Question

Hey guys forgot mathematical induction I knew in class 9..............

If y is a number obtained from x by rearranging the digits of x , and provided that :
x+y = 10^200
Prove that x is divisible by 10.
Note that x and y are natural numbers.

Well guess what I wanna give 49 points for it .
If you spam moderators get the points.
[Be wise ........................]
dont copy from google otherwise moderators will be benefited..
Any class 12 or 11 student or a good student plz help me if you know anything about mathematical induction.

Answer 1

Given :

x + y = 10²⁰⁰

First assume that :

x = a + 10 b + 100 c + ..................... something

  = a + 10 ( b + 10 c + .................. something )

[ a is the last digit of x ]

Let 10 ( b + 10 c + ........... something ] be k

So : x = a + 10 k ......................(1)

Now y is formed by rearranging the digits of x

y = s + 10 a + 100 b +........... something

y = s + 10( a + 10 b + .......something )

s is the last digit of y

Let a + 10 b + ........... something be l

So : y = s + 10 l ..................(2)

Add both (1) and (2) we get :

x + y = a + 10 k + s + 10 l

But x + y = 10²⁰⁰

10²⁰⁰ = 10 k + 10 l + a + s

==> a + s = 10²⁰⁰ - 10 ( l + k )

==> a + s = 10 ( 10¹⁹⁹ - ( l + k ) )

==> a + s = 10 × something

This tells us that 10 | a + s

10 | last digit of x + last digit of y

Now if x has last digit 0 then only 10 | x

So we have to prove a = 0 or s = 0

Note that a is the last digit and so is s

So : a > 10 and s > 10

As 10 | a + s

This means that a + s = 10

If we subtract 1 from a ( if there exists the last digit ) :

x + y - 1 = 10²⁰⁰ - 1

            = 9999999999......................200 times

Sum of digits = 9 + 9 + ............. 200 times

                      = 200 × 9

So :

x + y has 200 × 9 + 1 ( sum of digits )

x + y = 10²⁰⁰

So sum of digits = 1

This is a contradiction regarding the digit a

This means the digit a does not exist.

There is no last digit  ! last digit = 0

x  = a + 10 k

         = 0 + 10 k

x = 10 k

Hence 10 | x

please mark it as the branliest!!!!

Answer 2

HEY ! You already know the answer . Then why asking ?

x + y = 10^200

Let last digits be s and t .

s + t + 10 * something = 10^200

= > s + t = 10^200 - 10*something

= > s + t is divisible by 10 .

s + t can only be 10 . But hey if s + t is 10 , then y + x should be divisible by 10 . But no they are not equal to 10 . s + t = 10^(200 - something ) where we have something = 199 ( which is not possible ) .

So s = 10 , t = 0 ?

This cannot be possible as s and t are the last digits and hence they should be less than 10 . They should be 0,1....9 .

So there exists no last digit .

x = 10 * second last digit + 100 * third last digit + ....

= > x is divisible by 10 off course .

I think I proved it . But you already know . So I jumped steps .