Where Algebra Fails

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willowtree
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Re: Where Algebra Fails

Post by willowtree » Mon Oct 26, 2015 11:05 pm

Paidion wrote:
I understood what you were trying to do by making the elements increase as in the original series. And you were on the right track!

So now I must do what I promised. The experts said that while algebra works with infinite series that converge, such as 1/2 + 1/4 + 1/8 +..., it does not work with series that diverge such as 3/2 + 9/4 + 27/8 +... They did not state WHY algebra doesn't work with divergent series. So I still do not know why. If you should find out, please post the answer. I would very much like to know!
Thanks. I have always had an interest in numbers,and so puzzles like this catch my attention. I can understand the diminishing series arriving at a value of 1 or close to it. I expect the diverging series will result in a value of infinity. But I will leave that for people smarter than myself.

Thanks for the discussion.

Graeme
If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Tue Oct 27, 2015 9:40 am

Since you enjoy numbers, Graeme, you might like this little number trick. First I'll do the trick with you, and later explain how to do it.

Pick a positive integer from 1 to 100. Let's say you pick 24. Divide it by 3, then by 5, and finally by 7. The remainder when you divide it by 3 is 0. By 5 it's 4. By 7 it's 3.
Post the remainders in that order: 0,4,3. I will then tell you that you chose 24.

So select your number and post your three remainders and we'll see what I can do. Then after you learn the trick, I'll post three remainders as see whether you can determine my number.
Paidion

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willowtree
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Re: Where Algebra Fails

Post by willowtree » Wed Oct 28, 2015 9:25 pm

Paidion wrote:Since you enjoy numbers, Graeme, you might like this little number trick. First I'll do the trick with you, and later explain how to do it.

Pick a positive integer from 1 to 100. Let's say you pick 24. Divide it by 3, then by 5, and finally by 7. The remainder when you divide it by 3 is 0. By 5 it's 4. By 7 it's 3.
Post the remainders in that order: 0,4,3. I will then tell you that you chose 24.

So select your number and post your three remainders and we'll see what I can do. Then after you learn the trick, I'll post three remainders as see whether you can determine my number.
A. Here are my three remainders 1,2,4. FInd out what number this represents.

B. I have just picked 3 new remainders and will try to figure this out, - 2.4.6, and let you know if I can determine what number this is. I will send you the answer and my reasoning in a little while.

Graeme
If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

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willowtree
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Re: Where Algebra Fails

Post by willowtree » Wed Oct 28, 2015 10:32 pm

willowtree wrote:
Paidion wrote:Since you enjoy numbers, Graeme, you might like this little number trick. First I'll do the trick with you, and later explain how to do it.

Pick a positive integer from 1 to 100. Let's say you pick 24. Divide it by 3, then by 5, and finally by 7. The remainder when you divide it by 3 is 0. By 5 it's 4. By 7 it's 3.
Post the remainders in that order: 0,4,3. I will then tell you that you chose 24.

So select your number and post your three remainders and we'll see what I can do. Then after you learn the trick, I'll post three remainders as see whether you can determine my number.
A. Here are my three remainders 1,2,4. FInd out what number this represents.

B. I have just picked 3 new remainders and will try to figure this out, - 2.4.6, and let you know if I can determine what number this is. I will send you the answer and my reasoning in a little while.

Graeme
Oops, looks like neither of these combinations work. I made an error on the first one and just picked random remainders for the second. I am sorry about that. Here is my calculation for the second combination.

Calculation for remainders (3) 2,(5) 4, (7) 6
Let the two digits of the final answer be x and y
Using the (5) remainder, calculate y. Since all multiples of 5 end in either 0 or 5, then with the remainder of 4, y must be either 4 or 9. So the final answer must be x4 or x9
Using the (7) remainder of 6, subtract this from the two possible values of y to get 8 or 3 ( x4 -6 =?8 or x9 - 6 =?3)
Find a multiple of 7 between 1 and 100 that has a final digit of either 8 or 3. Answers = 28 or 63 . Add back the remainder to get 34 (28 + 6) or 69.
Using the (3) remainder of 2, find which number satisfies this result. Deduct 2 from 34 (=32) and 69 (67) and see if these divide by three. Neither do.

If I had picked a remainder of 1, then 34 would have satisfied the answer, or if I had picked a remainder of 0, then 69 would have been correct.
Result, there is no number between 1 and 100 that satisfies the pattern of these remainders 2,4,6
Test for 34 34 = 7 x 4 + 6, 34 = 5 x 6 + 4, 34 = 11 x 3 + 1 , not 2
Test for 69 69= 7 x 9 + 6, 69 = 13 x 5 + 4, 69 = 23 x 3 + 0, not 2

Maybe I had better stick with theology.

Graeme
If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Thu Oct 29, 2015 11:48 am

Hi Graeme,

Actually, I don't understand what you are doing. You don't "pick" a remainder. You pick a positive integer from 1 to 100, divide it by 3, by 5, and by 7 and post the three remainders that you get.

I presume you did this, so that when you divided your number by 3, you got a remainder of 1. When you divided your number by 5, you got a remainder of 2, and when you divided your number by 7, you got a remainder of 4. On that basis, I say that the number you picked was 67.

When you divide a positive integer from 1 to 100 by 3, by 5, and by 7, you never get a negative remainder. But you have -2, 4, and 6 as remainders. How do you get a remainder of -2?
Paidion

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Thu Oct 29, 2015 12:02 pm

Hi Graeme,

Having reread your post, it seems that you "picked" the integers 2,4,6 at random as remainders, and then tried to find out the number that would generate these remainders when divided by 3, 5, and 7.
The fact is, there is no integer from 1 to 100 that satisfies the remainders 2, 4, and 6. However 104 does.

It seems you are struggling for a method to go from the remainders to the number selected. Please test your method on these remainders: 2,3,2
I actually picked a number between 1 and 100 and divided my number by 3, and got a remainder of 2. When I divided it by 5, I got a remainder of 3, and when I divided it by 7, I got a remainder of 2.
Paidion

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willowtree
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Re: Where Algebra Fails

Post by willowtree » Thu Oct 29, 2015 10:16 pm

Paidion wrote:Hi Graeme,

Having reread your post, it seems that you "picked" the integers 2,4,6 at random as remainders, and then tried to find out the number that would generate these remainders when divided by 3, 5, and 7.
The fact is, there is no integer from 1 to 100 that satisfies the remainders 2, 4, and 6. However 104 does.

It seems you are struggling for a method to go from the remainders to the number selected. Please test your method on these remainders: 2,3,2
I actually picked a number between 1 and 100 and divided my number by 3, and got a remainder of 2. When I divided it by 5, I got a remainder of 3, and when I divided it by 7, I got a remainder of 2.
I apologize for my efforts yesterday. The minus sign before the 2 was supposed to be read as a dash so all the remainders read as positive. The number I had in mind was 67 as you calculated. And yes the remainders 2 4 6 were just picked as random numbers to work the puzzle without a prior result in mind. I sent this off before I worked it through and found to my embarrassment that the result was out of range and did not work.

The remainders 2 3 2 , above, provide an answer of 23.

In my calculation I worked in the order: remainder 5, then 7 then 3. I think it is easier to work: 5, then 3 then 7. Since the sum of the digits of all numbers that divide evenly by three are also divisible by three, it makes the solution a little easier.

Graeme
If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

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Re: Where Algebra Fails

Post by Paidion » Fri Oct 30, 2015 2:41 pm

Congratulations, Graeme! You have developed a method for finding my number 23.

Here is the method I use. I even wrote a computer program in BASIC to do the work for me.

For any remainders a,b,c, use the formula 70a + 21b + 15c. To make it easier subtract 105 whenever possible.

Suppose we use this procedure to calculate the number which yields the remainders I gave you: 2,3,2

We have 70×2 + 21×3 + 15×2

But 70×2=140. That's more than 105, so subtract 105 and get 35.
21×3=63. Add the 35 to it and get 98.
15×2 =30. Add the 98 to it and get 128. That's more than 105, so subtract 105 and get 23.

I still don't understand how to carry out your procedure. But is it harder or easier to work than this one?
Paidion

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Re: Where Algebra Fails

Post by willowtree » Sat Oct 31, 2015 1:40 pm

Paidion wrote:Congratulations, Graeme! You have developed a method for finding my number 23.

Here is the method I use. I even wrote a computer program in BASIC to do the work for me.

For any remainders a,b,c, use the formula 70a + 21b + 15c. To make it easier subtract 105 whenever possible.

Suppose we use this procedure to calculate the number which yields the remainders I gave you: 2,3,2

We have 70×2 + 21×3 + 15×2

But 70×2=140. That's more than 105, so subtract 105 and get 35.
21×3=63. Add the 35 to it and get 98.
15×2 =30. Add the 98 to it and get 128. That's more than 105, so subtract 105 and get 23.

I still don't understand how to carry out your procedure. But is it harder or easier to work than this one?
I like the formula and imagined there would be a simple way to calculate the number. But I cannot figure out some of it. I understand the use of the figures 115,21, and 105. I haven't figured out why you use 70 and not 35 for the first part of the formula, except that it works. I recall the days when I wrote in BASIC as well and had fun doing it. The BASIC I wrote in was very, very,' basic' and I have not kept up with the modern versions.

Here is a more detailed explanation of how I solved the puzzle. I will use the remainders 2,3,2 to arrive at the number 23

1. There are 100 possible answers and the task is to eliminate all but one.

2. Solve the test of the multiplier (5). We know that any number that is a multiple of 5 ends with either 0 or 5 (5,10,15,etc). But because there is a remainder, then the 0 and 5 must be increased by that remainder. So for the remainder of 3, the answer must end in either 3 (0 + 3) or 8 (5 + 3). So now we know that the solution to the puzzle must have a last digit of either 3 or 8 and our options are limited to these numbers 3,13,23,33,43,53,63,73,83,93 and 8,18,28,38,48,58,68,78,88,98. There are no other numbers between 1 and 100 that pass the test of the multiplier (5).

3. Solve the test of the multiplier (3). We know that any number that is divisible by 3 also has a multiple of 3 for the sum of its digits. So, for example, 12, 72, 1212, are all divisible by 3 because in each case the sum of the digits that make up the number are also divisible by 3 (e.g. 1+2+1+2 = 6 and that is divisible by 3). [Incidentally this is also true for the number 9 as well. And 3^n.]

4. Because there is a remainder of 2 we have to factor that in as well. So instead of looking for digits that add up to 3,6,9,12,15 or 18, we must add to each of these the remainder and look for digits that add up to 5,8,11,14,17,20.

5. Of the list in point 2 only the following qualify 23 (2 + 3 = 5 etc), 53,83,38,68,98. So our list of candidates is now down to 6

7. Solve for the test of the multiplier (7). I do not know of any test for seven other than trial and error. But the easy way to do this is to deduct the remainder of 2 from each candidate (in point 5) and see if the result is divisible by 7. So this adjusted list ( above less 2 ) now becomes 21,51,81,36,66,96, for which we determine that only 21 is divisible by 7. Then we add back the 2 to get our final result 23.

8. Test the result 5 x 4 +3 = 23, 3 x 7 + 2 = 23, 7 x 3 + 2 = 23.

At first I processed the results in this order multiplier 5, then 7 then 3.
If I had done this on this example I would have determined that of the list in point 2, to find the possible results for the remainder of seven, I would have subtracted 2 from each to spot any numbers that divide by 7. This would have given me 21, 91 and 56, or finals of 23, 93 and 58. For the multiplier of three, (sum of the digits), only 23 works.

The results are the same. If I had the formula I would have used it. It looks easier, even if I don't know how it works.

Graeme
If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

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Paidion
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Re: Where Algebra Fails

Post by Paidion » Sat Oct 31, 2015 6:52 pm

Thanks, Graeme. I think you are quite skilled in manipulating numbers to solve the problem as you did!

I want to share with you a math trick that you can do with a friend, a trick that is rather impressive. The impressive part can be solved algebraically, but is not obvious:

Number 10 lines on a piece of paper. On the back of the paper write the number 1.61, lightly in pencil but don’t let your friend know that you have done so. Now ask your friend to write any two positive integers on the first two lines. To avoid large numbers, it might be wise to ask your friend to pick integers which are less than 50. For example your friend might write:

1. 27
2. 42

Give your friend a calculator and ask him or her to write the sum of the numbers on line 3.

1. 27
2. 42
3. 69

Now ask you friend to write the sum of lines 2 and 3 on line 4, the sum of lines 3 and 4 on line 5, and so on until all ten lines are completed. If done correctly, your friend should have:

1. 27
2. 42
3. 69
4. 111
5. 180
6. 291
7. 471
8. 762
9. 1233
10. 1995

Now tell your friend that you and he (or she) will have a race. Your friend will add the ten numbers on the calculator, while you add them on a piece of paper.

While your friend is adding, you simply take line 7 and multiply it by 11. You will get:

—471
471
--------
5181

Now lay your paper down and say, “Okay, I’ve finished adding them”. Your friend may not believe it. He will continue adding with the calculator, and if he does it correctly, he too will get 5181.

Now for the really impressive part! Ask your friend to take use the calculator to divide his number in line 10 by his number in line 9. In this case it will be 1995 ÷ 1233. Then ask him to write down the first three digits of his answer. Now have him turn the paper over and look at the other side. He will see the very number he wrote down, that is, 1.61
Paidion

Man judges a person by his past deeds, and administers penalties for his wrongdoing. God judges a person by his present character, and disciplines him that he may become righteous.

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