Tried: Seems the ten-digit number ends with $240$ or $640$ or $840$ (Is not true, there are more ways the number could end)

$8325971640,$ $8365971240,$ $8317956240,$ $8291357640,$ $8325971640,$ $8235971640,$ $1357689240,$ $1283579640,$ $1783659240,$ $1563729840,$ $1763529840,$ $1653729840,$ $7165239840,$ $7195236840,$ $2165937840,$ $9283579640$

  • $\begingroup$ any multiple of 27720 is a multiple of all those numbers up to 11. Not sure what else they re asking $\endgroup$ – Will Jagy Oct 27 '18 at 0:11
  • $\begingroup$ maybe I understand. Divisibility by 9 is automatic here. The digits add up to 45. For 11, we need to choose a group of four digits and a group of five digits, so the sums differ by a multiple of 11, yet add to 45. So, 28+17 = 45 or 39+6 = 45. We cannot use the second one because four digits ad up to bigger than 6, so five (distinct) digits (including the highest 10^9 place) add to either 17 or 28. $\endgroup$ – Will Jagy Oct 27 '18 at 0:21
  • $\begingroup$ 100 place a and 10 place b, we need $100a + 10b \equiv 0 \pmod 8,$ or $4a+2b \equiv 0 \pmod 8,$ or $2a + b \equiv 0 \pmod 4. $ The choices become 120, 320, 520, 720, 920; 240, 440, 640, 840 where 440 has a repeat. then 160, 360,560,760,960; 280, 480, 680, 880 which is a repeat $\endgroup$ – Will Jagy Oct 27 '18 at 0:26
  • $\begingroup$ Found the answer 2438195760, 3785942160, 4753869120 and 4876391520 $\endgroup$ – user608997 Oct 27 '18 at 0:48
  • $\begingroup$ those four are divisible by all the numbers from 1 through 18, but not 19. $\endgroup$ – Will Jagy Oct 27 '18 at 1:03

Suppose the number is of form $N=jihgfedcba$. We may write:


A: Any number such as following forms are divisible by 2, 4, 5 and 8:

$(2k)(40)$ such as $240, 440, 640, 840 . . .$

$(2k+1)(20)$, such as $120, 320, 520, 720, . . .$

B: Whatever the value of g is, the term $\frac{10^6-1}{9}$ is divisible by $77$.

C: For 7 we consider the remainder of $10^n-1$ when divided by 7:

$T=.....10, 10^2, 10^3, 10^4, 10^5, 10^6, 10^7, 10^8, 10^9$

$R_{10^n}=...3,..2,.. 6,.. 4,.. 5,... 1,.. 3,.. 2,.. 6$


We can make following relation for divisibility by 7:

$a+b+c+d+e+f+g+h+i+j+(2)b+(1)c+(5)d+(3)e+(4)f+(0)g+(2)h+(1)i+(5)j≡ mod 7$

D: For 11 we just consider the remainder of $10^n-1$ for odd n because for even n,. $(10^n-1)$ is divisible by 11 :

$T= .......10,....10^3,...10^5,...10^7,..10^9$



So we can make following relation for divisibility by 11:

$a+b+c+d+e+f+g+h+i+j+(-2)b+(-2)d+(-2)f+(-2)h+(-2)j≡ mod 11$

So we have following system of Diophantine equations:

$a+b+c+d+e+f+g+h+i+j+(-2)b+(-2)d+(-2)f+(-2)h+(-2)j≡ mod 11$

$a+b+c+d+e+f+g+h+i+j+(2)b+(1)c+(5)d+(3)e+(4)f+(0)g+(2)h+(1)i+(5)j≡ mod 7$

The sum $a+b+c+. . .i+j= \frac{9(9+1)}{2}=45$ is divisible by 3 and 9. This system of equations indicates that the question can have numerous solutions, to find one for example take $cba=840$ which is divisible by 2, 3, 4, 5, 7 and 8, That is we assume $a=0$, $b=4$ and $c=8$ and look for other digits as follows, we have:

$45+4\times2+8\times1+5d+3e+4f+2h+i+5j≡ mod 7$


$61+5d+3e+4f+2h+i+5j≡ mod 7$

$45-2\times 4-2(d+f+h+j)=37-2(d+f+h+j)≡ mod 11$

Suppose $37-2(d+f+h+j)=11$$d+f+h+j=(37-11)/2=13$

Suppose $d=1, . f=2,.h=3,. and,..j=7$ then we have:

$61+5+3e+8+6+i+35=115+3e+i≡ mod 7$

Let $115+3e+i=21\times 7=$$3e+i=32$$e=9$ and $i=5$

The only number which remains is 6 for g, so one solution can be:


  • $\begingroup$ That is hard to read. The naming of the digits as a,b,c is not very useful, $a_0, a_1,\ldots$ make more sense. Formulas like $61+5d+3e+4f+2h+i+5j≡ mod 7$ have no mathematical meaning. It seems you try to use dots to align values, but one should Latex arrays or similar structures to display this. $\endgroup$ – miracle173 Oct 27 '18 at 17:01

If the digit representation of such number is $ \langle d_9 d_8 d_7 d_6 d_5 d_4 d_3 d_2 d_1 d_0 \rangle$, where $$ \langle d_9 d_8 d_7 d_6 d_5 d_4 d_3 d_2 d_1 d_0 \rangle:=\Sigma_{i=0}^{9}10^id_i$$ then we know that $d_0=0$ because $$10\mid \langle d_9 d_8 d_7 d_6 d_5 d_4 d_3 d_2 d_1 d_0 \rangle.$$

The sum $$d_9 +d_8+ d_7+ d_6+ d_5+ d_4+ d_3+ d_2+ d_1 +0$$ is $$0+1+2+3+4+5+6+7+8+9=45,$$ so $$9 | \langle d_9 d_8 d_7 d_6 d_5 d_4 d_3 d_2 d_1 0\rangle.$$

Because $$8\mid \langle a_2 a_10\rangle.$$ we also know that $$4\mid \langle a_2a_1\rangle \tag{2}$$

So we have $$t\mid \langle d_9 d_8 d_7 d_6 d_5 d_4 d_3 d_2 d_1 0 \rangle, \; \forall t \in \{2,3,4,5,6,8, 9, 10\}$$ if $(2)$ holds.

If $$11\mid \langle d_9 d_8 d_7 d_6 d_5 d_4 d_3 d_2 d_1 d_0 \rangle$$ then for the alternate sum holds
$$11 \mid d_9 +d_8- d_7+ d_6- d_5+ d_4- d_3+ d_2- d_1+0$$

The alternate sum is between $$-9-8-7-6-5+1+2+3+4=35$$ and $$9+8+7+6-1-2-3-4-5=30.$$ But we know that the alternate sum is divisible by $11$ and it is the sum of $5$ odd and $5$ even numbers, so it is odd. Therefore the alternate sum is in $ \{-33,-11,11\}.$

How to construct a solution?

  1. Set $a_0=0$
  2. Start with a possible value for $\langle a_2a_1\rangle$ such that

    • $4\mid \langle a_2a_1\rangle$
    • $a_1 \ne a_2$
    • $a_1 \ne 0$
    • $a_2 \ne 0$
  3. The remaining digits $\{1,2,3,4,5,6,7,8,9\}\setminus\{d_1, d_2\}$ partion into two sets, set $\cal{O}=\{d_3, d_5, d_7, d_9\}$ that contains $4$ elements at the odd indexed positions and set $\cal{E}\{d_4, d_6, d_0\}$ that contains the $3$ elements at the even indexed positions.

  4. If $$\Sigma_{d \in \cal{E}}-\Sigma_{d \in \cal{O}}+a_2-a_1 \in \{-33,-11,11\} \tag{1}$$ we are done, otherwise select an element $\cal{e} \in \cal{E}$ and $\cal{o} \in \cal{O}$, remove $\cal{e}$ from $ \cal{E}$ and $\cal{o}$ from $ \cal{O}$ and add $\cal{e}$ to $ \cal{O}$ and $\cal{o}$ to$ \cal{E}.$ The left hand side of $(1)$ is incremented by $2(\cal{o}-\cal{e}).$ Repeat this step until $(1)$ is satisfied or if you are bored.
  5. If $(1)$ holds then assign the elements of $ \cal{O}$ to $d_9, d_7, d_5, d_3$ in an arbitrary way and assign the elements of $ \cal{E}$ to $a_8, a_6, a_4$ also in an arbitrary way. Now $$t\mid \langle d_9 d_8 d_7 d_6 d_5 d_4 d_3 d_2 d_1 0 \rangle, \; \forall t \in \{2,3,4,5,6,8, 9, 10,11\}$$ holds.


The smallest valid value for $\langle a_2 a_1 \rangle$ is $12.$ The values $00, 04, 08, 20$ are not valid because they contain $d_0.$ The numbers $44$ and $88$ are not valid because $d_2=d_1$, so the number cannot be a permutation.

So $d_1=6$ and $d_2=1$, we set $ \cal{O}=\{2,3,4,5\}$ and $ \cal{E}=\{7,8,9\}$. The left hand side of $(1)$ is $-14+24+1-6=-5.$ Now we shift $7$ to $\cal{O}$ and $4$ to $\cal{E}.$ This decreases the LHS of $(1)$ by $6$ to $-11$ and we are done. So we have $$\cal{O}=\{2,3,5,7\}$$ $$\cal{E}=\{4,8,9\}$$ $$\langle d_2 d_1 d_0 \rangle =160 $$ and can construct the number $$ 2435879160$$ $\square$

We can generate $4!\cdot 3!=144$ different numbers from our sets $\cal{O}$ and $\cal{E}.$ There is a good chance that about $\frac{1}{7}$ of these 144 numbers are divisible by $7$, these are about $20$ numbers. If there is no such number we can construct other numbers by repeating steps 2 to 5.

Here number $ 2435879160$ is already divisible by $7.$


We just need a number divisible by $5,7,8,9,11$ and everything else is automatic. Divisibility by $9$ isn't a concern, as the digits already sum to $45$ and $9\mid45$. The number must end with $0$ since it's even and divisible by $5$. The last three digits must be divisible by $8$, so they're some multiple of $040$ (of course $040$ isn't actually a valid candidate, since it repeats $0$ twice). Divisibility by $11$ means the alternating sum of the digits must be a multiple of $11$. Divisibility by $7$ means the first $9$ digits is a multiple of $7$. Now, I think the most efficient method is to write a program taking into account all of these parameters to find a numerical answer.

  • $\begingroup$ I don't think that one should use a program to solve this problem. $\endgroup$ – miracle173 Oct 27 '18 at 16:24
  • $\begingroup$ @miracle173 Well, I guess that is the easy way out. Since I posted my answer sirous and you have posted pen-and-paper solutions, but of course it's quite tedious. $\endgroup$ – YiFan Oct 27 '18 at 16:26

COMMENT.- It is clear that the required number, $N$, must be a multiple of $2^3\cdot3^2\cdot5\cdot7\cdot11=27720$ so we must have $N=27720x$ where $x$ is such that $N$ have ten (distinct) digits. It follows after a calculation$ $ that if there is solution then $x$ is an integer such that$$36076\le x\le360750$$ In other words, $x$ is a number belonging to a set of $324675$ integers.


The number which are divisible by $8$ is also divisible by $2$ and $4$.

The number which are divisible by $9$ is also divisible by $3$.

The number of which is divisible by $6$ is also divisible by $2$ and $3$.

The number which are divisible by $10$ is also divisible by $2$ and $5$.

Also also the number we expect is divisible by $11$ and $7$.

So the number is in the form $=P×2^{3i}×3^{2j}×5^k×7^m×11^n$, Where $i$, $j$, $k$, $m$, & $n$ are positive any positive integer and $P$ is any positive integer integer.Using this condition we will produce a required ten digit number.

  • 2
    $\begingroup$ This is not correct. You need something in the form of $k \times 2^3 \times 3^2 \times 5 \times 7 \times 11$. What you have is only a subset (and a quite restrictive one of that) of the numbers that fulfill the divisibility criteria described. $\endgroup$ – Jean-Luc Bouchot Oct 27 '18 at 2:38
  • $\begingroup$ @Jean-LucBouchot yes Sir. Absolutely. I forget to include this. Anyhow Thanks much. $\endgroup$ – Avinash N Oct 27 '18 at 2:45
  • $\begingroup$ Why did I get downvote? Then mentioned me where the mistake occurs. It will help me to correct myself. Thank you. $\endgroup$ – Avinash N Oct 27 '18 at 5:38
  • $\begingroup$ I voted this down. @Jean-LucBouchot already showed you what was wrong. And it is still wrong. e,g 16*9*5*7*11 is not contained in your set. And even if you replace it by Jean's correct formula, how does this answer the question? How do you use this formula to create such a number or to proof that such a number does not exist? $\endgroup$ – miracle173 Oct 27 '18 at 6:11
  • $\begingroup$ @miracle173 oh yes. There is a mistake. Thanks much. Any help to fix this issue? It is greatly appreciated... $\endgroup$ – Avinash N Oct 27 '18 at 12:00

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