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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,455

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

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
- Posts: 90,455

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

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
- Posts: 90,455

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

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
- Posts: 90,455

16^3 + 18^3+ 25^3 + 26^3 + 29^3 = 67518

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

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: 90,455

Hi;

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

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

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

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: 15,657

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
- Posts: 90,455

You want to use the command in your posts in the long form?

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

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
- Posts: 90,455

DigitRoot[n_] := If[Mod[n, 9] == 0, 9, Mod[n, 9]] silly but works.

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

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: 90,455

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

68508

68508/post#=**99**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,455

68607

6 + 86 + 0 + 7 = 99

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

68706

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,455

68805

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

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

68904

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,455

16^3+2 24^3+26^3+27^3=69003

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

69102

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,455

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

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,915

69300

69+30+0=**99**

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