# The Naked Scientists Forum

### Author Topic: Enigma  (Read 2263 times)

#### syhprum

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##### Enigma
« on: 06/09/2009 19:43:16 »
This weeks NEW SCIENTIST Enigma problem asks you to evaluate the check sum of

(2^100+3^200)^300   With Mathmatica it is easy to evaluate this statement (which runs into hundreds of digits), does anyone skilled in the use of Mathmatica Know how to produce a check sum (digit sum) from this as adding up all the digits is rather labour intensive.
« Last Edit: 07/09/2009 09:19:08 by syhprum »

#### glovesforfoxes

• Sr. Member
• Posts: 372
• Matthew 6:21
##### Re: Enigma
« Reply #1 on: 06/09/2009 22:56:38 »
naughty! you're trying to get your free £25!

i won't tell.

#### syhprum

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##### Re: Enigma
« Reply #2 on: 07/09/2009 06:48:29 »
The answer consists of one digit between 0 and 9 to get a share of the £25.00 pound prize I need only to send in 10 entries !
I don't think this is a good way to make money.

#### syhprum

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##### Enigma
« Reply #3 on: 07/09/2009 10:28:35 »
Eezzy pezzy with Mathmatica once you get a hold on the syntax

DigitSum = Total[IntegerDigits[(2^100 + 3^200)^300]]

128 458

DigitSum = Total[IntegerDigits[(128458)]]
28
DigitSum = Total[IntegerDigits[(28)]]
10
DigitSum = Total[IntegerDigits[(10)]]
1
But Mathmatica is an expensive program which is not available to everyone so any ideas how to do it without?, casting nines or such like.
I will not be claiming any share of the prize but feel free to do so

This question will self destruct in seven days but anyone interested in these problems can go to.

http://www.newenigma.com/enigma/view/1561/Some-Check#content

« Last Edit: 08/09/2009 17:08:21 by syhprum »

#### Soul Surfer

• Neilep Level Member
• Posts: 3345
• keep banging the rocks together
##### Enigma
« Reply #4 on: 09/09/2009 11:47:29 »
I am presuming that there is a simple solution other than the computer blast approach and the working will have to be shown rather than just the answer.

It is probably based on the properties of numbers notably those of the numbers 2 and 3 and/or modulo processes.

I have not yet solved the problem but suggest an approach might be using

Any number divisible by 3 always has digits that add up to a number divisible by 3

This means that the checksum must end up being 3 6 or 9

That suggests that the above solution must be wrong.

Any number divisible by 2 is always an even number but checksums of even numbers do not have to be even numbers so that does not get you anywhere unless some orther fact forces the checksum to be even which would give you a simple and direct answer which would be 6.

Can anyone fault this logic or supply the final fact?

« Last Edit: 09/09/2009 11:51:26 by Soul Surfer »

#### syhprum

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##### Enigma
« Reply #5 on: 09/09/2009 12:03:39 »
Working does not have to be shown, take a look a the URL that I provided, simple back of an en volope solutions are shown there as well as computer solutions using a freeware program PYTHON
I showed a four line MATHEMATICA solution but the last three lines can be done in your head also if you understand the syntax better only one line is needed.

#### syhprum

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##### Enigma
« Reply #6 on: 11/09/2009 04:06:41 »
28,628 Digits is correct but not the check sum !

Take a look at the link I provided numerous ways of solving the problem are shown.
« Last Edit: 11/09/2009 04:10:55 by syhprum »

#### The Naked Scientists Forum

##### Enigma
« Reply #6 on: 11/09/2009 04:06:41 »