Calculator Games: Pythagorean Triples

 Pythagorean Triples

The Calculator Fun and Games book probably didn’t cover Pythagorean Triples because back in the mid 80s, pocket calculators didn’t have an x² button.  Also, I didn’t start studying trigonometry and geometry until the late 80s, five or so years after I had the book, so the lack of an x² button on my calculator didn’t even bother me.  The square root button was exciting enough, although it was a little boring because if you hammered it enough it always led to an answer of 1.

However, the concept still holds and it’s possible to study Pythagorean Triples with a basic calculator and a notepad and pen (although a scientific calculator will be helpful).

Pythagoras once decreed that for a right-angled triangle, a² + b² = h².  The square of the hypotenuse, h, is equal to the sum of the squares of the other two sides.  The hypotenuse is the side of the triangle which is opposite the right angle.  It's not shown on the diagram below, because I couldn't work out how to add it to the picture ;-)  This is a key result in geometry.

Furthermore, we can find an easy solution to a² + b² = h² since 3² + 4² = 5² (9 + 16 = 25).  Integer values of a, b and h (sometimes called c) are called Pythagorean triples.

The challenge is:  can you find any more?

Since Pythagoras did his maths thousands of years ago, many people have studied his work and found plenty more triples.  To list a few:

(3, 4, 5)

3²+4²=9+16=25=5²

(5, 12, 13)

5²+12²=25+144=169=13²

(7, 24, 25)

7²+24²=49+576=625=25²

(8, 15, 17)

8²+15² = 64+225 = 289 =17²

(9, 12, 15)

9²+12²=81+144=225=15²

Others include (20, 21, 29) and (12, 35, 37). 



GENERATING PYTHAGOREAN TRIPLES

It’s possible to generate a set of triples using Euclid's formulae:

a = m²−n²

b = 2mn

c = m²+n²

where m and n are positive integers.  For example, m = 4 and n = 2 gives:

a = 12
b = 16
c = 20

Which is a scaled-up version of the 3,4,5 triple (each number is 4* larger than the 3,4,5 triple).  This isn’t a unique triple, it’s just one that’s four times larger than an existing one.

Some observations of Pythagorean Triples:

It is not possible to have a Pythagorean triple that contains all prime numbers. 

At least one of the numbers must be even (b = 2mn).  m = 2 and n=1 yields 3,4,5.

There are an infinite number of Pythagorean triples (since m and n can have any values).  Many of these are duplicates of other triples, just scaled up (for example, 12,16,20 is a scaled up version of 3,4,5).  

So I’m going to search for ‘unique’ triples that aren’t just multiples of smaller ones, and for this, I suspect that m and n must not have common factors.  For example, m = 4 and n = 2 can be simplified to m = 2 and n = 1 (divide by two), which is how (12,16,20) is related to (3,4,5).  (Wider reading on Pythagorean triples led me to discover that the accepted term is not ‘unique’ but ‘primitive’ triples.  You get the idea.  I have reworded the rest of my article to call them primitive, in line with the rest of the mathematical world).

m = 5, n = 1 gives (10,24,26) or (5,12,13) – which brings me to another observation: 

Primitive Pythagorean triples generally contain at least one prime number. 

In my list above, I gave (9,12,15) as an example, but that’s just another version of (3,4,5), in this case, scaled up by 3.  By definition,(and I know I haven’t defined them very well) primitive triples must contain three numbers that have no common factors that would enable them to be simplified.  Prime numbers guarantee this, but are not essential – for example, (16, 63, 65) is a triple with no primes.

An easy way to generate primitive triples is to use the m, n method, setting m to an even number and n to 1.  This means that a = m²−n², and c = m²+n² will only have a difference of two.

m	n	a	b	c	a2 + b2	c2
2	1	3	4	5	25	25
4	1	15	8	17	289	289
6	1	35	12	37	1369	1369
8	1	63	16	65	4225	4225
10	1	99	20	101	10201	10201
12	1	143	24	145	21025	21025
14	1	195	28	197	38809	38809
16	1	255	32	257	66049	66049
18	1	323	36	325	105625	105625
20	1	399	40	401	160801	160801
22	1	483	44	485	235225	235225
24	1	575	48	577	332929	332929
26	1	675	52	677	458329	458329
28	1	783	56	785	616225	616225
30	1	899	60	901	811801	811801

Prime numbers highlighted in bold.  Rows of green triples contain no prime numbers.

So: to continue our investigation: what happens if we set n = 2?  Do we still get an abundance of primitive triples?  We do, but only when m is odd.  When m is even and n=2, the values of a,b,c are all divisible by 4 (not shown below).  When m is divisible by 4, then a,b,c are all divisible by 8.

Here are the results for n=2 and m is odd: a series of primitive triples, some containing primes.

m	n	a	b	c	a2 + b2	c2
3	2	5	12	13	169	169
5	2	21	20	29	841	841
7	2	45	28	53	2809	2809
9	2	77	36	85	7225	7225
11	2	117	44	125	15625	15625
13	2	165	52	173	29929	29929
15	2	221	60	229	52441	52441
17	2	285	68	293	85849	85849
19	2	357	76	365	133225	133225
21	2	437	84	445	198025	198025
23	2	525	92	533	284089	284089

Prime numbers highlighted in bold.  Rows of green triples contain no prime numbers.

When n=3, we can generate primitive triples when m is an even number that is not divisible by 3 (m = 4, 8, 10, 14, 16, 20…).  When m and n are both odd, then a,b,c are divisible by 2 (unless m is divisible by 3, when a,b,c are all divisible by 9).  This leads to a complicated repeating pattern, which is easier to show than explain: 

m	n	a	b	c	Notes
4	3	7	24	25	Primitive
5	3	16	30	34	m,n both odd, a,b,c divisible by 2
6	3	27	36	45	multiples of 3, all divisible by 9
7	3	40	42	58	m,n both odd, a,b,c divisible by 2
8	3	55	48	73	Primitive
9	3	72	54	90	m,n, odd and divisible by 3, a,b,c divisible by 18
10	3	91	60	109	Primitive
11	3	112	66	130	m,n both odd, a,b,c divisible by 2
12	3	135	72	153	multiples of 3, all divisible by 9
13	3	160	78	178	m,n both odd, a,b,c divisible by 2
14	3	187	84	205	Primitive
15	3	216	90	234	m,n, odd and divisible by 3, a,b,c divisible by 18
16	3	247	96	265	Primitive
17	3	280	102	298	m,n both odd, a,b,c divisible by 2
18	3	315	108	333	multiples of 3, all divisible by 9
19	3	352	114	370	m,n both odd, a,b,c divisible by 2
20	3	391	120	409	Primitive
21	3	432	126	450	m,n, odd and divisible by 3, a,b,c divisible by 18
22	3	475	132	493	Primitive
23	3	520	138	538	m,n both odd, a,b,c divisible by 2
24	3	567	144	585	multiples of 3, all divisible by 9
25	3	616	150	634	m,n both odd, a,b,c divisible by 2
26	3	667	156	685	Primitive

The investigation (and the pattern-finding) continues, as we try n=4.  Here’s the result – it’s possible to find primitive Pythagorean triples for every case where m = odd again. 

m	n	a	b	c	Notes
5	4	9	40	41	Primitive
6	4	20	48	52	m,n both even: a,b,c divisible by 4
7	4	33	56	65	Primitive
8	4	48	64	80	m,n both divisible by 4, a,b,c divisible by 16
9	4	65	72	97	Primitive
10	4	84	80	116	m,n both even: a,b,c divisible by 4
11	4	105	88	137	Primitive
12	4	128	96	160	m,n both divisible by 4, a,b,c divisible by 32
13	4	153	104	185	Primitive
14	4	180	112	212	m,n both even: a,b,c divisible by 4
15	4	209	120	241	Primitive
16	4	240	128	272	m,n both divisible by 4, a,b,c divisible by 16
17	4	273	136	305	Primitive
18	4	308	144	340	m,n both even: a,b,c divisible by 4
19	4	345	152	377	Primitive
20	4	384	160	416	m,n both divisible by 4, a,b,c divisible by 32

We find that m and n must have no common factors in order to generate primitive triples.  This means m and n must be ‘relatively prime’ (an expression I learned during my wider reading into this project).  

This is starting to go beyond the scope of a calculator, and into spreadsheet territory, but it's still fascinating, and can still be done with a simple pocket calculator (or an app on your phone, naturally), even without an x² button.

Summary of Pythagorean Triples:

It is not possible to have a Pythagorean triple that contains all prime numbers

It is possible to have a primitive Pythagorean triple that contains no prime numbers; a, b and c will be relatively prime, and these are generally generated by m and n being relatively prime too.

There are an infinite number of triples, and an infinite number of primitive triples

Other Calculator Fun and Games articles:

Sums of digits and Multiples of 9
Snakes and Ladders (Collatz Conjecture)
Crafty Calculator Calculations (numerical anagrams with five digits)
More Multiplications (numerical anagrams, four digits)
Over and Out (reduce large numbers to zero as rapidly as possible)
Calculator Games: Front to Back
Calculator Games: Up, up and away with Ulam sequences
Calculator Games: The Kaprekar Constant

Calculator Games: Reverse Digits and Subtract

REVERSING DIGITS: IS THERE A PATTERN?

This one comes from CoPilot, but is again in the style of Calculator Fun and Games, and is treated slightly differently, as "One Thousand and Eighty Nine" in the book.

Copilot:

Pick a number and try to form a specific pattern using the digits, for example, “Change 123 to 321 using only addition and subtraction.”

Book:
Take a three digit number, and reverse the digits.  Subtract the smaller from the larger.  Then, take the result of this subtraction, and add it to its digits reversed; you will always get 1089.

The Copilot idea is not a perfect suggestion for a game (why subtract when you could just add a smaller number?) but it has potential for an investigation.

Let’s take the example as written:  change 123 into 321.

321-123 = 198

This means that we can change 123 into 321 just by adding 198.  And 198, for those who’ve been reading my series on calculator fun and games, immediately stands out as a multiple of 9.  The digits sum to 18, which is a multiple of 9 (and its digits sum to 9 as well).

Let’s try another number and see if this is a fluke:  Let’s try 456 and 654.
654-456 = 198

And another, with non-consecutive digits:

781 – 187 = 594, also a multiple of 9 (5 + 9 +4 =18)

In fact, let’s try a whole range of random numbers, with reversed digits, making sure to subtract the smaller of the pair from the larger.

872 – 278 = 594 (reversed = 495, and 495 + 594 = 1089)
512 – 215 = 297 (reversed = 792, and 792 + 297 = 1089)
503 – 305 = 198 (reversed = 198, and 198 + 891 = 1089)
902 – 209 = 693 (reversed = 396, and 396 + 693 = 1089)

There's a full solution to the 1089 phenomenon online (there are many; and here's another that's simpler to follow but less detailed); I won't add anything further to it here.

And four-digit numbers:
8112 – 2118 = 5994 
7205 – 5027 = 2178
(if you reverse and sum these results, you get 10989 and 10890; there isn't a unique answer).

And five-digit numbers:
98765 – 56789 = 41976  (41976 / 9 = 4664, in case you were wondering, as I was)
85127 – 72158 = 12969  (also divisible by nine; 12969 / 9  = 1441)

Why do we always get 1089?

Well, let's take a four-digit number ABCD, where A, B, C,and D are its digits. The value of the number can be expressed as 

$1000A+100B+10C+D$.

When you reverse the digits, the number becomes dcba, which can be expressed as:
$1000D+100C+10B+A$.

Now, when we subtract the smaller number from the larger, we get (assuming ABCD is the larger; it works either way)

• The difference is: $(1000A+100B+10C+D)-(1000D+100C+10B+A)$
• Simplifying this, we get: $999A+90B-90C-999D$

• Factoring out common terms, we get: $999(A-D)+90(B-C)$
• Notice that both $999$ and $90$ are multiples of 9.
• Therefore, the entire expression $999(A-D)+90(B-C)$ is a multiple of 9.

And since the result is a multiple of 9, it shows why the result of the subtraction is always a multiple of nine (for four-digit numbers).  The same logic can be applied to three-digit numbers (where it's easier), or to numbers with more digits too. 

In the case of three digits, the result will be simpler.  For a number abc and its reverse, cba, the b will cancel and the result is always be 9 * (11(a-c)) or 99(a-c).

And it stands as a useful little trick to show on a calculator!

Previous Calculator Fun and Games articles:
Snakes and Ladders (Collatz Conjecture)
Crafty Calculator Calculations (numerical anagrams, five digits)
More Multiplications (numerical anagrams, four digits)
Over and Out (reduce large numbers to zero in as few steps as possible)

Calculator Fun and Games 4: Sums of Digits and Multiples of 9

This post isn't taken directly from Calculator Fun and Games, but it is definitely a puzzle of the sort that Ben Hamilton could have included in his book.  It's very straightforward, and can be demonstrated easily with some examples.

One (or more) players each write down a six- (or seven-, or eight-) digit number on a piece of paper.  Using the calculator, they sum the digits and write down the total.  They then sum those digits (using the calculator), and repeat until they have a single-digit answer.  Largest number wins.

Let's try an example: I'll start with 987654 (regular readers will recognise this from my previous blog post on Calculator Fun and Games, called Over and Out).

987654

9 + 8 + 7 + 6 + 5 + 4 = 39
3 + 9 = 12
1 + 2 = 3

My opponent tries a different approach:

123456

1 + 2 + 3 + 4 + 5 + 6 = 21
2 + 1 = 3

So it's a tie.

Time for a tie-breaker:  I go completely random and punch six digits into my calculator:

946305

9 + 4 + 6 + 3 + 0 + 5 = 27
2 + 7 = 9

My opponent needs to find something special just to draw this round:

813524

8 + 1 + 3 + 5 + 2 + 4 = 23
2 + 3 = 5 

So I win, because 9 is greater than 5.

Analysis

The result for any random number is largely random.  You can't identify the final sum-of-digits just by looking at a number, especially for numbers where the first sum of digits is going to be more than 9.

However, there is a definite pattern in the sums of digits of sequences of numbers which is probably not surprising if you think about it (or use a spreadsheet, as I did to analyse some long sequences of numbers).

The second sum - i.e. the final iteration of the summing - follows a very simple pattern: 1-9, then 1-9 over and over again.  This extends across hundreds (from 899 to 900) and by the same process, extends across thousands (1999 - 2000) and so on.  This means that each digit from 1 to 9 is represented equally as the final sum.  The probability of ending up with a 1 or a 7 or an 8 is equal - each digit has a probability of 1/9 of being the final sum.

And providing you don't memorise a number (or a collection of numbers) which sum to a final answer of 9, this makes it an interesting and fun game for young calculator users.

Not that there are many of them around these days! :-)

Note on the spreadsheet:

To break down a three-digit number to its individual digits (which can then be summed), I used these formula.  Place the three-digit number in cell A1.

Hundreds:  
B1 = INT((A1 - MOD(A1, 100)) / 100)
Takes the remainder of dividing A1 by 100 and subtracts this from A1, then divides this by 100 to provide a digit.

Tens:
C1 = INT((A1 - MOD(A1, 10)) / 10) - 10 * INT(A1 / 100)
Divides A1 by 10 to give a two-digit integer, then subtracts 10 times the hundred digit.  The trickiest of the calculations.

Units:
D1 = MOD(A1, 10)
yields the remainder when A1 is divided by 10 - i.e. the units

Further extension:  asking an AI bot for a series of maths puzzles that can be solved with a calculator, it replied with this:

Magic Number Game:

• Choose a number, multiply it by 9, and then add the digits together until you get a single digit. This digit will always be 9. Try it with different numbers and see if it works.

So:  is it true that all multiples of 9 eventually sum to 9?

I have known for a long time that the first sum of multiples of 9 is always 9, and can explain why intuitively, but not comprehensively.

1 * 9 = 9
2 * 9 = 9 + (10 - 1) which means that the tens column is incremented by 1, and the units column is reduce by 1.  Therefore the sum of the tens and units will remain unchanged at 9.

9
18
27
36

and so on.

However, this breaks down after reaching 90
9 * 11 = 99
--> 9 + 9 = 18
--> 1 + 8 = 9

...and further iterations of the sums of digits are needed.  I've known this as a useful way of determining if a number is a multiple of 9, but I can't prove why it's always the case except by demonstrating it.  And my A-level mathematics teacher told me that demonstrating for some cases is hardly a proof for all. 

More mathematically advanced brains than mine have already shown why the sums of digits of multiples of nine are always nine - I won't pretend to understand it, but it has been done. 

Further reading

Sums of digits of nine are always multiples of 9

Other Calculator Fun and Games articles:

Snakes and Ladders (Collatz Conjecture)
Crafty Calculator Calculations (numerical anagrams, five digits)
More Multiplications (numerical anagrams, four digits)
Over and Out (reduce large numbers to zero in as few steps as possible)

Calculator Fun and Games 3: Over and Out

Calculator Fun and Games 3:  Over and Out

This visit to Calculator Fun and Games is one of my favourites, and features a game that I’ve played more than any of the others in the book (even the ‘spell words with your calculator’ game).  It’s called Over and Out, and has a loose cricket theme.  

The rules are simple:  you start with a six-digit number (no zeroes allowed, no duplicated digits), and the aim is to get it down to zero in as few steps as possible, and the target is six (the number of balls in an over in a game of cricket).  The steps are addition, subtraction, multiplication and division, and you can add, subtract, multiply or divide by any one-digit or two-digit numbers.

The example in the book goes like this:

583621

1)       583621 + 19 = 583640

2)       583640 / 40 = 14591

3)       14591 – 11 = 14580

4)       14580 / 45 = 324

5)       324 / 12 = 27

6)       27-27 = 0

Mission accomplished, game over!  583621 is reduced to zero in six steps.

The hints in the book include generally working on subtracting and dividing, but adding or multiplying where necessary to give you a change to divide by a larger number – as we saw in the example here, the first step of adding 19 enables us to divide by 40 in the second step.

The largest six-digit number that has no zeroes or duplicates is 987654 and it presents a challenge!  Here’s how I would have approached it as a young child – the ‘adult’ mathematician will appear later.

987654

1)       It’s even, so I can divide by 2.  And why not?
987654 / 2 = 493827 which is odd, so I can’t divide by 2 again.

2)       493827 – 27 = 493800 which I can divide by 50, or even 80.  I’ll go with 50 for now (because I’m thinking as a child, and I'm not sure how old I would have been when I realized I could divide by 80).

3)       493800 / 50 = 9876  Now we’re talking!  Getting our number down to four digits so quickly puts us in a really good place.  I can see that we can divide this number by four, so that’s my next step.

4)       493800 / 4 = 2469  isn’t divisible by 2, 3 or 5, so I’m struggling to find an easy way down.  I’ll subtract the 69 and it should be easier from there.

5)       2469 – 69 = 2400.  Still quite high, but divisible by 80.

6)       2400 / 80 = 30

7)       30-30 = 0

So it took me seven steps, although I was starting from the highest six-digit number allowed in the game.

Over the years, I’ve played this game quite often when I have a spare minute and a calculator.  In fact, I got so advanced with it, that I started using a spreadsheet to keep track of my results (although I’ve since lost it).  Then, just for good measure, I wrote a macro that would work out the optimal route down from any digit to zero.  It works on one simple principle:  the fastest way down is to divide.  Find the largest factor for any number, and divide by it.  If it’s prime (or its lowest factor is higher than 99) then calculate the highest multiple of 99 below it, and subtract down to it.

Let’s take 987654 as our example, and go through how the macro worked:

1)       987654 – highest factor below 99 is 97.  Divide by 97.
(to determine this, the macro divides 987654 by every number between 2 and 99 and checks if there is a remainder after the division.  If there’s no remainder, then store the number in a variable.  The largest value for that variable after checking all the numbers is the one to use.  In other words, check if (987654/x) = INT( 987654/x).
987654 / 97 = 10182

2)       10182 – highest factor is 6.  Divide by 6.
10182 / 6 = 1697

3)       1697 is prime.  Highest multiple of 99 below 1697 is 17 * 99 = 1683, so we will subtract 14.
(to calculate this: round down the value of (1697/99) to the nearest integer).
1697 – 14 = 1683

4)       1683 is divisible by 99, we set it up that way.
1683 / 99 = 17

5)       17 – 17 = 0

The macro is brutally strong, and took the highest possible value for the game (987654) to zero in five steps.

So let’s change the rules of the game, and look at seven-digit numbers.  Can the macro crunch a seven-digit number down to zero in six steps?  I’ll go first, as an adult mathematician, using every trick I can think of (without using a spreadsheet!).

9876543

1)       9876543 + 57 = 9876600
I prefer this instead of subtracting 43, because I know I can divide the answer by 60, whereas dividing 987500 will only divide by 50.  I’ll do a quick check on my calculator to see if I can divide by 80 (I can’t, fractions aren’t allowed), and hence divide by 60 for my next step.

2)       9876600 / 60 = 164610
So I have a six-digit number, but at least it’s low and it’s divisible by 10.  Easy next step (for the human, anyway)

3)       164610 / 10 = 16461
Now what? I can add 39 and then divide by 50, or subtract the 61 and divide by 80.  I’ll take the second option.

4)       16461 – 61 = 16400

5)       16400 / 80 = 205

6)       205 is visibly divisible by 5, so I’ll do that, get down to less than 100 and subtract what’s left.
205 / 5 = 41

7)       41 – 41 = 0

Seven steps for a seven-digit number seems good to me!  Let’s see what the macro would do:

9876543

1)       Largest factor below 100 is 3.
9876543 / 3 = 3292181 which is prime.

2)       Since 3292181 is prime, find the multiple of 99 which is below it:  33254 * 99 = 3292146.  Second step is to subtract down to this number:
3292181 – 35 = 3292146

3)       This is now divisible by 99:
3292146 / 99 = 33254 (at this stage, the macro is at around twice the value I was at; I had 16461 after my third step).

4)       33254 is divisible by 26:
33254 / 26 = 1279

5)       1279 is prime, so find a multiple of 99 nearby. 12 * 99 = 1188, so I need to subtract down to 1188.
1279 – 91 = 1188

6)       1188 / 99 = 12

7)       12 – 12 = 0

So even my macro needs to take seven steps with this seven-digit number.

Looking further, and with some assistance, I can see that I could improve the human answer.  I reached 164610 after two steps, and some mathematical assistance shows that this is actually divisible by 90 and by 93.  Dividing by 90 gives me 1829, while dividing by 93 gives me 1770.  Let’s follow this path – I’ll recap first:

1)       9876543 + 57 = 9876600

2)       9876600 / 60 = 164610

3)       164610 / 93 = 1770

4)       I can see by eye that 1770 is divisible by 10, but it’s also divisible by 59. 
1770 / 59 = 30

5)       30 – 30 = 0

So my brute-force macro (which I programmed, I’ll take the blame/credit for it) took seven steps by trying to divide by 99 each time, whereas another path can reach the end in five steps.  The macro has a short-sighted strategy – what’s the best ‘next step’ - compared to trying to find a number like 9876600 which can be divided by 60, 93 and 59 to get down to less than 100 in three further steps (in fact 9876600 has 192 factors and 30 of them are below 100, including 1). I will hold my hands up and confess that this was a fluke, I didn’t plan to find such a ‘good’ number with so many factors. 

A follow-up to the macro coding (which I have not pursued) is to take the initial number and work out how to operate on it to get the most factors below 100.  9876600 has 192 factors; 9876500 has only 24, the best next step would be to divide by 50:

1)       9876543 – 43 = 9876500

2)       9876500 / 50 = 197530

3)       197530 / 10 = 19753 which is prime.  Need to find another number within +- 99 of 19753 which has the best sequence of factors (and I need a new algorithm to do that).  For now, let’s use the ‘divide by 99’ even though we know it’s sub-optimal.
19753 / 99 = 199.5 so let’s find 200 * 99 = 19800.  Let’s add up to 19800

4)       197530 + 79 = 19800

5)       19800 / 99 = 200

6)       200 / 50 = 4

7)       4 – 4 = 0

Still stuck on seven steps!

And so you can see why I have whiled away many hours on this particular problem.  It just takes one person and a calculator (and I was playing this when calculators were a feature on watches, not portable phones) and like all good maths puzzles, can be expanded indefinitely.  I could get a spreadsheet and a macro to throw nine-digit or ten-digit numbers around, getting down to zero in minimal steps; or I could use my calculator and take the human approach, and then compare the difference!

Follow-ups are to work out the best way to treat prime numbers.  Anything else can be divided down to something smaller, but the main challenge is what to do when there’s no more dividing available.  At the moment, it follows the “divide by 99 next time”, but it might be worthwhile finding larger factors so that a series of divisions is possible instead of just 99.  That’s a challenge for another day – I’m still happy playing with my calculator and my macro!

Further Reading

An algorithm which determines odd or even to reduce a number to zero, although it only divides by two... instead of looking for the largest factor

Other Calculator Fun and Games articles:

Snakes and Ladders (Collatz Conjecture)
Crafty Calculator Calculations (numerical anagrams, five digits)
More Multiplications (numerical anagrams, four digits)