Elementary Arithmetic Problem In 1988, in a French math competition for middle school grades, the following problem was given:
Complete this multiplication so that all the digits from 0 to 9 appear:
$... × .. = ....1$
I am stumped. Of course the last digits of the two numbers are 3 and 7 but this is as far as I can go!
I am curious if there is a logical way to solve this problem without too much trial and error.
For the record, after significant experimentation my 3rd grade daughter has managed to obtain one solution, but this was pure luck.
 A: I think the trick must be "casting out $9$'s."  Add the digits of the number.  If the result is greater than of equal to $9$, add the digits of the result, and so on, until you get a number less than $9$.  This gives the congruence class of the original number modulo $9$.
Let $x$ be the congruence class of the two-digit number, $y$ that of the three-digit number, and $z$ that of the five-digit number.  We know $$xy\equiv z\pmod9$$  Since the sum of the digit from $0$ to $9$ is $45$, we also know
$$x+y+z\equiv0\pmod9\\
xy\equiv-x-y\pmod9\\
(x+1)(y+1)\equiv1\pmod9$$
As you pointed out the last digits of the numbers on the left-hand side must be $3$ and $7$, which means that the number on the right-hand side must be at least $20451$.
If we know that the product of two numbers is congruent to $1\pmod9$ then the congruence classes of those numbers can only be:

*

*$1$ and $1$

*$2$ and $5$

*$4$ and $7$

*$8$ and $8$
I think you can put all this together to find the solution in fairly short order.  I'll give some examples of the kind of reasoning I have in mind.
First, suppose that the two-digit number ends in $3$.
$13$ isn't big enough to get a product of $20451$.
The digital root of $23$ is $5$ so $x+1=6$ and there's no possible value for $y+1$.
$33$ is inadmissible, so the two-digit number is $43$ or greater.  The digital root of $43$ is $7$ so $x+1=8$ and $y+1=8$, so $y=7$.  The first two digits of the three-digit number must sum to $9$.  The $1,3,4,7$ have already been used so the first two digits can only be $9$ and $0$ in that order.  However,$$43\cdot907=39001,$$ which doesn't work, so the two-digit number isn't $43$
The two-digit number isn't $53$ because then we'd have $x+1=9$ and there's no possible value for $y$.
If the two-digit number is $63$ then $x+1=1$, so $y+1=1$ and $y=0$.  The first two digits of the three-digit number must sum to $2$ or $11$.  If they sum to $2$ the number is $207$ which isn't big enough, so they sum to $11$.  We have used $1,3,6,7$, so the only possibility is $2$ and $9$.  The three-digit number must be $297$ or $927$.  The first is too small, since $$63\cdot297<63\cdot300=18900<20451$$  The second gives $$63\cdot927=58401$$ which is the answer.
One can continue in this manner test $73,83,93$ and the possibilities where the two-digit number ends in $7$ to show that the above is the only solution.
A: The rule of nines:  $jkl \equiv j+ k + l \pmod 9$ so
So if we have $abc\times de = fghi1$ and $a,b,....,f,g,h,i,1$ are the digits from $0,....,9$ then $abc + de +fghi1 \equiv 0 \pmod 9$
And if $abc \equiv j\pmod 9$ and $de \equiv k \pmod 9$ we have $fghi1 \equiv jk \equiv -(j+k)$.
Or $(j+1)k \equiv -j$ and $(k+1)j \equiv -k$
Possible values for $j,k$.
$(0,0)$, $(1,4)$, $(3,6)$, $(4,1)$, $(6,3)$
Furthermore
$(a+b+c)(d+e) \equiv f+g+h+i+1 \equiv 0+1+2+3+4+5+6+7+8+9 - (a+b+c+d+e)\pmod 9$
$ad +bd +cd + ae +be +ce \equiv -a-b-c-d-e \pmod 9$
And as $\{c,e\} = \{7,3\}$ we have
$ad + bd +cd +ae +be + 21 \equiv -a-b-d -10\pmod 9$ so
$ad + bd + cd +ae + be +a+b+d \equiv 5\pmod 9$.
If $c=7;e=3$ we have
$ad + bd + 8d+3a+3b +a+b\equiv (a+b)(d+4)-d \equiv 5\pmod 9$
$(a+b)(d+4) \equiv 5+d\pmod 9$
Now $a+b+7\equiv a+b - 2\equiv j$ and $d+3\equiv k$ for $j,k$ above.
We can have
1)$a+b\equiv 2$ and $d= 6$
2)$a+b\equiv 3$ and $d= 1$ (impossible as $1$ is accounted for)
3)$a+b\equiv 5$ and $d= 3$ (ditto $3$)
4)$a+b\equiv 6$ and $d= 7$ (ditto $7$)
5)$a+b\equiv 8$ and $d= 0,9$.($d=0$ is impossible as $de$ is two digits)
Case 1: $a+b\equiv 2$ and $d=64
$(a+b) \equiv 2$.  As $a,b\ne 1,3,7,6$ we have $a+b=11$ and $a,b=2,9$
$(297,927)\times 63 = 18711, 58401$.
$927\times 63 = 58401$ is a working solution.
Case 2: $a+b \equiv 8$ and $d=9$.  As $a,b\ne 1,3,7,9;a\ne b$ we have $a+b=8$ and $a,b = 2,6$
$(267, 627)\times 93$ do not work.
If $c=3,e = 7$ we have
$ad + bd + cd +ae + be +a+b+d \equiv 5\pmod 9$.
$ad + bd + 3d + 7a + 7b + a+b+d\equiv 5\pmod 9$
$(a+b)d + 4d +8(a+b) \equiv (a+b)(d-1) -5d \equiv 5\pmod 9$ so
$(a+b)(d-1)\equiv 5(d+1)\pmod 9$.
$a+b+3 \equiv j$ and $d+7\equiv d-1\equiv k$ for $j,k=(0,0), (1,4), (3,6)$, (4,1), (6,3)$ above.
Which all lead to contradictions

*

*$a+b\equiv 6$ and $d=1$ but $6*0\not \equiv 5*2$

*$a+b\equiv 8$ and $d=5$ but $8*4\not \equiv 5*6$

*$a+b\equiv 0$ and $d=7$ but $0*6\not \equiv 5*8$

*$a+b\equiv 1$ and $d=2$ but $1*1 \not \equiv 5*3$

*$a+b\equiv 3$ and $d=4$ but $3*2\not \equiv 5*5$
A: If the numbers are $x$ ($3$ digits), $y$ ($2$ digits) and $z$ ($5$ digits) then we have the following deductions:

*

*$x$ and $y$ end in $3$ and $7$.

*$x$ and $y$ start with $4,5,6,8$ or $9$ (otherwise $z$ is too small).

*$y \mod 9$ cannot be $2, 5$ or $8$ (otherwise $x + y + xy \not \equiv 0 \mod 9$).

This leaves $7$ possibilities for $y$:
$43, 63, 93, 57, 67, 87, 97$
and a total of $13$ possibilities for the pair $x,y$. Then I think you have to check each of these $13$ possibilities individually.
If the condition that $z$ ends in $1$ is removed then there are $9$ solutions, and $z$ can end in $0, 1, 4$ or $8$. The $9$ solutions break down as follows:

*

*$z$ ends in $0$ : $4$ solutions

*$z$ ends in $1$ : $1$ solution

*$z$ nds in $4$ : $1$ solution

*$z$ ends in $8$ : $3$ solutions

