A queer question for smart people =]

Click here · 2007-07-20 03:04:43

Well here it is...

Show that:

27 x 23^n + 17 x 10^2n is divisible by 11 for all positive integers of n.

2//

1999 numbers are placed along the cricumference of a cirlce. when any four successive numbers are added, the total is always 28. what are these 1999 numbers? find all possible solutions

I don't even understand the question >.<

John E. Franklin · 2007-07-20 04:05:26

2.)
27 combinations is the answer if clockwise/counter-clockwise is forgotten
41 combinations is the answer if clockwise and counter-clockwise are DIFFERENT.
See way below for how I did it.

Nineteen Hundred Ninety-Nine is a prime number and I think that is not what they are talking about. I think the 1, 9, 9, 9 is an example of 4 numbers that add to 28.
Now 9199 and 9919 and 9991 are all the same if they go in a circle is what they mean, I think.
So now use other numbers like 8 8 8 3, which add to 27.
And use 5 + 10 + 6 + 6.
Just keep working on all of them until you get tired and then form another strategy.
Think small.

ZERO
0 0 0 0

ONE
1 0 0 0

TWO
2 0 0 0
1 1 0 0, 1 0 1 0

THREE
3 0 0 0
2 1 0 0, 2 0 1 0
1 1 1 0

FOUR
4 0 0 0
3 1 0 0, 3 0 1 0
2 2 0 0, 2 0 2 0
2 1 1 0, 2 1 0 1
1 1 1 1

FIVE
5 0 0 0
4 1 0 0, 4 0 1 0
3 2 0 0, 3 0 2 0
3 1 1 0, 3 1 0 1
(And for ten, 1234 is same as 4321 if clockwise/counter-clockwise is forgotten)
(Have you taken probability yet, that might help too.)
2 2 1 0, 2 0 2 1
2 1 1 1

SIX
6 0 0 0
5 1 0 0, 5 0 1 0
4 2 0 0, 4 0 2 0
4 1 1 0, 4 1 0 1
(There is some recursion here, but I can't see the general case.)
3 3 0 0, 3 0 3 0
(Like next I would break down the 3 in 3 spots, but the 4th circle link messes it up perhaps)
3 2 1 0, 3 0 2 1 (Because we got two cases here: 2 1 0 and 0 2 1 which would have been the same if it were a circle with 3 numbers)
WOOPS, AND FORGOT 3 2 0 1
3 1 1 1
2 2 2 0
2 2 1 1, 2 1 2 1

SEVEN
I'll keep working on this and edit it later, bye!!!

TWENTY-EIGHT
28 0 0 0
27 1 0 0, 27 0 1 0
26 2 0 0, 26 0 2 0
26 1 1 0, 26 1 0 1
25 3 0 0, 25 0 3 0
25 2 1 0, 25 2 0 1, 25 0 2 1
25 1 1 1
24 4 0 0, 24 0 4 0
24 3 1 0, 24 0 3 1, 24 3 0 1
24 2 2 0, 24 2 0 2
24 2 1 1, 24 1 2 1 (Note that the one's can be viewed like zeros, just ignore them sort of)
23 5 0 0, 23 0 5 0 (Same idea as line above, but with zeros)
This is easy and interesting, don't you think??

Hey I just thought, maybe we are only supposed to use
the digits 0 to 9.
Let's do that, it will be faster...
9 9 9 1, 9 1 9 1
9 9 8 2, 9 8 9 2
9 9 7 3, 9 7 9 3
9 9 6 4, 9 6 9 4
9 9 5 5, 9 5 9 5
cool!! continue...
9 8 8 3, 9 8 3 8
9 8 7 4, 9 8 4 7, 9 4 8 7 (Pretend 7 is a zero and ignore it, put the other 3 numbers in the middle)
9 8 6 5, 9 8 5 6, 9 5 8 6
Wow! Are we done yet???
9 7 6 6, 9 6 7 6
Woops I forgot one...
9 7 7 5, 9 7 5 7
I messed up the order of last two lines, sorry.
THIS ONE 9 6 6 6 IS TOO SMALL, NOT 28, SO IGNORE AND CONTINUE.
So go to 8's now, 9's are all done I think.
8 8 8 4
8 8 7 5
8 8 6 6
8 7 7 6
7 7 7 7
Yeah!!!!! We did it!!!!
(Now don't forget about clock-wise and counter-clockwise?

This will make some additonal ones if it is not forgotten.
Just turn them around backwards and see if they are different in a circle.
27 combinations is the answer if clockwise/counter-clockwise is forgotten
41 combinations is the answer if clockwise and counter-clockwise are DIFFERENT.

27 + 14 is 41
So there are 14 ones that if you turn them around the other rotation, I consider them different.
Can you find 14 too.
At first I thought it was 16, but I made a mistake.

Last edited by John E. Franklin (2007-07-20 05:07:18)

Ricky · 2007-07-20 04:40:43

27 x 23^n + 17 x 10^2n is divisible by 11 for all positive integers of n.

This is easy once you know numbers modulo n. I'm going to assume that you do...

First theorem is that:

It should be fairly easy to see that this is true. Because n^2 = n*n which is really 1*1. Come up with a similar theorem about -1 (Or rather in this case, 10). Remember, the 2n in this power is very important, and it's crucial to see why.

Once this is done, simply multiply each of these by 5 and 6, add then, and you got your answer.

mathsyperson · 2007-07-20 04:48:16

Define a starting point somewhere on the circle, so that we have a "1st" number (situated between the 1999th and the 2nd).

Calling the first five numbers a, b, c, d, e we know that a+b+c+d = 28.
However, moving once along, we also know that b+c+d+e = 28.

Therefore, a+b+c+d=b+c+d+e and so a=e.
Generalising this, it is easy to see that any numbers that are 4 away from each other must match.

The 4th number is the same as the 8th number is the same as the 12th number is the same as the ... is the same as the 1996th number. But going along once more, we've gone all the way around the loop and back to the 1st. Going along once more again, we get to the 5th. The important discovery here is that #4 is the same as #5.

Generalising this, any number is the same as the number next to it.
This means that all numbers on the circle are equal to each other, and so the only solution is to fill it with sevens.

JaneFairfax · 2007-07-20 08:16:54

Click here wrote:

Show that:

27 x 23^n + 17 x 10^2n is divisible by 11 for all positive integers of n.

Use induction on n, noting that

Last edited by JaneFairfax (2007-07-20 08:17:56)

krassi_holmz · 2007-07-30 12:26:16

Or we may use some non-trivial way-using generating functions instead.
First we find the generating function of

:

Now the generating function of the expression will be:
, which is clearly divisible by 11. And we can continue this to obtain:

Last edited by krassi_holmz (2007-07-30 12:46:47)

Math Is Fun Forum

#1 2007-07-20 03:04:43

A queer question for smart people =]

#2 2007-07-20 04:05:26

Re: A queer question for smart people =]

#3 2007-07-20 04:40:43

Re: A queer question for smart people =]

#4 2007-07-20 04:48:16

Re: A queer question for smart people =]

#5 2007-07-20 08:16:54

Re: A queer question for smart people =]

#6 2007-07-30 12:26:16

Re: A queer question for smart people =]

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