Mathematics

[18goyaan]

@Jian I recommend the book ‘Fermat’s Last Theorem’ by Simon Singh. It was the book prize for the 2018 JMO (look at the UKMT website if you don’t know what I mean).

[The-letter-wriggler]

Number Theory deals with the mathematics of the integers and a lot of the time BIG numbers.

Here is a sample (taken from the book Keys To Infinity by C. A. Pickover 1995.)

Narcissistic numbers “numbers in love with themselves,” Armstrong numbers, or perfect digital variants have fascinated many.
This is a number having each of it’s digits raised to the power of the number of the count of it’s digits, and the sum equal to the number.

123 (3 digits) would give 1^3+2^3+3^3 = 36 but 36 is not equal to 123 so is not a narcissistic number.

The highest number found in base 10 is this incredible 39 digit number:
115132219018763992565095597973971522401 (39 Digits to the power 39)

Here I have expanded the work-out…
====================================
1^39 = 1
1^39 = 1
5^39 = 1818989403545856475830078125
1^39 = 1
3^39 = 4052555153018976267
2^39 = 549755813888
2^39 = 549755813888
1^39 = 1
9^39 = 16423203268260658146231467800709255289
0^39 = 0
1^39 = 1
8^39 = 166153499473114484112975882535043072
7^39 = 909543680129861140820205019889143
6^39 = 2227915756473955677973140996096
3^39 = 4052555153018976267
9^39 = 16423203268260658146231467800709255289
9^39 = 16423203268260658146231467800709255289
2^39 = 549755813888
5^39 = 1818989403545856475830078125
6^39 = 2227915756473955677973140996096
5^39 = 1818989403545856475830078125
0^39 = 0
9^39 = 16423203268260658146231467800709255289
5^39 = 1818989403545856475830078125
5^39 = 1818989403545856475830078125
9^39 = 16423203268260658146231467800709255289
7^39 = 909543680129861140820205019889143
9^39 = 16423203268260658146231467800709255289
7^39 = 909543680129861140820205019889143
3^39 = 4052555153018976267
9^39 = 16423203268260658146231467800709255289
7^39 = 909543680129861140820205019889143
1^39 = 1
5^39 = 1818989403545856475830078125
2^39 = 549755813888
2^39 = 549755813888
4^39 = 302231454903657293676544
0^39 = 0
1^39 = 1
===============================================
Total = 115132219018763992565095597973971522401

=================================================
Highest found and shown in their respective base.
=================================================
Base(2): 1 [1 only obviously]
Base(3): 122 (3 Digits) [5 others lower]
Base(4): 3303 (4 Digits) [11 others lower]
Base(5): 14421440424444 (14 Digits) [17 others lower]
Base(6): 105144341423554535 (18 Digits) [30 others lower]
Base(7): 12616604301406016036306 (23 Digits) [59 others lower]
Base(8): 11254613377540170731271074472 (29 Digits) [62 others lower]
Base(9): 104836124432728001478001038311 (30 Digits) [58 others lower]
Base(10): 115132219018763992565095597973971522401 (39 Digits) [88 others lower]
Base(11): 12344AA12A721803422912A8AA4963568083A268456A4 (45 Digits) [45 others lower]
Base(12): 15079346A6B3B14BB56B395898B96629A8B01515344B4B0714B (51 Digits) [51 others lower]

Martin Gardner said it has been proven that narc. numbers are finite, for instance they cannot have more than 58 digits in base 10 (do you understand why?)

If you want to…
As an exercise for yourself find some of those in base 10.
To start with find the only four 3 digit numbers (base 10).

[Jian]

@18goyaan I have already read it! It was fun! Any more recommendations?

[18goyaan]

Just check out the UKMT books, while they are not in depth in literature and history they really help build knowledge in how to solve mathematical problems. Also, The Simpsons and Their Mathematical Secrets by Simon Singh.

[Person314]

Is there any way to find narcissistic numbers without brute force checking all (b-1)*b^(n-1) cases (where b is the base and n is the number length)? Because if that is the only way to find them then surely as n becomes large it just becomes a computing problem. Unless their is some underlying pattern or reason why it happens, aren’t they really nothing more than a mild curiosity? Still interesting though and thanks for the post.

[The-letter-wriggler]

@Person314
I only posted as a sample in number theory.
I didn’t put up the post with the intention of entering into a discussion as I have very little time to myself at this time. So Sorry on that count.

I would say a lot of N.T. is done out of curiosity. It is how to obtain a means to an end that drives the maths on, to find what is possible with the integers and how to implement a solve, and yes a lot of the time it just becomes a computing problem.

It is the very question that you asked that one needs to seek an answer to!

Unless one is doing math for a specific purpose, like in engineering, science and computing to obtain a real world answer, then other math is subsidary like Pure Mathematic for instance but they all help drive forward the advancements in math workings and understanding.

If anyone is interested in N.T. there are a lot of those types of questions to be found among the currently 737 over at:

https://projecteuler.net (it is free and a great source for math lovers, my opinion)

I myself have managed to solve many of them, some are hard, and for some I do not quite understand what the problem is asking! But I really like to have a go!

“Now how do I program my math package to solve number 737 ‘Coin Loops’?”…

[Kford-academy]

Can you prove the following statement?

For any positive integer n:
1^3 + 2^3 + 3^3 + … + n^3 = (1 + 2 + 3 + … + n)^2

[Kford-academy]

OK. I’ve waited long enough now, so here are my proofs for the following equation being true for any positive integer n:
1^3 + 2^3 + 3^3 + … + n^3 = (1 + 2 + 3 + … + n)^2

—

There are various ways to prove this one, but here are three proofs that I thought of.

PROOF ONE

This particular proof includes two separate proofs via induction, and is probably not for the easily baffled!

1. The statement is certainly true for n = 1, as 1^3 = 1^2 (both sides equal 1).

2. Assume that the statement is true for some positive integer m (so 1^3 + 2^3 + 3^3 + … + m^3 = (1 + 2 + 3 + … + m)^2). To prove that it is true for m+1, we go as follows:
1^3 + 2^3 + 3^3 + … + m^3 + (m + 1)^3 = (1 + 2 + 3 + … + m + (m + 1))^2
(1 + 2 + 3 + … + m)^2 + (m + 1)^3 = (1 + 2 + 3 + … + m)^2 + 2(1 + 2 + 3 + … + m)(m + 1) + (m + 1)^2
m^3 + 3m^2 + 3m + 1 = 2(1 + 2 + 3 + … + m)(m + 1) + m^2 + 2m + 1
m^3 + 2m^2 + m = 2(1 + 2 + 3 + … + m)(m + 1)
(m^2 + m)(m + 1) = 2(1 + 2 + 3 + … + m)(m + 1)
m^2 + m = 2(1 + 2 + 3 + … + m)

So all we need to do to prove the first statement is to prove that ‘n^2 + n’ equals ‘2(1 + 2 + 3 + … + n)’ (because of the law of induction). This goes as follows:

1. This statement is certainly true for n = 1, as 1^2 + 1 = 2*1 (both sides equal 2).

2. Assume that the statement is true for some positive integer m (so m^2 + m = 2(1 + 2 + 3 + … + m)). To prove that this statement is true for m+1, we go as follows:
(m + 1)^2 + (m + 1) = 2(1 + 2 + 3 + … + m + (m + 1))
m^2 + 2m + 1 + m + 1 = 2(1 + 2 + 3 + … + m) + 2(m + 1)
(m^2 + m) + (2m + 1 + 1) = 2(1 + 2 + 3 + … + m) + 2(m + 1)

We previously assumed m^2 + m = 2(1 + 2 + 3 + … + m), so:
2(1 + 2 + 3 + … + m) + (2m + 1 + 1) = 2(1 + 2 + 3 + … + m) + 2(m + 1)
2m + 1 + 1 = 2(m + 1)
2m + 2 = 2m + 2

This is obviously true, so n^2 + n = 2(1 + 2 + 3 + … + n). It follows that 1^3 + 2^3 + 3^3 + … + n^3 = (1 + 2 + 3 + … + n)^2.

PROOF TWO

If you don’t like using induction, this proof simplifies each side into a nice, simple expression (and let’s be honest, who thinks it would be simpler to work with expressions with undefined lengths rather than – at the most – quartic expressions?) and proves that both sides are equal to one another. (Note: more on quartic expressions later…)

1. The sequence 1, 8, 27, 64, … (i.e. 1^3, 2^3, 3^3, …) is a cubic sequence (for obvious reasons), and 8, 27, 64, 125, … (i.e. 2^3, 3^3, 4^3, …) certainly is too. The sequence 1^3, 1^3 + 2^3, 1^3 + 2^3 + 3^3, … (the left hand side of the equation as n is equal to 1, 2, 3, and so on) has a difference of 8, 27, 64, 125, (etc) and so is a quartic sequence (i.e. the nth term is an^4 + bn^3 + cn^2 + dn + e). In fact, this particular sequence is (n^4 + 2n^3 + n^2)/4 (I leave this for you to prove), which can be factorised down to n^2 * (n + 1)^2 / 4.

2. The sequence 1, 2, 3, 4, … is a linear sequence (obviously), and therefore 2, 3, 4, 5, … is as well. This means that, because 1, 1+2, 1+2+3, 1+2+3+4, … (i.e. the right hand side of the sequence, without the square symbol) has a difference of 2, 3, 4, 5, and so on (where n is equal to 1, 2, 3, 4, (etc) respectively), it is a quadratic sequence. Again, I leave you to prove this (should you wish!), but it is equal to (n^2 + n)/2, or n(n+1)/2 when factorised.

3. So we get:

n^2 * (n + 1)^2 / 4 = (n(n+1)/2)^2

The right hand side is equal to n^2 * (n+1)^2 / 2^2, or n^2 * (n + 1)^2 / 4, exactly the same as the left hand side. So the original statement is true.

PROOF THREE

Still need more? OK, one more attempt. This proof is a variation on the first proof, but only needs one piece of induction:

1. The statement is certainly true for n = 1 and n = 2, as 1^3 = 1^2 (equals 1) and 1^3 + 2^3 = (1+2)^2 (equals 9).

2. Assume that the statement is true for some positive integers m and m+1 (so 1^3 + 2^3 + 3^3 + … + m^3 = (1 + 2 + 3 + … + m)^2 and 1^3 + 2^3 + 3^3 + … + m^3 + (m+1)^3 = (1 + 2 + 3 + … + m + (m+1))^2). You can use these two statements to get another expression (by subtracting one away from the other), as follows:
(m+1)^3 = (1 + 2 + 3 + … + m + (m+1))^2 – (1 + 2 + 3 + … + m)^2
(m+1)^3 = (m+1)((1 + 2 + 3 + … + m + (m+1)) + (1 + 2 + 3 + … + m))) [difference of two squares]
(m+1)^3 = (m+1)(2(1 + 2 + 3 + … + m) + (m+1))
(m+1)^2 = 2(1 + 2 + 3 + … + m) + (m+1)

3. To prove that the statement is true for m+2, we go as follows:
1^3 + 2^3 + 3^3 + … + m^3 + (m+1)^3 + (m+2)^3 = (1 + 2 + 3 + … + m + (m+1) + (m+2))^2

We already assumed 1^3 + 2^3 + 3^3 + … + m^3 + (m+1)^3 = (1 + 2 + 3 + … + m + (m+1))^2, so we take this away from the above equation:
(m+2)^3 = (1 + 2 + 3 + … + m + (m+1) + (m+2))^2 – (1 + 2 + 3 + … + m + (m+1))^2
(m+2)^3 = (m+2)((1 + 2 + 3 + … + m + (m+1)) + (1 + 2 + 3 + … + m + (m+1) + (m+2))) [difference of two squares]
(m+2)^2 = (1 + 2 + 3 + … + m + (m+1)) + (1 + 2 + 3 + … + m + (m+1) + (m+2))
(m+2)^2 = 2(1 + 2 + 3 + … + m + (m+1)) + (m+2)
(m+2)^2 = 2(1 + 2 + 3 + … + m) + 2(m+1) + (m+2)
(m+2)^2 = 2(1 + 2 + 3 + … + m) + (m+1) + (m+1) + (m+2)
(m+2)^2 = 2(1 + 2 + 3 + … + m) + (m+1) + (2m+3)

We already assumed (m+1)^2 = 2(1 + 2 + 3 + … + m) + (m+1), so:
(m+2)^2 = (m+1)^2 + (2m+3)
m^2 + 4m + 4 = m^2 + 2m + 1 + 2m + 3
m^2 + 4m + 4 = m^2 + 4m + 4

This is obviously true, so the initial statement is also true.

So there you have it. Not one, not two, but three proofs of the statement given in the question. Surely you are satisfied now?!

[Mattyrat2027]

One of my favourite calculations is 111,111,111×111,111,111 (111,111,111^2) because it equals 12,345,678,987,654,321, or without the breaks, 12345678987654321. This echoes true to all other 11111….^2:

11×11=121
111×111=12321
1111×1111=1234321
11111×11111=123454321
111111×111111=12345654321
1111111×1111111=1234567654321
11111111×11111111=123456787654321
111111111×111111111=12345678987654321
Then things change:
1111111111×1111111111 = 1234567900987654321

Also something I came across:
1×9+2 = 11
12×9+3 = 111
123×9+4 = 1,111
1234×9+5 = 11,111
12,345×9+6 = 111,111
123,456×9+7 = 1,111,111
1,234,567×9+8 = 11,111,111
12,345,678×9+9 = 111,111,111
123,456,789×9+10 = 1,111,111,111
1,234,567,900×9+11 = 11,111,111,111

Do I love recurring ones…

[Author]

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