Exponent Slash, #1

Mathegocart · 2016-07-19 07:20:45

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

2. Express $d99cf3151cf8b5fd07dacb5a1e81c28dd0c22613.png$ as the product of 2 factors, where neither of them is 1. Do not express these 2 factors explicitly.

3. What does $4270153977b88c146f9cdfb3155db286e02ea467.png$ equal?

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

5. 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.

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

Last edited by Mathegocart (2016-07-21 02:09:08)

bobbym · 2016-07-19 09:27:01

Hi;

Mathegocart · 2016-07-19 09:55:12

bobbym wrote:

Hi;

Correct.

bobbym · 2016-07-19 12:59:29

Hi;

thickhead · 2016-07-19 16:14:30

The answer is not unique. There are 63 different ways.

Last edited by thickhead (2016-07-20 04:05:57)

thickhead · 2016-07-19 16:22:11

Mathegocart · 2016-07-21 02:07:08

thickhead wrote:

Thickhead, could you explain the process on the inequality? I used a bit of Cauchy Schwarz on this.

Last edited by Mathegocart (2016-07-21 02:08:44)

Mathegocart · 2016-07-21 02:10:10

bobbym wrote:

Hi;

Both answers are reasonable solutions, also thickhead has a good soultion as well.

Mathegocart · 2016-07-21 02:12:24

thickhead wrote:

6 also is a solution, and 6 is not a cube. $gif.latex?%5Cfrac%7B6%21%7D%7B6%5E%7B2%7D%7D$ = 20

Last edited by Mathegocart (2016-07-21 02:12:49)

thickhead · 2016-07-21 02:46:10

Mathegocart wrote:

thickhead wrote:
6 also is a solution, and 6 is not a cube. http://latex.codecogs.com/gif.latex?%5Cfrac%7B6%21%7D%7B6%5E%7B2%7D%7D = 20

I just wrote offhand. I had an obvious flash about cubes.Unless n is specified it becomes difficult to make exhaustive list.

bobbym · 2016-07-21 02:52:50

Hi;

thickhead · 2016-07-21 03:47:12

bobbym is right. I had not thought properly.After knowing it it looks very simple. A composite number means at least two factors which make up for one "n" the other n being provided by the number itself. Perhaps I tried 4 and seeing it not correct went for 8 and made it a cube. Very unscientific thinking on my part.

thickhead · 2016-07-24 22:41:11

Last edited by thickhead (2016-07-24 23:20:14)

thickhead · 2016-07-25 04:51:22

Last edited by thickhead (2016-07-25 15:28:10)

thickhead · 2016-07-26 23:57:50

Why is Mathegocart silent ?Is he too busy with other problems? One more thing instead of G.P. if the 3 numbers are in A.P. what is the probability for first 1000 natural numbers or even first n natural numbers?

bobbym · 2016-07-27 01:40:23

Hi;

f the 3 numbers are in A.P. what is the probability for first n natural numbers?

zetafunc · 2016-07-27 03:27:15

Last edited by zetafunc (2016-07-27 05:02:26)

thickhead · 2016-07-27 04:02:20

bobbym · 2016-07-27 06:05:17

You are correct, further simplification is possible. Normally, I do not automatically simplify all the way. There are good reasons for this but in this case they do not apply, further simplification was called for.

Mathegocart · 2019-01-03 02:20:48

Could you guys explain how 1, 2, 3, 4 and 5 are done? I'm rather confused in regards with number theory.

alter ego · 2019-01-04 00:12:56

Hi Mathegocart,

At the time I didn't try these. I can show two now and I'll have a think about the others. No one method for all.

Q3. You'll need to know about geometric series and sum to infinity.

The series can be divided up as

5/13 + 5/13squared+ 5/13cubed + ....

+ 50/13squared + 50/13cubed + .....

+500/13cubed +.........

etc

Each line is a geometric series. I'll abbreviate each by (a,r) where a is the first term and r the common ratio.

(5/13 1/13) + (50/13squared, 1/13) + (500/13cubed 1/13) + .......

Using the sum formula we get

5/13 x 13/12 + 50/13 x 13/12 + 500/13 x 13/12 + ......

= 5/12 x 1/13 x (1 + 10/13 + 100/13squared + ......)

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

5/12 x 1/13 x 1 x 13/3 = 65/36

More follows,

Bob.

Last edited by alter ego (2019-01-04 00:20:18)

alter ego · 2019-01-04 00:39:23

Q4.

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 these are all smaller.

So what if n is not Prime? It must have a decomposition into factors all of which are smaller and so contained somewhere in the factors of the factorial. But 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 sounds a bit clumsy so I'll illustrate with an example.

Suppose n = 147 = 3 x 7squared

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

So all composite numbers have the property.

Bob

Last edited by alter ego (2019-01-04 00:42:47)

Mathegocart · 2019-01-04 05:16:48

alter ego wrote:

Q4.

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 these are all smaller.
So what if n is not Prime? It must have a decomposition into factors all of which are smaller and so contained somewhere in the factors of the factorial. But 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 sounds a bit clumsy so I'll illustrate with an example.
Suppose n = 147 = 3 x 7squared
146! = 3 x 7 x 14 x other numbers so 147 divides it.
So all composite numbers have the property.
Bob

Any idea for 5? thickhead says (sigma of stuff)/(some number choose 3), presumably that is the number of all three-number combinations, though my mind is totally lost on the numerator. Could you explain the numerator? I have to admit, I was spitballing when I saw made these problems and I didn't really have a clue on how to solve these. Bob? As in Bob Bundy?

Last edited by Mathegocart (2019-01-04 05:19:40)

alter ego · 2019-01-04 06:49:28

Hi Mathegocart

Yes, as in Bob Bundy. Someone once said that joining twice was against the rules but it isn't. I wouldn't encourage everyone to do it ... sometimes spammers who've got banned do it so it is suspicious.

If you look up my earliest posts using this name you'll find my reasoning in a problem about a professor of logic. Then, some years later I was asked whether only moderators could do a particular thing and it was helpful to have a non mod account to test it out.

Today I have a reason for using alter ego which I don't want to disclose publicly. It's to do with forum security. Please don't speculate. In a week ask again and I'll explain in a private message.

Q5 has me stumped at present. Pick any two numbers and they will be in GP. But the next term may not even be an integer so I'm stuck as to how to proceed.

I'm also not getting anywhere with Q1 and Q2 at the moment. Lots of pages of algebra but no results yet. I feel that, in both one needs to start with a binomial expansion.

Bob

alter ego · 2019-01-04 11:33:25

Q1.

300^3 + 1 = 301^3 - 3x300^2 -3x300 = 301^3 -3x300(300+1)

=301^3 - 30^2 x 301

Divide by 301

301^2 - 30^2 = (301 - 30)(301 + 30) = 271 x 331

Bob

Math Is Fun Forum

#1 2016-07-19 07:20:45

Exponent Slash, #1

#2 2016-07-19 09:27:01

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#3 2016-07-19 09:55:12

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#4 2016-07-19 12:59:29

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#5 2016-07-19 16:14:30

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