A few questions

bobbym · 2012-12-04 14:51:17

Hi Agnishom;

Will you please explain this line?

Agnishom · 2012-12-05 01:00:26

Thanks

Question 5:
The 64 squares of an 8×8 chessboard are filled with positive integers in such a way that each
integer is the average of the integers on the neighbouring squares. (Two squares are neighbours
if they share a common edge or a common vertex. Thus a square can have 8, 5 or 3 neighbours
depending on its position).
Show that all the numbers are infact equal

bobbym · 2012-12-05 09:50:13

Hi Agnishom;

Agnishom · 2012-12-05 14:41:31

Hmm... I see

Question 6: Prove that the tens digit of any power of 3 is even. [e.g. the tens digit of 3^6 = 729 is 2].

anonimnystefy · 2012-12-05 22:03:15

You can just start writing out the last two digits of all powers of 3. It should evetually get into a loop. Yiu can then note that all powers of three have an even ten's digit.

scientia · 2012-12-05 22:40:10

Agnishom · 2012-12-06 01:06:10

Hi anonimnystefy;
Would the loop be enough for a proof?

Hi scientia;
Would you please explain a little? Especially, the last two lines

scientia · 2012-12-06 01:29:41

Agnishom wrote:

Hi scientia;
Would you please explain a little? Especially, the last two lines

The last two digits of are the last two digits of . Hence the ten's digit of (call it N) is the last digit plus any carry over from . If , is a single-digit number so there is no carry-over: N is just the last digit of . If , so the carry-over is 2: N is the last digit of .

scientia · 2012-12-06 02:00:30

Here's another proof, by induction; perhaps you might find it easier to follow.

First note that the tens' digit of is 0, which is even.

Suppose the tens' digit of

is even, say, . If the its last digit is , we can write

where

is a multiple of 100 and (powers of 3 can only end with these digits).

Now multiply through by 3.

The tens' digit of

is thus the last digit of plus the carry-over from .

(i) If s = 1 or 3, there is no carry-over: the tens' digit is the last digit of

.

(ii) If s = 7 or 9, the carry-over is 2 because

or , so the tens' digit is the last digit of .

In either case, the tens' digit is even. Hence, by induction, the tens' digit of

is even for all .

Agnishom · 2012-12-06 03:49:28

I think I like this proof more! Thank You

Agnishom · 2012-12-06 13:45:14

Question 7: Determine the set of integers n for which

Last edited by Agnishom (2012-12-06 14:56:39)

bobbym · 2012-12-06 15:21:01

Hi Agnishom;

Agnishom · 2012-12-07 00:53:47

In the first place how do I know that we have to multiply the sides by 4?
I mean how do we get that idea?

By the way, what do you mean by 'y'?

Last edited by Agnishom (2012-12-07 00:55:36)

bobbym · 2012-12-07 04:00:55

In the first place how do I know that we have to multiply the sides by 4?
I mean how do we get that idea?

Whether that is a stroke of genius or just a load of hooey, I am not sure. Perhaps when you try to complete the square or collocate for the RHS it becomes easier to see.

That y is a k, sorry. I have changed it.

Agnishom · 2012-12-07 04:27:41

And how did you manage to find the unique solution in the last few lines?

bobbym · 2012-12-07 04:35:18

The solution comes from this line here.

The squares only have a few numbers that they can end in. This is another way of saying mod 10. The subtraction of two squares has only one choice that can end in a 7.

anonimnystefy · 2012-12-07 05:20:57

Another thing you could do is apply the difference of squares formula and set the factors you get equal to factors of 7.

Agnishom · 2012-12-07 15:51:42

bobbym wrote:

The solution comes from this line here.
The squares only have a few numbers that they can end in. This is another way of saying mod 10. The subtraction of two squares has only one choice that can end in a 7.

I understand upto that much but what should we be doing after that?

anonymnistefy wrote:

Another thing you could do is apply the difference of squares formula and set the factors you get equal to factors of 7.

Will you please explain a little bit?

bobbym · 2012-12-07 15:58:41

Agnishom · 2012-12-07 16:07:15

Hmm...I see. Thanks.

How do you know that those are the only solutions? There could be other numbers which ends with 6.

bobbym · 2012-12-07 16:09:19

Hi;

That is what we get when we solve the two equations. The mods only told us that one term had to end in 6 and the other in 9, it doesn't tell us how many. Solving the equations give the only solutions.

I can show you another way that maybe is easier for you.

Agnishom · 2012-12-07 16:23:27

Oh! Please do so

bobbym · 2012-12-07 16:49:17

Hi;

We can use the difference of two squares rule to factor it like this:

The factors of 7 are 7 * 1 and -7 * -1

You now have to solve 4 sets of 2 equations.

Same answer as before.

This way is easier to understand at first but I wanted to show you how it was done with mods first. For these MO problems a very good understanding of the properties of mods is required.

noelevans · 2012-12-07 17:13:54

Here's a thought:

In general the difference of two squares can be written as

2 2 2 2 2 2
(x+i) - x = x + 2xi + i - x = 2xi+i = i(2x+i).

Setting this equal to 7, the only solution for i and x both greater than zero is x=3 and i=1.

2 2 2 2
So this (x+i) - x = 4 - 3 = 16-9 = 7.

So the (2k)^2 = 16 making k=2 or -2.
And the (2n+19)^2 must equal 9 so 2n+19=3 or -3
Hence 2n+19 = 3 or 2n+19=-3
2n = -16 or 2n = -22
n = -8 or n = -11

phrontister · 2012-12-07 18:08:59

Hi noelevans,

Have you seen the 'useful symbols' above the menu headings near the top of the page?

With the superscript '2', this

2 2 2 2 2 2
(x+i) - x = x + 2xi + i - x = 2xi+i = i(2x+i).

can be turned into that

(x+i)² - x² = x² + 2xi + i² - x² = 2xi+i² = i(2x+i).

Math Is Fun Forum

#51 2012-12-04 14:51:17

Re: A few questions

#52 2012-12-05 01:00:26

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