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.

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 $3*2\not \equiv 5*5$
