Mark Said:

HARD middle school(?) math problem!!?

We Answered:

Funny, any possible n seems the binary expression of another number. I dont still know if its important but all the powers of 10 that are perfect squares correspond, if considered binary expressions, to perfect squares.

ie: 10^6 is the binary expression of 2^6.

I realised that a square doesn't have to have a binary expression that is a square if considered a decimal number,

ie 25 = 11 001 (2) and 11 011 is not a square.

I don't think that the contrary is possible but I don't still have a proof for this intuitive statement.

****************** ********* *****************
If x is sum 10^k x_k, where x_k is a digit, then

x^2 = sum 100^k (x_k)^2 + 2 sum x_i x_j 10^(i+j), i different from j

********************* ************ ************
Step 1: From 10^(2k) to 10^(2k+1)

+++++++ +++++++++++ ++++++++++++++
A PREVIOUS REMARK

There are no perfect squares between 10^(2k) and 10^(2k+1) or the counter-example would be very easy to find. So, I will work with k that are not the first couple of natural numbers.

Now, lets try to prove that there are no perfect squares between 10^(2k) and 10^(2k+1). I will accept that there are none until a certain k_o, which can be proved by inspection of the first possible n.
++++++++++ +++++++++ +++++++++++++

There are no two perfect squares that are consecutive positive integers. So, since 10^(2k) is a perfect square, then 10^(2k) + 1 is not one.

In fact, * if * there is a number n that is an even perfect square, say 1010....00010, then n+1 is not a perfect square, either. Hence, an odd n can only be a square if n-1 is not one, that is even.

Once I have found a perfect square, p^2, I know how many numbers arent perfect squares: all the ones that are less than (p+1)^2, the difference between them grow with p

(p+1)^2 - p^2 = 2p+1, these are consecutive odd integers.

Since 10^(2k) is a perfect square, then not only 10^2 + 1 can't be a perfect square, neither 10^(2k) + 10 nor 10^(2k) + 11 can. This reasoning can be extended and prove that some of the possible n aren't perfect squares.

............... ................... ............
For n = 10^2k, then p = 10^k and 2p+1 = 2*10^k + 1, which means that all the n that are less than (10^2k + 1)^2 cant be squares. *And* there are no possible n between 10^2k + 2*10^k+1 and 10^2K + 10^(k+1)
-----------------------------------
FOR ANY 10^2K, NONE OF THE NUMBERS THAT ARE BETWEEN IT AND 10^2K+10^(K+1) CAN BE A PERFECT SQUARE.
-------------------------------------
Example 1: There are no perfects squares between 1 000 001 and
1 010 000.

We can predict what the perfect squares just by adding consecutive odd integers:
--------------------------------------…
10^2k
10^2k + 2*10^k + 1
10^2k + 4*10^k + 4
10^2k + 6*10^k + 9
10^2k + 8*10^k + 16
10^2k + 10*10^k + 25 = 10^2k + 10^(k+1) + 25

The last digits, these are the perfects squares from 1, 2, 3, 4 and 5. There is no way we'll get only 0 and 1 unless we go on until I get to 10^2

10^2k + 6*10^k + 36
..............................
10^2k + 10*10^k + 100 = 10^2k + 2*10^(k+1) + 100

Lots of 0 and 1 but there's a 2...

------------- ------------------ -------------------
The perfect squares' end is ciclic:

00*, 01*, 04, 09, 16, 25, 36, 49, 64, 00*, 21, 44, 69, 96, 25, 56, 89, 24, 61, 00*, 41, 84, 29, 76, 25, 76, 29, 84, 41, 00*, 61, 24, 89, 56, 25, 96, 69, 44, 21, 00*, 81, 64, 49, 36, 25, 16, 09, 04, 01*, 00*
-------------- ----------------- ------------------

From now on the perfect square at the end of n wont only have 0 and 1 until we get to 10^4 .... 90 perfect squares later!

Example 2:
10 000 200 001
10 000 400 004
10 000 600 009
10 000 800 016
10 001 000 025
10 001 200 036
10 001 400 049
...........................
10 002 000 100
10 002 200 121
10 002 400 144
..........................
10 003 800 361
10 004 000 400
..........................
10 020 010 000

HENCE, THERE WONT BE ANY PERFECT SQUARES AT LEAST BETWEEN 10^2K + 1 AND 10^2K + 2*10^(K+2) + 10^4 (k >3
since this is the square from 10^k+100)

In fact, we wont get an end only composed by 0 and 1 unless we have a power of 10. Otherwise, we would have found this square before because its less than 10^2k

So, only the numbers that end in a power of 10 can be perfect squares. None of the other ones are. So, lets study these ones

(10^k + 10^j)^2, where k is greater than j

10^(2k) + 210(j+k) + 10^(2j). But these ones have a 2 in the middle.

This proves all the numbers *until* the square that is in the end gets to the position where the 2 is.

Example 3:

1 000 000
1 002 001
................
1 018 081
1 0'20'100 = 10^6 + 20*10^3 + 10^2
.................
1 060 900
1 062 961

From now on the situation is different: the next square is1 064 000 + 1024 but the digits 4 and 1 are added and this changes things in a drastic way: a square like 40000 could eventually be the end of an n

............... ...................... ........................
So: There won't be any square between 10^(2k) and (10^k+ p)^2 if p^2 has less than k+1 digits.

10^2k+ 2*10^k + p^2

Example:

Numbers like 1012 won't be squares
(10^4+12)^2 = 10^8 + 24*10^4 + 12^4 = 100 000 000+ 240 000 + 144

But numbers like 1400 eventually could
(10^4+1400)^2 = 10^8 + 2*1400*10^4 + 1400^2

100 000 000
. 28 000 000
... 1 960 000

In this case, 8 and 1 are added together and they could eventually generate more 0 or 1 than the other kind of number.

(Still working on it)
--------------------------------------…
Step 2: From 10^2k+1 to 10^(2k+2)

What about the numbers that are odd powers of 10? These aren't perfect squares, either, we have proved these ones too. What about the previous square?

Example:

[V(10^7)] = [V10] V10^6 = [V10] * 10^3

V10 = sqrt 10 and [x] is the integer function, as in [8.4] = 8

So, we know that, between this number and the following square there are no possible n. This proves a lot more numbers.

Not still sure that repeating this procedure until we get to the next power of 100 is always possible but I tend to agree with the statement to be proved in an intuitive way.

(I'll go on)

Ana

To Ray: I will think of your argument, maybe it will complete somehow what I already proved

To manjyomesando1: Your answer is surelly an answer and a quite good one. Now, why do you state this?

0.1 * 10^(2k) < n < (1/9) * 10^(2k)
10^(2k-1) < n < [10^(2k)]/9

Example:

k = 4
0.1 10^8 < n < (10^8)/9
10 000 000 < n < 111 111 1111

I think that maybe 0.01 10^2k would be better, then you'd have all the n between two powers of 100.

Why should a square only be between these powers of 10? Why exclude [10^(2k/)]9

I think that you should have written

0.1 * 10^(2k) < n <= (1/9) * 10^(2k)

I like your reply but for this point.

Ana

Dora Said:

need help on what middle school math book to buy?

We Answered:

Check saxon.com for math books.
Also check parentteacher store site for information.

Louise Said:

i forgot my math book at school and i need to do my homework from there?

We Answered:

type the name of the book into google. there is likely a website for the classzone books and it is possible to see some textbooks online so look around the webpae. otherwise just call a friend to get the assignment.

Jay Said:

What is the answer to the math book Scott Foresman-Addison Wesley Middle School math course 2?

We Answered:

first:WHAT
2nd:dont ask a stupid question
3rd:give more info
4th:ask someone who cares
5th:dont give the pages..give the equation/s..not everyone has the book

Terrence Said:

WDYT of the book: Math Doesn't Suck: How to Survive Middle-School Math Without ... Breaking a nail?

We Answered:

Maybe middle-school girls generally need some extra help with math, I don't know. But "breaking a nail"? Who came up with that, Cosmo?

Troy Said:

Where can I find the full text of a middle school math book online?

We Answered:

I don't think the full text is online. Textbooks have websites to go along with the text that have review and practice tests and things like that, but the entire text I don't think so.


Do you know the names of publishers?
Like Prentice Hall, McGraw-Hill, Glencoe