Airband Radio Aerials: Maths in Action

I've been interested in aircraft and airshows for over 40 years - anything military or civil, and I've blogged in the past about how to use a spreadsheet to track down where to watch the Red Arrows fly past on their transit flights.  You didn't think that post was about geometry without some real-life applications?  What is this - "Another day I haven't used algebra"?

Anyway - I've been particularly interested in the Red Arrows and their air-to-air chatter, and the communications between pilots and air traffic control.  Yes, I take my airband radio along to airshows and to airports, and listen to the pilots request and receive clearance to take off or land.  Getting to airports is more of a challenge than it used to be - my children aren't as interested as I am in the whole thing, and standing at the end of a runway in poor weather isn't as much fun as it sounds!

So, I've started developing my home-based receiver.  In other words, I spent my birthday money on an airband antenna and an extension cable to connect it from outside (cold and sometimes rainy) to my desk (warm and inside) so that I can listen to pilots flying nearby.

Now: nearby is a relative term.

From Stoke on Trent, I've been able to pick up pilot transmissions from about 35 miles away, on the southern edge of Manchester Airport.  That's with a very basic antenna, set on my garden gatepost and about two metres off the ground - not bad for a first attempt.

My dad, on the other hand, has been tracking radio transmissions for decades.  His main areas of interest are long wave (around 200 kHz), medium wave (500-1600 kHz), and TV (UHF, 300 Mhz to 3GHz).  

Airband falls into the Very High Frequency range, around 100-200 MHz.

Here comes the maths:
All radio transmissions travel at the speed of light, c = 2.998 * 10^8 ms-1.
c = f w

Where f (sometimes the Greek letter nu, ν) is the frequency, and w (usually the Greek letter lambda, ƛ) is the wavelength.

So, if we know the frequency range that we want to listen to, we can calculate the wavelength of that transmission.  And this is important, because the length of the antenna (or aerial) that we need will depend on the wavelength.  Ideally, the aerial should be the length of one full wavelength, for maximum reception effectiveness.  Alternatively, a half-wavelength or a quarter-wavelength can be used.

So:  we know the speed of light, c = 2.998 * 10^8 ms-1
And we know the frequency of the transmissions we want to receive, which is around 118 MHz.

c/ν = ƛ

ƛ =   2.5 metres

Which is feasible for an external, wall-mounted aerial.  Can you see where this is going?

Exactly.  And here it is:  

It's just over two metres from end to end, with a feed at the midpoint.  This is the Mark One; the Mark Two will be the same aerial but even higher up, and closer to vertical (with a bracket that will enable it to dodge the eaves of the roof!

Calculator Games: Ulam Sequences: Up, Up and Away!

 Up, Up and Away With Ulam Seqeunces

This article in the ongoing series of ‘mathematical puzzles you could investigate with a calculator’ (that’s why I just call it ‘Calculator Fun and Games’) is the Ulam Sequence.   It’s an interesting mathematical sequence named after the Polish-American mathematician Stanisław Ulam. It’s a sequence rather than a series, and it generates some strange and interesting results, which seem almost random.

How to Generate an Ulam Sequence

Take with two numbers (specifically, positive integers).  A good place to start is with a =1 and b=2.  To find the next number in the sequence, find the smallest integer that can be written as the sum of two distinct earlier numbers in just one way.  Continue with the next number, and the next.

For example, let’s start the Ulam Sequence with the numbers 1 and 2.  These are the first two terms in the sequence.

The next term is 3 (since 1+2=3).
After that comes 4 (since 1+3=4).

The next terms is not 5.  We can write 5 as 1+4 and as 2+3 using the terms that we’ve generated already.

The next term is 6 (since 2+4=6), and we can only write this in one way using our terms.

We can write 7 as 4+3 and as 6+1, so we skip 7.

The next term is 8 (since 2+6=8).

So, the beginning of the sequence is: 1,2,3,4,6,8,… (and it continues
1,2,3,4,6,8,11,13,16,18,26,28,36,38,47,48,53,57,62,69,72,77,82,87,97,99,102)

Note that 5, 7, 9, 10, 12, 14, 15, 17, 19, 20, 21 and 22 can be obtained in multiple ways using the terms before them.

However:  23 is not in the sequence because it cannot be obtained using the previous terms at all!    24 can be written as both 16+8 and 18+6, while 25 is not obtainable.

The sequence grows in an irregular, almost random pattern.  Let’s see what happens when we start with 1 and 3 instead of 1 and 2.

4 = 1 + 3 only
5 = 1 + 4 only
6 = 5 + 1 only
7 can be written as 4+3 and 6+1
8 = 5+3
9 is 4+5 and 6+3
10 = 6+4
11 is 6+5 and 8+3

The first 20 terms for the 1,3 Ulam Sequence are:

1,3,4,5,6,8,10, 12,17, 21, 23, 28, 32,34,39,43,48,52,59 and 63.

The Ulam Sequence is an interesting example of how simple rules can lead to complex and intriguing mathematical structures, which makes it ideal for calculator (or spreadsheet) exercises.  For example, here’s a comparison of the Ulam sequences for 1,2 compared with 1,3 (as I’ve calculated above) and then 1,4 and 1,5.  Interestingly, the 1,5 sequence does not race ahead of the 1,2 sequence as I had originally expected.

Term	Ulam (1,2)	Ulam (1,3)	Ulam (1,4)	Ulam (1,5)
1	1	1	1	1
2	2	3	4	5
3	3	4	5	6
4	4	5	6	7
5	6	6	7	8
6	8	8	8	9
7	11	10	10	10
8	13	12	16	12
9	16	17	18	20
10	18	21	19	22
11	26	23	21	23
12	28	28	31	24
13	36	32	32	26
14	38	34	33	38
15	47	39	42	38
16	48	43	46	40
17	53	48	56	41
18	57	52	57	52
19	62	59	66	57
20	69	63	70	69

There’s a balance between the ability to leap to larger numbers (1,5) initially – from 1 to 5 – and the need to fill in more numbers between 5 and 10 (because there are very smaller numbers that can be made in multiple ways).

A quick comparison of the Ulam Sequences for (2,b) is even more interesting.  We have to start with 2,3 since 2,1 is the same as 1,2 above, and 2,2 will only produce the even numbers (which is cute but dull).  In fact, any even number paired with 2 will produce uninteresting results!

Let’s compare 2,3 and 2,5:  These grow at a slower rate compared to the 1,b sequences.  Interestingly, they contain far fewer even numbers than the 1,b sequences; in fact 2,5 only contains the even numbers 2 and 12 in the first 20 terms (with no indication that there are any more even numbers further along the sequence).

Term	2,3	2,5
1	2	2
2	3	5
3	5	7
4	7	9
5	8	11
6	9	12
7	13	13
8	14	15
9	18	19
10	19	23
11	24	27
12	25	29
13	29	35
14	30	37
15	35	41
16	36	43
17	40	45
18	41	49
19	46	51
20	51	55

So there’s plenty of scope for investigation with a spreadsheet for the larger numbers.  For example, I haven’t found anybody else list the Ulam sequence for 10,11… so here it is:  the Ulam Sequence for (10,11)

10, 11, 21, 31, 32, 41, 43, 51, 54, 61, 62, 65…

After huge initial leaps of +10 or +11 between consecutive terms, the growth rate of the sequence starts to slow down.  There is only one term in the 20s, then two in the 30s, 40s and 50s, then three in the 60s.

Further reading:
Wolfram has lists and links for many of the 1,b and 2,b Ulam sequences

Game review: Pocket Mars

The game Pocket Mars is, in our experience, more of a puzzle than a game. We found it difficult to win (by getting all our colonists to Mars). The challenge wasn't beating the other players, it was achieving the win conditions at all.  Any thoughts of competition were dwarfed by the challenge of getting to the finish line by any means possible.

In fact, it was so difficult that we only played it twice before returning it to the charity shop from whence it came. 

In our opinion, a game needs to have a high level of player interaction.  My race to win should come at the detriment of your chances of winning.  Snakes and Ladders has no interaction, but at least it's easy and you can feel like you're competing.  However, a good game for us isn't Snakes and Ladders, where we have no interaction at all, and it's just you and the dice, then me and the dice, and we could add dozens of players without affecting the game at all.  No.  Player interaction is key - I want to win because I beat you, not just because I rolled bigger numbers or drew better cards compared to you.

And Pocket Mars feels like it has no player interaction at all - you take a turn, I take a turn.  We launch our men to Mars, we take turns to draw cards, and we see who can do the best.  It is, at best, a complicated puzzle where we each battle the rules and restrictions of the game in order to get the best outcome we can, and then compare against the other players.  The game talks about sabotaging the opponents, but this didn't happen when we played, and the game lacked any 'fun'.  It was too difficult to make any real progress in the game, and this meant that the return on effort was too high and we promptly gave up.

Nope, this game was not for us, which was a shame because it's been produced to a very high level of quality.  The box, cards and pieces all looked and felt great, but I feel that they could have been reorganised into a game that would be a lot more fun.

Calculator Games: Front to Back

This puzzle is not from the Calculator Fun and Games book, but another one that would be suitable (if you had time).  Early indications and preliminary reading indicate that this one could take some time to complete, but let's wade in anyway.

The game would be described in this way:

Take a number (three digits to start with)
Reverse the digits to form a new number
Add the two numbers together (this makes a change from subtracting...!)
If the new number is not a palindrome, continue reversing digits and adding.

Let's start with 456

456+654 = 1110

1110+0111 = 1221 so it's a palindrome.

Let's try 782

782+287 = 1069

1069 + 6901 = 10670

10670 + 7601 = 18271

18271 + 17281 = 35552

35552 + 25553 = 61105

50116 + 61105 = 111221

111221 + 122111 = 233332  which is a palindrome.

And let's try a smaller three-digit number, 165

165 + 651 = 816
816 + 618 = 1434
1434 + 4341 = 5775 which is a palindrome.

This is called the Lychrel process, and it's not named after a famous mathematician.  It makes a change!  It was named by a man called Wade van Landingham in 2002, who created the name as a rough anagram of his girlfriend's name, Cheryl.  Unlike most mathematical concepts, this one is not hundreds of years old - this also makes a pleasant change!

There is one number which has not yet been found to form a palindrom after many, many iterations of the Lychrel process, and that's 196.  Innocuous, isn't it?

196 + 691 = 887

887 + 788 = 1675

1675 + 5761 = 7436

7436 + 6347 = 13783

13783 + 38731 = 52514

52514 + 41525 = 94039

94039 + 93049 = 187088

187088 + 880781 = 1067869

1067869 + 9687601 = 10755470

10755470 + 7455701 = 18211171

18211171 + 17111281 = 35322452

35322452 + 25422353 = 60744805

60744805 + 50844706 = 

And at this point, my calculator says "Enough!"  I can't get all the digits any more, and this number still isn't reaching a palindrome.

My spreadsheet goes a little further:

60744805 + 50844706 = 111589511

111589511 + 115985111 = 227574622

227574622 + 226475722 = 454050344

454050344 + 443050454 = 897100798

897100798 + 897001798 = 1794102596

And then starts throwing "#VALUE!" messages at me, without reaching a palindrome.

The general definition for a Lychrel number is one that does not reach a palindrome in fewer than 500 iterations. This is easier to measure compared to 'never reaches a palindrome', and that means that the Lychrel numbers (more than 500 iterations) include 295, 394, 493, 592 and 689.

Some numbers immediately become palindromes after one iteration - these are trivial, commonplace and not very interesting!  For example, 110 + 11 = 121, and any other number where the units value is zero, and the hundreds and tens are both less than five.  The longer ones are definitely more interesting, because there's no obvious pattern (and it reminds me of the Collatz conjecture, which I'll be revisiting soon).  Larger numbers which need more than 500 iterations include 10538, 10553 and 10585.

So: can you find numbers which reach a palindrome before they make your calculator (or your spreadsheet) explode?

Calculator Fun and Games: Premium Prime Numbers

 PRIME NUMBERS

 Calculator Fun and Games is a maths puzzle book that’s worth its salt: it has a section on prime numbers, as any good maths book does.

Prime numbers are only divisible by themselves and 1, they have no other factors.  1 is not a prime number (by definition), but 2 is, making 2 the only even prime number.

Calculator Fun and Games lists the prime numbers up to 1000, just for good measure.  But is there any relationship between them?  Are they connected?  Are there ways of finding (or generating) them?  And can all this be done with a humble calculator?  Maybe one day.

However, the book points out an interesting fact:  if you

Take a prime number (greater than 3)
Square it
Add 14
Divide by 12

Then the remainder is always 3.

For example:
5² = 25
25 + 14 = 39
39/12 = 3, with a remainder of 3

Let’s take a larger example: 577
577² = 332929
332929 + 14 = 332943
332943 / 12 = 27745 remainder 3

Alternatively, x² + 11 always divides exactly by 12.  I’m not sure why Ben Hamilton decided on adding 14… maybe he likes the remainder 3?

Does this apply in reverse?  If I take a number, and subtract 11 and take the square root, do we always get a prime? 

No.  Even if the square root is an integer, it doesn’t mean that the starting number was a prime.  For example, 92-11 = 81, and the square root of 81 is 9.  Only a subset of a = (SQRT(b-11)) will give a as a prime.



TWIN PRIMES

Twin primes are primes with a difference of two; for example, 11 and 13, or 17 and 19.  These are rarer than prime numbers, but it still seems that there will be an infinite number of twin primes.  An example of a larger twin prime is 971 and 973; while the largest known twin primes have 388,342 digits:  they are:  2996863034895×2^1290000±1.

Not a number you’d fit on your calculator.  The largest prime number which will fit on an eight-digit calculator (with a good old-fashioned LCD display) is 99,999,989.

Looking at all those nines, I’d like to play Over and Out with it, and see how long it would take to get it down to zero! 😊

Other recent 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 Games: The Kaprekar Constant

The Kaprekar Constant: 6174

This is one of those problems that doesn't come from Calculator Fun and Games, but would be an interesting offshoot, and can certainly be done with a calculator, pen and pencil.

Take any four digit number.  It must contain at least two different digits (i.e. 9933 is permitted, but 8888 is not).  Sequence the digits in descending order (e.g. 2536 becomes 6532), and ascending order (2536 becomes 2356).  Subtract the smaller number from the larger, then repeat the process of finding the descending and ascending numbers, and subtracting.  To quote Calculator Fun and Games (and countless vague Maths GCSE questions):  What do you notice?

First example:

1874
Rearrange:  8741-1478 = 7263
Rearrange: 7632 - 2367 = 5265
6652 - 2566 = 4086
8640 - 0468 = 8172
8721 - 1278 = 7443
7443 - 3447 = 3996
9963 - 3699 = 6264
6642 - 2466 = 4176
(7641 - 1467 = 6174, which goes round in a loop)

Some wider reading shows that 6174 is the Kaprekar constant, and it is the end result for all four-digit numbers, and that all numbers terminate there after no more than seven steps.  Never one to shirk a challenge (especially if it involves a calculator), I tried out some more numbers:

9954 --> 5553 --> 9981 --> 8870 --> 8532 --> 6174

9732 --> 7533 --> 6174

1110 --> 9900 --> 9810 --> 9621 --> 6174

I repeated this several times, with upwards of 30 different numbers, and in order to summarise my results, I have drawn out a network of key paths to 6174.  I've consistently used a colour scheme - dark green for 6714, light green for numbers which lead directly to 6174; yellow for numbers which need two steps, light blue need three steps, and so on.

Firstly:  pathways for numbers which require all seven steps to reach 6174.  I've found four numbers that require seven steps, and they follow the paths below; three of them go through 9954, and they all go through 8532.  My work on the Kaprekar constant tried to focus on numbers which either need all seven steps, or resolve to 6174 in one step.

And interestingly, five six-step numbers that all follow the same pathway:

And finally, a collection of one-step and two-step numbers which have no parents:

One step:  6642, 8532, 9621, 7533, 7531, 6200, 8754 and 8752.  Bold numbers have no parents - no other numbers (that I have found) go through them to 6174.  7533 has six parents, but none of these parents have any parents of their own.

These findings are all based on my own work, and I fully acknowledge that they could be incorrect due to my incomplete research.


Extension:  a five-digit Kaprekar process... another time!

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.

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 the 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:

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:

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

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)
Sums of digits and Multiples of 9