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**ZHero****Real Member**- Registered: 2008-06-08
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1. A 50-digit natural number is completely divisible by 13. If all the digits, except the 26th digit, are 1 then find the 26th digit.

If two or more thoughts intersect, there has to be a point!

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**mathsyperson****Moderator**- Registered: 2005-06-22
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Why did the vector cross the road?

It wanted to be normal.

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**ZHero****Real Member**- Registered: 2008-06-08
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**mathsyperson:** WOW!! Catchy!

2.

If two or more thoughts intersect, there has to be a point!

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**mathsyperson****Moderator**- Registered: 2005-06-22
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Why did the vector cross the road?

It wanted to be normal.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,435

1.

ZHero wrote:

mathsyperson:WOW!! Catchy!

I could only come up with

.*Last edited by phrontister (2010-06-24 15:12:19)*

"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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**ZHero****Real Member**- Registered: 2008-06-08
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**phrontister:** UNIQUE! It took me a while understanding why you divided the power of 10 by a 9 but got it now! Comps must be your Favorite Toys!!

If two or more thoughts intersect, there has to be a point!

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**ZHero****Real Member**- Registered: 2008-06-08
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**mathsyperson:** you saw it through again!

If two or more thoughts intersect, there has to be a point!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,435

ZHero wrote:

Comps must be your Favorite Toys!!

Yes...my scientific calculator and my pocket BASIC computer. I know where the buttons are...still trying to suss out what they do!

"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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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Seems like one day you'll be able to work upon them blindfolded!

3. Which is greater?

If two or more thoughts intersect, there has to be a point!

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**ZHero****Real Member**- Registered: 2008-06-08
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4. Suppose N is an n-digit positive integer such that

(a) all the n-digits are distinct.

(b) the sum of any three consecutive digits of N is divisible by 5.

How many such N's exist? What is the maximum value of 'n'?

If two or more thoughts intersect, there has to be a point!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
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Is 4. (a) correct, Zhero? I don't understand it as is, and you might have to explain it to me. I would have thought that "n" should read "N" there. Or maybe, "all of N's digits are distinct".

Also, I'm having trouble understanding Q3.

"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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**ZHero****Real Member**- Registered: 2008-06-08
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You got it! In 4 (a), the actual statement "all the n-digits are different" actually means that "all digits of N are different".

You might have considered it as "1-digit N, 2-digit N, 7-digit N etc." where all "n"s are different. Instead, consider it as "4-digit N: 1234, 3456, 4567" where "all the 4-digits" of N are distinct.

For 3, its a Power Tower. 2 is raised by 1001 two's and 3 is raised by 1000 three's. You don't start counting from the "base" of the exponent in counting 1001/1000.

Thus, 2² has 2 raised to 2, 1 times.

Is the explanation good?

If two or more thoughts intersect, there has to be a point!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,435

Thanks for the clarifications, ZHero. I think I get 'em - I think (sic).

4. "How many such N's exist?"

*EDIT*:

"What is the maximum value of 'n'?"

*Last edited by phrontister (2010-06-29 06:39:41)*

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Awesome job!

I'll need to verify the answer to first part and for the second part perhaps "n" is standing upon its head?

If two or more thoughts intersect, there has to be a point!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
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ZHero wrote:

...for the second part perhaps "n" is standing upon its head?

Oops! Tricky blighter...doing handstands while I wasn't looking! I've stood him back up on his feet again!

*Last edited by phrontister (2010-06-27 02:46:12)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,435

Hi ZHero!

Have you had a chance yet to verify my answer to the first part of Q4? Just curious as to whether or not I got it right. Ta.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Hi phrontister!

Sorry for the delay! I did not forget bout it but just was a little bit messed up on somethings. I gave the problem to two of my workmates and they came up with

Your solution is flawless!

You Rock!

If two or more thoughts intersect, there has to be a point!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,435

Hi ZHero,

Thanks for that...I'd been hoping I'd got it right.

Do you know how your workmates came up with their results? Is there a good logical method I missed?

*Last edited by phrontister (2010-07-02 20:44:42)*

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

phrontister wrote:

I'd been hoping I'd got it right.

By all means! When I asked my colleagues this questions and found a mismatch, I told them that the correct answer has to be

for the reasons you can understand!The method they used was....

If two or more thoughts intersect, there has to be a point!

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,435

Hi, ZHero! Thanks.

Finding n's maximum value is neat! Good logic...I'd never have thought of that. My method was to record the length of the highest-value number by comparing the results during the calcs.

Re finding all Ns...from your info I can't see how the initial 3/6 and 4/5 numbers are established before going cycling with them, but unless there's an astounding logical solution I guess I don't *really *have to know.

Nice problem, btw!

*Last edited by phrontister (2010-07-03 23:58:28)*

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Hi phrontister!

There's no logical method known to me to find those 3, 4, 5 and 6 digit Ns.

Your computer program is certainly better and faster too!

If two or more thoughts intersect, there has to be a point!

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**nerissa22****Member**- Registered: 2010-11-08
- Posts: 2

i need a math tutor..please help me...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,700

Hi nerissa22;

Welcome to the forum. Please post your math problems in the Help me section. You will get all the help you want.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**manpreet singh saluja****Member**- Registered: 2012-04-23
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ZHero wrote:

Seems like one day you'll be able to work upon them blindfolded!

3. Which is greater?

please give a solution to this problem.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,700

Hi manpreet singh saluja;

So call 3^3 = a and 2^2^2 = b. Now,

call 3^a = a1 and 2^b = b1, now we can continue

Call 3^a1 = a2 and 2^b1 = b2. We can continue recursively like this until we work our way down the tower. So

Welcome to the forum!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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