Conan Edogawa

Mathematics Challenges v2

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1.Express 8f3344ddc83b6ba70f7ea2524dfa8e916e464fae.png
as the product of 2  factors, where neither of them is 1 or itself.

2. What does4270153977b88c146f9cdfb3155db286e02ea467.png equal?

3. Find all integers where n! is a multiple of n^2.

4. If 3 numbers are chosen at random from first 1000 natural numbers,what is the probability that the three numbers are in G.P.? All numbers are distinct.

5. prove that 7aa736fda62b8cc084a32d1c3f0130a101b1fd30.png is true for all non negative integers.

Edited by Donald J Trump

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23 minutes ago, Donald J Trump said:

1.Express 8f3344ddc83b6ba70f7ea2524dfa8e916e464fae.png
as the product of 2 factors, where neither of them is 1.

true for all non negative integers.

Found the loophole. :P 

Answer: (300^3 + 1) * 1/301

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13 hours ago, SilverCode said:

Integers, my boy.

Not necessarily. I interpreted it as factorize. :P Anyways, do you know how to do these? Especially that 4th one. It gives me the shivers.  

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On 10/8/2016 at 5:21 PM, IntegrationNation said:

1.Express 8f3344ddc83b6ba70f7ea2524dfa8e916e464fae.png
as the product of 2  factors, where neither of them is 1 or itself.

2. What does4270153977b88c146f9cdfb3155db286e02ea467.png equal?

3. Find all integers where n! is a multiple of n^2.

4. If 3 numbers are chosen at random from first 1000 natural numbers,what is the probability that the three numbers are in G.P.? All numbers are distinct.

5. prove that 7aa736fda62b8cc084a32d1c3f0130a101b1fd30.png is true for all non negative integers.

Can you post the solutions please? My math IQ peaked in 12th grade and dropped from there. 

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22 hours ago, Master Flap said:

Can you post the solutions please? My math IQ peaked in 12th grade and dropped from there. 

1. 271 * 333

2. 65/36

3. All composite numbers including 1 but excluding 4.

4. 6deb1074e6deb753d2735ea3daeaff1.gif

5.  Use Halder's inequality.

 

Edited by IntegrationNation

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11 hours ago, IntegrationNation said:

1. 271 * 333

2. This was a poorly constrained problem... 197x 1583619814549797419363188797991331 is one, though there are 63 solutions.

3. 65/36.

4. All composite numbers inclusing 1 but excluding 4.

5.  6deb1074e6deb753d2735ea3daeaff1.gif

6. Use Halder's inequality.

Can you explain how you arrived at these answers?

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On 1/3/2019 at 2:55 AM, Master Flap said:

Can you explain how you arrived at these answers?

1. Factor the cubic into 

CodeCogsEqn.gif

which equals

CodeCogsEqn (1).gif

divide by 301..

and you get

CodeCogsEqn (3).gif

2. Utilize geometric sequences and a summation to infinity.

The series can be grouped as..

5/13 + 5/13^2+ 5/13^3 + ....

        + 50/13^2 + 50/13^3 + .....

                         +500/13^3 +.........

Each line is a geometric sequence, and I'll abbreviate them with (a, r), where a is the first term and r is the common ratio.

CodeCogsEqn (4).gif

Using the sum formula we get

CodeCogsEqn (5).gif

CodeCogsEqn (6).gif

This last part is also a geometric series (1, 10/13) so we get

CodeCogsEqn (7).gif.

3. If you divide both by n, then the question becomes when is (n-1)! divisible by n.

If n is any prime, this cannot happen, as none of the factorial factors can contain it because it's prime and those are all smaller.

So what if n isn't prime?  It must decompose into factors all of which are smaller and so contained somewhere in the factors of the factorial. What if such a factor occurs twice?  It can be further factored using numbers we haven't yet used, so n will divide the factorial.

This might sound a little confusing, so..

Suppose n = 147 = 3 x 72.

146! = 3 x 7 x 14 x ...., so 147 divides it.

So all composite numbers have the aforementioned property.

As for the others, I'm a little fatigued. I'll post the other answers later tommorow. 

Edited by IntegrationNation
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53 minutes ago, IntegrationNation said:

What do you think?

I didn't go through the answers yet, I will later. 

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On 1/19/2019 at 11:41 AM, Shawnic said:

I didn't go through the answers yet, I will later. 

You've had the better part of a year to think through it.. thoughts?

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On 1/19/2019 at 10:16 AM, Conan Edogawa said:

1. Factor the cubic into 

CodeCogsEqn.gif

which equals

CodeCogsEqn (1).gif

divide by 301..

and you get

CodeCogsEqn (3).gif

2. Utilize geometric sequences and a summation to infinity.

The series can be grouped as..

5/13 + 5/13^2+ 5/13^3 + ....

        + 50/13^2 + 50/13^3 + .....

                         +500/13^3 +.........

Each line is a geometric sequence, and I'll abbreviate them with (a, r), where a is the first term and r is the common ratio.

CodeCogsEqn (4).gif

Using the sum formula we get

CodeCogsEqn (5).gif

CodeCogsEqn (6).gif

This last part is also a geometric series (1, 10/13) so we get

CodeCogsEqn (7).gif.

3. If you divide both by n, then the question becomes when is (n-1)! divisible by n.

If n is any prime, this cannot happen, as none of the factorial factors can contain it because it's prime and those are all smaller.

So what if n isn't prime?  It must decompose into factors all of which are smaller and so contained somewhere in the factors of the factorial. What if such a factor occurs twice?  It can be further factored using numbers we haven't yet used, so n will divide the factorial.

This might sound a little confusing, so..

Suppose n = 147 = 3 x 72.

146! = 3 x 7 x 14 x ...., so 147 divides it.

So all composite numbers have the aforementioned property.

As for the others, I'm a little fatigued. I'll post the other answers later tommorow. 

Oh yeah :P 

Like I said, I'm out of touch with playing around with math

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