2
$\begingroup$

If a nine-digit number is formed using each of the digits $1,2,3,...,9$ exactly once, for $n = 1,2,3,...,9$, $n$ divides the first $n$-digits of the number, Find the number.

I can think that the fifth digit will be $5$, as there is no other option, and also that I do not have to think of making the no. divisible by $9$, as the sum of its digits will always be equal to $45$. I have also applied the divisibility rules of $2,4,6,8$ and can say that the second, fourth, sixth and eighth should be even, which implies that others are odd. But I am unable to pile up all the findings leading to the solution.

$\endgroup$
  • $\begingroup$ This is remarkably similar to one of Conway's Pi Day pizza problems. $\endgroup$ – Chris May 19 '17 at 4:55
  • $\begingroup$ Yes, you are right $\endgroup$ – ami_ba May 19 '17 at 5:04
  • $\begingroup$ If we label each of the digits by position, then so far I get $3|(d_1 d_2 d_3)$, $4|(d_3 d_4)$, $6|(d_5 d_6 d_7)$, and $8|(d_7 d_8)$. Not sure if this is the right direction to go in.... $\endgroup$ – Chris May 19 '17 at 5:09
  • $\begingroup$ ofc. I also got these, but after that? These are just math. rep. of what I said $\endgroup$ – ami_ba May 19 '17 at 5:11
  • $\begingroup$ Well, for one thing there are few two-digit numbers divisible by $8$, so I'm looking at that right now. $\endgroup$ – Chris May 19 '17 at 5:13
5
$\begingroup$

The fifth digit is $5$ as you said.

The second, fourth, sixth, and eighth digits are even, as you said.

The fourth digit must be $2$ or $6$ since the third digit is odd.

The fourth, fifth, and sixth digits must sum to a multiple of $3$, so the sixth digit must be $8$ or $4$ respectively (depending on whether the fourth digit is $2$ or $6$ respectively).

The 3-digit number formed by the sixth, seventh, and eighth digits must be divisible by $8$. Since this looks like "$8ab$" or "$4ab$" where $a$ is odd, the eighth digit must be $6$ or $2$, whichever is not used by the fourth digit.

So, the second digit must be $4$ or $8$, whichever is not used by the sixth digit.

So far, we have two choices: $$x_1 4 x_3 2 5 8 x_7 6 x_9$$ $$y_1 8 y_3 6 5 4 y_7 2 y_9$$

We just need to put $1$, $3$, $7$, and $9$ in the blanks to make the first three digits add up to a multiple of $3$, and the last three digits add up to a multiple of $3$.

Let's try the first choice. The first three digits are either $147$ or $741$. In either case, the divisibility by $8$ forces the last three digits to be $963$. Finally, we check divisibility by $7$, to decide between $147$ or $741$ in the first three digits, but neither work.

So we are left with the second case. The divisibility by $8$ means the seventh digit $y_7$ is either $3$ or $7$. If it is $3$, we can try either $789$, $189$, $987$, or $981$ for the first three digits and see if the divisibility by $7$ holds; none of them work.

So we must have $y_7=7$, and we can try $183$, $189$, $381$ and $981$ for the first three digits; $381$ satisfies the divisibility by $7$. Thus the answer is $$381654729.$$

$\endgroup$

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.