A question from a 6th grader in China...who can do this?

Question

I have a niece living in China and she's in her 6th grade now. Here's one of her math homework:

In the series 11, 111, 1111, 11111, 111111, ......prove that no number in it is the square of an integer.

I know the square of any number ending with a 1 or 9 will be ending with 1. But what is next??

Thanks*10000000000000

Answer 1

let suppose one of the numbers is equal to m^2
m^2 ends in 1 so m must end in 1 or 9
so m must be of the form 10a+1 or 10a+9
so m^2 = 100a^2 + 20a + 1
... but the first term ends with two zeros and the second has to be an even number with a zero after, so the second place of m can't be 1
or m^2 = 100a^2 + 180a +81
... but then we need 18a +8 to give the second place one so 18a must end with 3, but it can't because its even
Therefore numbers of this shape can not be a perfect square of an integer

Answer 2

All these numbers are odd.
The first is of the form 8n+3 and the second is
of the form 8n+7. All the rest are of the form
8n+7 because they can all be written as 1000m + 111
= 8(125m+n) + 7.
We will show that any odd square is of the form 8n+1,
which will show none of these numbers can be a square.
Let's consider 4 cases:
A). 8k+1. (8k+1)² = 64k²+16k+1 = 8(8k²+2k) +1.
B). 8k+3. (8k+3)² = 64k²+48k+9 = 8(8k²+6k +1) + 1
C). 8k+5. (8k+5)² = 64k² +80k+25 = 8(8k²+10k +3)+1
D). 8k+7. (8k+7)² = 64l² + 112k + 49 = 8(8k²+14k+6) +1
So the result follows.

Answer 3

hi

square ending with 1 and 9 -> ending with 1
square ending with 2 and 8 -> ending with 4
square ending with 3 and 7 -> ending with 9
square ending with 4 and 6 -> ending with 6
square ending with 5 -> ending with 5


then
square ending with 1 and 9
(10*k +/-1)^2 =
= 100*k +/- 20*k + 1

two last digits

21
41
61
81
01

bye

Answer 4

If I understand your question correctly(which maybe I don't) cause it sounds to easy but can you just say that the square root of each of those numbers does not equal an interger. hahahhahahahah i guess that is not the format of proofs right?

Answer 5

to square any of the numbers you are effectively adding diadonal rows on ones together

111.....1111 X11......1111=

. 111......1111
. 111......1111
etc
etc

--------------------------
1........................21

so any of the numbers squared ends in..........21

hence none cane be squares of lesser ones