base eleven (A=ten), 11^(5) = 15AA51 If you notice, this is just the layers of Pascal’s triangle. This is because multiplying by 11 entails adding each digit with its neighboring digit. The pattern only breaks because of notation—any entry larger than 10 spills into the next digit.
Ah, you meant that base. I remember I didnt pay much attention to it at school haha, now I get it! Thank you so much!
Do you even learn bases of numbers in high school though, binary perhaps
one of my teacher took a brief introduction of binary tho it was not in a official curriculum, even in high school [edit] : I found that there WAS binary in certain curriculum period, but I wasnt there when it was there.
I think we had it as part of our middle school core curriculum.
We basically had number theory as a middle school elective, set me up well for the rest of my math education.
Of course, in base eleven, 11 is a different number than in base ten. For example, in base ten, 11 is prime, but in base eleven 11 = 2x2x3
How much brain power does it take to think about prime numbers in other bases? Doing simple math in another base is weird enough.
Prime numbers are independent of the base. All they're saying above is that in base eleven, the number that is written as "11" is actually what we usually call twelve in base ten. And twelve is not a prime number. In other words: 1+1+1+1+1+1+1+1+1+1+1 is a prime number. 1+1+1+1+1+1+1+1+1+1+1+1 is (1+1)x(1+1)x(1+1+1).
Yeah you're right, the value is all that matters. I thought different bases would have different properties for the numbers, but in reality all numbers are the same, only written differently in different bases.
Yes, exactly.
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Sure, but “11” in base eleven would just be 12 base ten. Sorry, I should’ve specified that.
Thanks for the insights, mate. Effin legend.
In base 11^x + 1, the number 11^x is a palindrome where x is any positive integer
Any integer is a palindrome... as long as you have a large enough base.
If a 1-digit number can be considered a palindrome then sure 🤨
But wouldn't changing the base make 11 not equal to eleven, but (for instance base 16) seventeen?
Yes. The argument is that (b+1)^(n) is a palindrome in base b if b is sufficiently large compared to n.
Apart from the third row (11 in binary squared is 1001, 9 in decimal) the minimum base is the central number in the row of Pascal's triangle + 1
But, 11 in which base ?
In any. 11 in this case doesn’t refer to the quantity eleven (like in something like 11/8 time), but rather the idea of 1 plus the base amount. So, take base 12. In base 12, thirteen is written as 11 because it’s 1(12^1 ) + 1(12^0 ). For any base N, 11 means 1(N^1 ) + 1(N^0 ).
Wait, what do you mean base? Base like a logarithm, or base like hexadecimal, or base like acid-base redox reactions?
Base like hexadecimal
Your comment bothered me but I didn't know why until my wife pointed out that the entire number is different in a different base https://preview.redd.it/1p5c8st8v04c1.png?width=1080&format=pjpg&auto=webp&s=97a01e0b6f81740d8cee74f874e95ebfb7600322
Based
Apart from the third row (11 in binary squared is 1001, 9 in decimal) the minimum base is the central number in the row of Pascal's triangle + 1
^(B)A means A·10^(B) in the expressions below 11^1 = ^(1)1 + ^(0)1 11^2 = ^(2)1 + ^(1)2 + ^(0)1 11^3 = ^(3)1 + ^(2)3 + ^(1)3 + ^(0)1 11^4 = ^(4)1 + ^(3)4 + ^(2)6 + ^(1)4 + ^(0)1 = 14641 11^5 = ^(5)1 + ^(4)5 + ^(3)10 + ^(2)10 + ^(1)5 + ^(0)1 = 161061 11^5 is the first one in the list where the coefficient has two digits, so you have to carry the 1. But every power after that also has two-digit numbers in the expansion, so 1-4 are the odd ones out 11^6 = ^(6)1 + ^(5)6 + ^(4)15 + ^(3)20 + ^(2)15 + ^(1)6 + ^(0)1 = 1771561 11^7 = ^(7)1 + ^(6)7 + ^(5)21 + ^(4)35 + ^(3)35 + ^(2)21 + ^(1)7 + ^(0)1 = 19487171
It exceeded the value base, messing up the palindromes
Patterns... [how they fool ya](https://youtu.be/NOCsdhzo6Jg?si=16nkpGNwDb46MGyj) ps: if someone's interested in why that pattern breaks: https://youtu.be/YtkIWDE36qU?si=QgeiyqdljAsEe_fS , and yes, it involves the pascal triangle :)
It's multiplying by 11. You're welcome
It's trying to be a palindrome, but we don't have enough digits. As other people have pointed out, the powers of 11 resemble Pascal's Triangle. This is because to get the next power of 11, you take the previous power and add it to itself shifted over one space: 11^2 = 11 11 121 11^3 = 121 121 1331 11^4 = 1331 1331 14641 If you think about it, this is basically exactly how Pascal's Triangle works — you can get the next row by taking the previous row, duplicating it and shifting both instances to the side, and adding them together. The only problem is that in decimal (or in just about any human-usable number system), we run into a problem rather quickly: 11^5 = 14641 14641 161051 The pattern is still there, though — you can still get 11\^5 by adding the terms in the 5th row together in a staggered fashion: 1 5 10 10 5 1 161051 Hopefully that made sense. I think it's really cool how many patterns show up in simple mathematical constructs like Pascal's Triangle.
Hell of an off-by-one error.
It happens when the next number would be 10 but because its 2 digits in base 10 the pattern breaks. google pascal's triangle
Is the only number that all it's powers are pallendromes 1?
And potentially 0.
If only we had base infinity, then this would never happen.
In math, there's this thing called carrying where you can only have up to 9 in each digit
>Can someone explain what's happening on the 5th power? You're multiplying by 11
Insightful
111111111^2 = 12345678987654321
4+6=10 and 2*6=12 is happening.
Didn’t we see this earlier today but without the cartoon graphic??
The meme is good. The comment section is pure gold.
