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 $153846$.

  • $\begingroup$ 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. $\endgroup$ – Theoretical Economist Dec 23 '16 at 2:35
  • $\begingroup$ No, I don't think that it necessarily terminates (although, it would be an unfair question if it didn't) $\endgroup$ – Michael Burr Dec 23 '16 at 2:36
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Dec 23 '16 at 2:40
  • $\begingroup$ Oh, so that's what the explanation meant... btw your answer is correct $\endgroup$ – suomynonA Dec 23 '16 at 2:41
  • 1
    $\begingroup$ 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... $\endgroup$ – 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.

  • $\begingroup$ Is there a formula or rule for this? $\endgroup$ – suomynonA Dec 23 '16 at 2:47
  • $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.