# Mathematics Challenges v2

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

2. What does 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 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 as the product of 2 factors, where neither of them is 1.

true for all non negative integers.

Found the loophole. Answer: (300^3 + 1) * 1/301

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4 hours ago, Master Flap said:

Found the loophole. Answer: (300^3 + 1) * 1/301

Integers, my boy.

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13 hours ago, SilverCode said:

Integers, my boy.

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

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3), anyone?

On 10/8/2016 at 8:25 AM, Master Flap said:

Found the loophole. Answer: (300^3 + 1) * 1/301

ninja'd.

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• 2 years later...
On 10/8/2016 at 5:21 PM, IntegrationNation said:

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

2. What does 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 is true for all non negative integers.

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

Unarchived

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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. 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. 6. Use Halder's inequality.

Can you explain how you arrived at these answers?

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• 3 weeks later...
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

which equals divide by 301..

and you get 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. Using the sum formula we get  This last part is also a geometric series (1, 10/13) so we get .

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

Can you explain how you arrived at these answers?

What do you think?

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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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• 2 weeks later...
On 1/19/2019 at 11:41 AM, Master Flap said:

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

Thoughts?

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• 5 months later...
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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• 1 month later...

@Shawnic.. thoughts?

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

1. Factor the cubic into

which equals

divide by 301..

and you get

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.

Using the sum formula we get

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

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

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