Question

N is the smallest natural number such that in decimal representation it ends with 6 and if we move last digit to the front of the number we get a number 4 times larger than the given number.Find the sum of digits of N .​

Answer 1

Step-by-step explanation:

n = 6 × 4=24

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Answer 2

Given : N is the smallest natural number such that  it ends with 6 and if we move last digit to the front of the number we get a number 4 times larger than the given number

To find :  smallest natural number Sum of digits

Solution:

Let say  Number  is  

N =  ABCD...........X6

As we do  no know  How many digits are in the Number

we move last digit to the front of the number

Hence

Number Becomes

6ABCD...........X

ABCD...........X6  *   4  =      6ABCD...........X

6 * 4  =  24      ( 2 is carried over)

Hence  X  must be 4

Repeating this process until we get 6  ....

4 *  4   = 16   + 2     =  18

Hence  Digit before X (4)   = 8     and carried over = 1

4 * 8 = 32 +  1   = 33  

Digit before 8  =  3     & Carried over  =  3

4 * 3 = 12  + 3  =  15  

Digit before 3   is  5   &  1 is Carried

4 *  5 = 20  +  1    = 21

Digit before 5  is   1   and  2 is Carried over

4 *  1= 4 + 2  = 6    (  Digit  we need at left Most )

Hence we get  615384  ( as Multiplication )  

and number before was 153846

Sum of Digit = 1 + 5 + 3 + 8 + 4 + 6  = 27

Sum of Digit of N = 27

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