Find the smallest natural number which has the following properties:
(a) Its decimal representation has 6 as the last digit.
(b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number.
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Let A be the remaining digits such that
4 ( 1 0 A + 6 ) = 6 × 1 0 k + A ⟹ A = 3 9 6 × 1 0 k − 2 4 ,
where k is the total number of digits of n .
Next we do case works from k = 1 , 2 , 3 , … and find that when k = 5 , A = 1 5 3 8 4 is an integer. Therefore, n = 1 5 3 8 4 6 .