# Find the smallest positive integer statisfying two conditions

Find the smallest positive integer satisfying both of the following requirements:
$(a)$ Its units digit is $6$
$(b)$ If the units digit $6$ is moved before the first digit, the new number is $4$ times the original number.

Well obviously the units digit of the answer is $6$, but I don't know how to proceed. The only explanation I received was "Just work out the digits one by one, starting from the right," which to me is ambiguous.

Start with the units digit being $6$. Then you know that the number is of the form $$\ast\ast\ast\ast\cdots\ast\ast6.$$ We know that $$6\ast\ast\ast\ast\cdots\ast\ast$$ should be $4$ times the original. Since $6\cdot 4=24$, the units digit of this new number is $4$. Therefore, this number is of the form $$6\ast\ast\ast\ast\cdots\ast4.$$ Hence, the original number is of the form $$\ast\ast\ast\ast\cdots\ast46.$$ Now, keep going like this.

I got an answer of $153846$.

• I was just trying this. Is there a way to show that this process will terminate? Other than actually doing it and finding that it does, I mean. – Theoretical Economist Dec 23 '16 at 2:35
• No, I don't think that it necessarily terminates (although, it would be an unfair question if it didn't) – Michael Burr Dec 23 '16 at 2:36
• This will always terminate if the problem has a solution. It is based on the repetition of the decimal expansion. If the problem does not have a solution (shame on the setter) it will loop without coming to a solution which indicates that fact. – Ross Millikan Dec 23 '16 at 2:40
• Oh, so that's what the explanation meant... btw your answer is correct – suomynonA Dec 23 '16 at 2:41
• You know that the last two digits are $46$, multiplying that by $4$ gives that the last two digits of $4$ times the original is $84$. Moving the $6$ back to the units place means that the last three digits are $846$. Continue... – Michael Burr Dec 23 '16 at 2:51

Hint: the formal way: write your original number as $10a+6$ and let $a$ have $n$ digits. Then $4(10a+6)=6\cdot 10^n+a$ Solve this for $a$ and you will get a fraction. Find the smallest $n$ that makes the fraction a whole number.

• Is there a formula or rule for this? – suomynonA Dec 23 '16 at 2:47
• If you did what I suggested you would see the denominator is $39=3\cdot 13$ The repeat of the decimal $1/39$ is $6$ places long because $39$ divides into $999\ 999$ which tells you the number you want is six digits long. – Ross Millikan Dec 23 '16 at 5:29

The hint means just what it says. Your know that your number is $\text{[some digits]}6$, and you should go ahead and start multiplying it by $4$ just as you would do multiplication of a long number by a single-digit number on paper by hand: $$\text{[some digits]}6\cdot4=6\text{[same digits]}.$$ First of all, since $6\cdot4=24$, the last digit of the product is $4$. So it's the last of unknown digits, and now you have $$\text{[some digits]}46\cdot4=6\text{[same digits]}4.$$ Keep on going until you get a $6$ in the beginning of the product, indicating that you're done.