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#176 Re: Exercises » Ver.. » 2013-07-12 14:27:05
hi Al-Allo
Yes, factors of 2 and 5 are the key to this. You seem to have got the terminating ones sorted out.
Now: why does 1/7 cycle with 6 digits recurring and all the others 2/7, 3/7 etc follow the same cycle?
Bob
Lol, the funny thing is, I had the exact problem asked with 1/7, 2/7, etc. before you asked me to search for it. I began it, but didn't finish it.
The number 142857 seems to have some interesting properties. Anyway, I'll continue searching 
#177 Re: Exercises » Ver.. » 2013-07-12 09:21:08
Hi Al-Allo;
I had the same idea about factors of 10.
You always were able to answer the question about my fraction.
You mean what I'm saying is correct ? (Just to be sure... xD) About your fraction, maybe I was able, it's just I didn't give it much thought, don't worry, I'll try to spend more time on it. I'll be adding my ideas on this topic anyway.
#178 Re: Exercises » Ver.. » 2013-07-12 08:55:03
Ok, well I'm not sure of this theory but let's go :
Just so you know, I was working with fractions like 1/2, 1/3, etc. So what is being said may not be completly true for other types of fractions ?
Also, I was using long divisions to work out the fraction.
In a fraction like 1/2, or any other(the denominator being higher), we know that it will always begin by 0.
(-------
2 ( 10
(
Answer 0,
As we can see, 10 (Which I'm totally aware is not the real value, but I wasn't sure which terminology to use if I had worked with decimals) has in it's prime factorization : 2 and 5. Every fractions will have to work with 10 and it's prime factors and this is exactly what will determinate if the decimal representation will be terminate or recurring.(Which will also depend on the divisor's prime factors.)
(-------
2 ( 10
( 10
Answer 0,5
Now, what conclusion can we make from this ? If we know that 10=2*5, we can say that a fraction will always? be terminate if it has divisor with a prime factorisation of 2,5 or both of them (Like 4,8,16,25,etc.) On the contrary, if you had 10 objects, 20 objects, 40 objects (All numbers whose prime factors are 2 and 5's or only one them) and you tried to complete it with an odd like 3, it would never work. So if we said 1/6
2*5/3*2, for the same reasons, 6, which is composed of two 3's, won't fit in.
What do you think ? I know that there is still a lot which is missing, and I don't know much about number theory, but am I on the right path ? (Btw, bobbym, I think that I will be able to answer your question if I know how terminate and recurring number appears)
And I only want critiques on what is said, no spoilers or anything else. (I haven't finished, some more could appear if I have new ideas)
#179 Re: Exercises » Ver.. » 2013-07-11 09:52:37
Ok, I'll also keep searching for that. I think that by understanding how to predict if they are terminate or not, it will help help me a lot in other related problems.
#180 Re: Exercises » Ver.. » 2013-07-10 12:00:32
I'll try to see.
Btw, here are the real numbers :
100-70=30
30-28=2
2,0-1,4=0,6
0,6-0,56=0,04
0,04-0,035=0,005
0,0050-0,0049=0,0001
It was my fault, sorry. (If you ever saw it ^^)
#181 Re: Exercises » Ver.. » 2013-07-10 11:50:17
WEll, for now I'm not sure. I need to get my imagination working on it....
Edit: I edited my numbers in my first post, a part of the calculations were wrong.
#182 Re: Exercises » Ver.. » 2013-07-10 11:36:26
Why would it be a little tedious with 1/99 ????
#183 Re: Exercises » Ver.. » 2013-07-10 11:25:06
What it is lacking is an argument of why this will always happen.
But isn't this correct :
100-70=30
30-28=2
2,0-1,4=0,6
0,6-0,56=0,4
0,4-0,35=0,5
0,50-0,49=0,1
Is it contradictable ?
#184 Re: Exercises » Ver.. » 2013-07-10 11:24:02
bobbym wrote:Yes, that is what the 3 dots at the end mean
1 / 3 = .333333333333...
When p is a prime, 1 / p will have a period of p - 1.
Wikipedia wrote:1/7 = 0.142857 ; 6 repeating digits
1/17 = 0.05882352 94117647 ; 16 repeating digits
1/19 = 0.052631578 947368421 ; 18 repeating digits
1/23 = 0.04347826086 95652173913 ; 22 repeating digits
1/29 = 0.0344827 5862068 9655172 4137931 ; 28 repeating digits
What about 5? Or 11? Or 13? They are not p-1 then.
That's what I also saw...
#185 Re: Exercises » Ver.. » 2013-07-10 11:13:08
Ok, but what do you think of my answer, should I elaborate, did I answer what was expected ?
#186 Re: Exercises » Ver.. » 2013-07-10 11:02:20
Other question, this is going to infinity, isn't it ? And could you elaborate what you just said?
#187 Exercises » Ver.. » 2013-07-10 10:54:51
- Al-Allo
- Replies: 24
Hi
While solving the two proceding problems, you may have discovered that quotient digits (and remainders) became periodic :
_______
7 | 100000..(twenty zeros)
|
|Answer :1428571428571...
Is it just a coincidence, or will this pattern repeat?
Well, no it's not a coincidence and yes the pattern will repeat. I'm not sure this is the complete reason, but anyway :
Let's take 100 to simplify things
_______
7 | 100
| answer : 14,2857...
100-70=30
30-28=2
2,0-1,4=0,6
0,6-0,56=0,04
0,04-0,035=0,005
0,0050-0,0049=0,0001
As we can see, a new cycle (Or a new patern in our quotient) begins when the digit "1" is the highest value in our number obtained after a substraction(The second period or cycle begins at 0,0001), and wether 7 for 10, 70 for 100 , etc. will always enter one time in it, and this patern continue on for an infinite time.(Well, If it doesn't, it goes for a long-long time, for sure)
Now, like I said, I feel it's lacking something. I don't want any help for now, besides answering this question : I'm not sure if the question in the quote is telling to find the exact cause to this problem, or it only wants a yes or no answer. WHat do you think ? Thank you
*sorry, I did an error in my calculations. Now it should be fine
#188 Re: Exercises » Other verification. » 2013-07-09 08:03:08
It's the same principle. Isn't it ?
#189 Re: Exercises » Other verification. » 2013-07-09 07:41:45
Ok, thank you!
#190 Exercises » Other verification. » 2013-07-09 06:08:04
- Al-Allo
- Replies: 5
Hi !
A boy claims that he can multiply any three-digit number by 1001 instantly. If his classmate says to him "715" he gives the answer immediately. Compute this answer and explain the boy's secret.
Ok, so the answer is obviously 715715. Also, the boy knows that his answer will always be in the hundred thousands value because :
1001*100=100100 (The lowest number having three digits)
1001*999=999999(The highest number having three digits)
Knowing that the value won't be one time in the thousands, another time in the hundreds,etc And knowing that its magnitude will always be in the hundred thousands, his trick is for sure to always work and won't have to worry about it changing. Now, as we can see, a certain pattern keeps showing up. The three digits number will always compose two times your answer.
1000*900=900900
So the boy can claim with insurance what the answer is without calculating anything. Now, why does the trick work ?
Let's work with 1001*715.
We can decompose 715 as : 700+10+5.
First, let's consider it this way:
1001*5=5005
1001*1=1001
1001*7=7007
Now we place them that way :
5005
1001
7007
____
Ok, so we know that a certain pattern keeps repeating, that is, to show TWO times the three digits number in our answer. Now, what I did earlier was to consider the 7 and 1 as a unit to show clearly what happens when we add their true value to these two digits (1*10=10 and 7*100=700)
5005
1001 *10
7007
____
If we want the trick to work, we must assure ourselves that only one of our digits (7,1 or 5) will compose each column of the decimal place.(Which will also include 0's, which changes nothing)
5005
10010
7007 Now, should we should by *100, because we need our numbers (the two 7's) to be moved two times to meet our criterias
_____
5005
10010
700700
_______
715715
Exactly what we wanted. Lastly, why in our number 715715 why should we get a 715 (the one in darker color) and not zero's or any other number? That's because of the "1" (unit) in the number 1001
WIthout it:
1000
715
____
715000 (we are missing the other 715)
Here's what's happening :
1
715
_____
715
715000+715=715715
And that's why the trick works. Did I answer the question correctly ? Or was I not hitting the core of the "trick" ?
Thank you !
#191 Re: Exercises » Verification » 2013-07-07 13:33:45
Well, I guess each person will give a different meaning to his existence 
Anyway, I'll go now. Bye Bye !
#192 Re: Exercises » Verification » 2013-07-07 13:22:26
I'm happy to hear these comments ! I'm even happier to know that a person like you exists to help me in needs !
#193 Re: Exercises » Verification » 2013-07-07 13:00:16
Ah, I don't know about that equation. (I'm still learning)
Thank you again ! I had the answer in my book, but I didn't want to see it,(and still haven't) just wanted it to be checked by someone who had experience ! Thanks !
#194 Re: Exercises » Verification » 2013-07-07 12:47:32
Ah... But you read my text??? And what do you mean by your technique ?
#195 Exercises » Verification » 2013-07-07 10:34:05
- Al-Allo
- Replies: 10
Hi, could you verify if my answer is correct ? I DON'T WANT THE ANSWER, I just want my work to be verified:
Several digits "8" are written and some "+" signs are inserted to get the sum 1000. Figure out how it is done (For example, if we try 88+88+8+88+8, we fail because we get only 280 insted of 1000.)
My answer(It may be long) :
1-First, the number 1000 is composed of 125 digits "8", because :
2-This limits us to 3 principal number in our sequence of numbers, which will help us build 1000.
8: He is composed of only one digit 8.
88: He is composed of 11 digit 8.
888: He is composed of 111 digit 8.
3-Now, there exists different ways to build the number 1000(Depending of the composition of your sequence of numbers.)
4-If your sequence of numbers is composed of the number 888(Which can only go one time in your sequence.), this results in this :
a) 125-111=14 numbers "8" left.
You could now continue your sequence by 88 because:
14-11=3 numbers "8" left. Then, finish off your sequence by 3 digits "8", because :
3-3=0 digits "8" left.
b) Or, you could have had 888 in your sequence, which would give :
125-111=14 chiffres '8' left.
And then, complete the sequence by 14 digits 8.
14-14=0 digits "8" left.
5)If your sequence contains one or many times the number 88, it should give :
a) Maximum of 88 composed in the number 1000: 11 times
11*88=968
digits "8" in total
125-121=4 digits '8'* left
To obtain 1000, we must add 4 times the digit 8
4-4=0 digits "8" left.
We can then conclude that there exists 11 ways possible to have 1000 with a sequence of numbers which contains at maximum 11 times the number 88 and at minimum 1 time the number 88.
Each time you will remove a "88" in your sequence, it will be needed to be replaced by 11 times the digit 8 so we can keep the answer equalling 1000.
Ex: 88+88+88+88+88+88+88+88+88+88+88+8+8+8+8=1000
Now, we will remove a "88"
88+88+88+88+88+88+88+88+88+88+8+8+8+8+8+8+8+8+8+8+8+8+8+8=8=1000
6- Finally, the last way of obtaining 1000 consists to write 125 times the number 8.
8+8+8+8...125 times
In total, you will have 14 ways of having one thousand.
Do you think I've figured out how it's done ? Or not ? No answers, just suggestions about what I wrote. Thank you!
#196 Re: Help Me ! » Quick question » 2013-07-04 09:45:43
To check your work you can use a calculator or google's calculator. They never make a mistake with the order of operations. If you do not get their answer you know you did not follow the order of operations correctly.
That's what I did, but I wanted a human approbation also ^^ Thank you again ! SOmetimes, I need to have more confidence in what I do. ( I hate when I lack it)
#197 Re: Help Me ! » Quick question » 2013-07-04 09:42:28
Ah! Ok, exactly what I thought. I just wanted to hear it from someone else, to assure myself I learned from my error. THank you again Bobbym !
#198 Help Me ! » Quick question » 2013-07-04 09:25:33
- Al-Allo
- Replies: 5
Here's an easy thing :
I did the error of doing this :
Which is clearly false.
I know what is the correct way of doing it, because the real answer is positive one.
But just to know, I should have done BEMDAS to have avoided that incorrect way of doing ? Right ? (It concerned the priority of operations?Just want to be sure I've learned correctly from my error)
Thank you !
#199 Re: Help Me ! » factoring a polynomial » 2013-07-04 03:51:04
Bye Bye ! Thank you again!
#200 Re: Help Me ! » factoring a polynomial » 2013-07-04 03:43:44
As the GCD we do not know yet but as a factor it is okay.
Ah ok. Still, as I see, I'm lacking knowledge of basic things xD Oh well, Atleast I can continue to learn about it.