Question

Hello!!!

Question:

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.


No Spam, Please!


Step By Step Explanation Needed

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 [ P.R.O.V.E.D ]

Hope it helps :-)

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Answer 2

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 [ P.R.O.V.E.D ]


Hope it helps :-)



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