Question

Is 22051946 a perfect square ?

Answer 1

You cannot use calculator.

Answer 2

it is not a perfect square

Answer 3

How did you get that ?

Answer 4

http://burningmath.blogspot.in/2013/09/how-to-check-if-number-is-perfect-square.html

Answer 5

22051946 = 2*11025973
so its odd * 2
cant be perfect

Answer 6

Great work guys. Thanks

Answer 7

Clever :) how about this question @BSwan
Is 22051946*2 a perfect square ?

Answer 8

no :)
but not the same reason hehe

Answer 9

Is there an easy proof for that digital root method ?

Answer 10

yep there is , mmm
if u show n^2 is not of the form 3m or 3m+1
then its not perfect
also there is a proof to know last digits for n^2 , wanna know it ?

Answer 11

1 , 4, 6, 9 and 0 can be the last digits of n^2 ...!

Answer 12

yes wanna prove it ?

Answer 13

So, all of them are basically of the form 3m or 3m+1

3(0) + 1 = 1
3(1) + 1 = 4
3(2) = 6
3(3) = 9
3(0) = 0

Answer 14

proof 1 :- 3k , 3k+1
proof 2 :- 0,1,4,5,6,9

they are independent

Answer 15

What kind of proof are you referring to @BSwan ? Is it something logical, like

n can have last digits as :

1 : n^2 = 1
2 : n^2 = 4
3 : n^2 = 9
4 : n^2 = 6
5 : n^2 = 5
6 : n^2 = 6
7 : n^2 = 9
8 : n^2 = 4
9 : n^2 = 1
0 : n^2 = 0

( I just noticed that I missed 5 in the previous comment)

Answer 16

number theory

Answer 17

I have read the proof of it before, but probably forgot how it goes. Will love to see how you proove it.
:-)

Answer 18

okk :)
let me get my notes , it will be better

Answer 19

Sure. Take your time.

Answer 20

so we start from here ok ?

\(\large a=\sum_{n=0}^{k} a_n 10^{10}\)

Answer 21

Okay.

Answer 22

10^n *

Answer 23

made a typo sry
\(\large a=\sum_{n=0}^{k} a_n 10^{n}\)

Answer 24

u know this cuz for any number a , digits a0,a1,a2....an cen be like this
\(\large a=a_010^0+a_110^1+a_210^2+...+a_n10^n\)
ok ?

Answer 25

so take 10 common from terms -a_0

\( a=10 M +a_0 \)
right ?

Answer 26

\[\large a = a_ka_{k-1}\cdots a_1a_0 =\sum_{n=0}^{k} a_n 10^{n} = 10M + a_0\]

Answer 27

take mod 10 for both sides
\(\large a\mod 10 = 10 M \mod 10 + a_0 \mod 10\)

Answer 28

that gives you last digit, so ?

Answer 29

mmm rash where is 10 +

Answer 30

yeah that gives u last digit for any number
\(\large a = a_0 \mod 10\)

Answer 31

nw , last digit for any number can be
\(a_0\) =0,1,2,3,4,5,6,7,8,9

Answer 32

\(\large a \equiv a_0 \mod 10\)

Answer 33

nw to check n^2

\(\large a = a_0 \mod 10\)
\(\large a^2 = a_0^2 \mod 10\)
see ?

Answer 34

Well, I need to learn modulo .. :(

Answer 35

Yes ! last digit can only be {0, 1, 4, 5, 6, 9}

Answer 36

how to prove the digital root stuff ?

Answer 37

http://burningmath.blogspot.in/2013/09/how-to-check-if-number-is-perfect-square.html

Answer 38

so we go again with
\(a_0\) =1 up to 9
and check \(a_0\mod 10\)
as u said before solver
0^2 mod 10 =0
.
.
.
.
9^2 mod 10 =1
they would be the only sol

Answer 39

Understood that last digit proof... I am trying to figure out how to prove restrictions on digital root

Answer 40

@mathslover I can teach you mods in less than 30 minutes :)

Answer 41

Will try some other time, ty @BSwan :)

Answer 42

Oh wait actually it looks easy to prove with your summation notation

Answer 43

nvm it looks complicated, wil try later >.<

Answer 44

@rational thanks for that. May be, tomorrow or day after that, we will talk about this :-) Will make a post then we will discuss. Thanks again.

Answer 45

sorry for laging ,i was having lunch :)
lets proof digital root :)

Answer 46

Interesting, how are you going to express that process mathically ?

Answer 47

to express lets say
\(\S_K= \large \sum_{n=1}^{k} a_n \mod 10^n\)
but im still thinking of a method to prove it