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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

66924

66 + 9 + 24 I think that idea works every time too.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

No...only sometimes.

67023

FromDigits[Reverse[{6, 7}]] + 0 + 23=**99**

Looks like for any number abcde, FromDigits[Reverse[{a, b}]] + c + de=**99**

*Last edited by phrontister (2013-10-06 12:41:58)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Do you have one where it does not?

2 x 15^3 + 19^3 + 26^3 + 33^3 = 67122

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Yes...eg, 15741, 42372, 99990. In fact, that approach (which is formula 5 below) gives only 81 solutions out of 909 possibles, while the other formulas give more.

I listed all 909 five-digit multiples of **99** and tested 5 formulas to see how many solutions I'd find. Here are the results:

Using 'abcde' for the 5-digit multiples...

1. a+bc+de=**99**: 855 solutions

2. FromDigits[Reverse[{a,b}]]+c+de=**99**: 855 solutions

3. ab+c+FromDigits[Reverse[{d,e}]]=**99**: 90 solutions

4. ab+cd+e=**99**: 90 solutions

5. ab+c+de=**99**: 81 solutions

- Formulas 1 & 2 gave the same results for each test number.

- Only 54 numbers didn't sum to **99 **with any of these formulas. Instead, they summed to 198 (2x99): all with formulas 1 & 2, 1 with formulas 3 & 4 and 10 with formula 5.

- Formulas 3 & 4 gave the same results for only 10 test numbers, with all of them being in the 90090 to 99000 range.

Conclusion: Using N = a 5-digit multiple of **99**, T = a multiple of 10000, and M = 99's multiplier, formulas 1 & 2 work for all N<T where M>0.01T.

------------------------------------------------------------------------------------------------------------

67221

FromDigits[Reverse[{6,7}]]+(2+2)!-1=**99**

*Last edited by phrontister (2013-10-07 02:50:34)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

Thanks for the work!

16^3+20^3+22^3+26^3+30^3=67320

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Thanks! It was a nice way to pass some spare time.

67419

n=Total[IntegerDigits[Total[IntegerDigits[67419]]]]; FromDigits[{n, n}]=**99**

Bedzzz!

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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16^3 + 18^3+ 25^3 + 26^3 + 29^3 = 67518

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

67617

n=DigitRoot[67617];FromDigits[{n, n}]=**99**

This pseudo code in M works for any correct entry in this thread just by substituting the multiple of **99 **for mine.

Prediction: One day someone will come up with a DigitRoot command.

*Last edited by phrontister (2013-10-07 11:37:26)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Hi;

What would you want DigitRoot[n] to spit out?

17^3+19^3+21^3+27^3+30^3=67716

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
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The single-digit end result of the iterative digital summing of n.

eg, n=67815=6+7+8+1+5=2+7=9...so I'd like DigitRoot[n]'s result to be 9.

There's some longish code on the net about it which I ignored for posting here and went for the pseudo code instead.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

What's wrong with M's command?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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You want to use the command in your posts in the long form?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Hi Bobby,

No, just in a short form. Ideally, I'd like a one-word command for that, but it doesn't exist, does it? What is the shortest alternative you know of?

68013 (catchup)

68112

6+81+12=**99**

anonimnystefy wrote:

What's wrong with M's command?

I made 'DigitRoot' up. Do you know of a command that will do what I want? (see my previous post)

*Last edited by phrontister (2013-10-07 13:25:52)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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DigitRoot[n_] := If[Mod[n, 9] == 0, 9, Mod[n, 9]] silly but works.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

Thanks, Bobby! That works well.

However, I've tried in my posts to avoid using numbers other than 99 and the post's answer - except for factors - and, inspired by your idea, I've now come up with this:

Mod[68310, **99**] + **99** = **99**

*Last edited by phrontister (2013-10-07 22:24:06)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

68508

68508/post#=**99**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

68607

6 + 86 + 0 + 7 = 99

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

68706

Rotate[FromDigits[{6+87*0,6}],\[Pi]]=**99**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

68805

13^3+14^3+16^3+30^3+32^3=68805

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

68904

6!/8+9+0x4=**99**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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16^3+2 24^3+26^3+27^3=69003

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

69102

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,347

15^3+20^3+2 21^3+34^3=69201

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,844

69300

69+30+0=**99**

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