Determine the smallest positive integer x, whose last digit is 6 and if we erase last digit 6 and put it in the left most of the number the new number becomes 4x.
Answers
Given : the smallest positive integer x, whose last digit is 6 and if we erase last digit 6 and put it in the left most of the number the new number becomes 4x.
To find : smallest natural number
Solution:
Let say Number is
x = ABCD...........N6
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...........N
ABCD...........N6 * 4 = 6ABCD...........N
6 * 4 = 24 ( 2 is carried over)
Hence N must be 4
4 * 4 = 16 + 2 = 18
Hence Digit before N (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
x = 153846
4x = 615384
153846 is smallest positive integer
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