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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

(1) If

Then value of is(2) If

. Then value of is*Last edited by jacks (2013-10-13 16:49:37)*

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**jacks****Member**- Registered: 2012-11-21
- Posts: 77

I have solved (1) one

19! contain factor of 9. so R.H.S must be divisible by 9.

So for divisibility by 9., sum of digit must be divisible by 9.

So

so a = 1 .

But I did not understand how can i solved (ii) one

Thanks

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,316

Hi jacks

There are a few thing you will need to do there. First, you should find the number of 0's with which 34! ends. You will find out that both a and b are 0. Then you use divisibility by 9 and 27 to get c and d.

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

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi bobbym

I did it of the top of my head and I remember a similar problem where the answer was found the way I described above, so I just guessed that was it. Looks like I'll have ti have a bit more thinking.

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

I am not seeing any quick way for either pair, are you?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
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I thought about using 4 divisibility rules, but I do not think that will do much.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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If we could get 4 congruences that could be solved with the CRT. I just do not see how.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
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Well a+b+c+d≡:7 (mod 9). And we can easily calculate it mod 11.

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**bobbym****Administrator**- From: Bumpkinland
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You mean

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,316

4, actually.

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**bobbym****Administrator**- From: Bumpkinland
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The numbers are 0, 3, 5, 2 which sum to 10. Which is 1 mod 9.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Actually, the problen is that there are 3 digits missing in post #1. ab basically stsnds for 64352 in that number.

*Last edited by anonimnystefy (2013-10-13 21:07:18)*

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**bobbym****Administrator**- From: Bumpkinland
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I am not following you.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Compare these two:

34!=295232799cd9604140847618609ab0000000

34!=295232799039604140847618609643520000000

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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That is true, I did not notice that.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I guess we'll have to wait for the OP to clear the question out.

But, either way, if it were like we initially thought, I don't know how we can get the answer...

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

I agree, I do not even have a question like that in my notes and an internet search came up dry.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,316

I've seen a similar one with 28!, but not this one.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Found only this:http://forum.math.uoa.gr/viewtopic.php?f=37&p=211113 which suggests that your answer is still the correct one.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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That post is like 3 years old but has the same username!

See you later am going offline for a bit.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,316

Hi

This one also turned up:http://mathforum.numberempire.com/index.php?topic=709.0. It's a bit more recent.

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**zetafunc.****Guest**

I remember doing this question in a BMO1 paper, I think.

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,316

Ah, thanks. Found it:

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,415

Hi;

I thought that they would not want any modulo-computation methods so I did not use them.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**