# Find the smallest insertable number

Let's say a number $$n$$ is insertable if for every digit $$d$$, if we insert $$d$$ between any two digits of $$n$$, then the obtained number is a multiple of $$d$$. For example, $$144$$ is not insertable because $$1474$$ is not divisible by $$7$$.

The question is the find the smallest insertable positive integer with at least two digits.

It is relatively easy to see that such a number have to be divisible by $$2520$$ (assuming it is at least $$4$$-digits long). I also ran a script to check all integers below 75,000,000,000 with no success (the issue might be my code).

Disclaimer. I do not know if such a number do exist.

• if the number ends with 000 then we have $2,4,5,8,10$ covered Sep 29, 2020 at 16:56
• if the number is a multiple of $9$ we also have $3,6$ and $9$ covered. Sep 29, 2020 at 16:57
• With this type of reasoning we may be able to find a satisfactory number, but not necessarily the smallest one.
– user601568
Sep 29, 2020 at 16:58
• I think this one works: $77777777000$ Sep 29, 2020 at 16:59
• Yes the big problem is $7$
– user601568
Sep 29, 2020 at 16:59

Let $$a_m....a_1a_0$$ be any insertable number. Then, for each digit $$d$$ we must have $$a_m..a_kda_{k-1}..a_0$$ is a multiple of $$d$$.

In particular, $$d| a_m..a_kda_{k-1}..a_0-a_m..a_ka_{k-1}d..a_0=10^{k-1}9(d-a_k)$$

Since $$7$$ is the only digit relatively prime with $$10$$ and $$9$$, we should concentrate on $$d=7$$.

$$d=7$$ implies that $$a_k=0,7$$ for all $$k$$.

Next, $$d=9$$ implies that the number must contain at least 9 sevens. Since it must end in $$000$$ the smallest possible example is indeed $$777777777000$$.

P.S. The above shows that any insertable number must have all digits $$0$$ and $$7$$, contain a multiple of $$9$$ number of $$7$$'s and end in three 0's. It is easy to check if the converse is also true, I think it is but I am too lazy :D

• I think only $2$ zeros suffice in retrospect Sep 29, 2020 at 17:29
• @JorgeFernández-Hidalgo $787777777700$ is not divisible by $8$. Sep 29, 2020 at 17:30
• oh yeah ${}{}$. Sep 29, 2020 at 17:30

We are going to characterize the numbers such that when you insert a $$7$$ in between the number is still a multiple of $$7$$.

Suppose the number has consecutive digits $$a$$ and $$b$$ such that $$a$$ is not the leftmost digit.

We can consider the number that is formed when you put the $$7$$ to the left of the $$a$$ and also the number that is formed when you put the $$7$$ between the $$a$$ and $$b$$. Notice the difference of these numbers is a multiple of $$7$$, the difference between these two numbers is $$9(7-a)$$ multiplied by a power of $$10$$. We conclude every digit must be a $$7$$ or a $$0$$ except for possibly the first and last ones. In our case we know that the final digit is a $$0$$. But if this happens then the first one must also be $$7$$ because otherwise the number won't work.

Therefore the number must only have zeros or sevens.

If the number ends in $$70$$ it wont work because $$780$$ is not a multiple of $$8$$. If it ends in $$700$$ it wont work because $$700$$ is not a multiple of $$8$$ (notice that there are at least $$9$$ digits so we can put the $$8$$ to the left).

With the condition that the number must be a multiple of $$9$$ we need at least $$7$$ zeros and the number must end in $$00$$. The smallest number is $$777777777000$$