## Multiple of 72

The seven-digit integer $2718AB6$ is a multiple of $72$ (where $A,B$ are digits). Determine all possible ordered pairs $(A,B)$.
Source: NCTM Mathematics Teacher, December 2005

SOLUTION
Since $3\times 2=6$ and $8\times 2=16$, we multiply $72$ by $03,13,23$, etc. and by $08,18,28$, etc. to find the possible $AB$ digits that would match $8AB6$
$72\times 03=216$
$72\times 13=936$
$72\times 23=1656$
$72\times 33=2376$
$72\times 43=3096$
$72\times 53=3816$
$72\times 63=4536$
$72\times 73=5256$
$72\times 83=5976$
$72\times 93=6696$
Only $216$ and $936$ are close match of $8AB6$.
Verification
$2718216=72\times 37753$
$2718936=72\times 37763$
We do the same multiplication of $72$ by $08,18,28$, etc.
$72\times 08=576$
$72\times 18=1296$
$72\times 28=2016$
$72\times 38=2736$
$72\times 48=3456$
$72\times 58=4176$
$72\times 68=4896$
$72\times 78=5616$
$72\times 88=6336$
$72\times 98=7056$
Only $576$ is a close match
Verification
$2178576=72\times 30258$
The possible ordered pairs $(A,B)$ are $(2,1),(9,3),(5,7)$.

Answer: $(2,1),(5,7),(9,3)$

Alternative solution
Since $72=8\times 9$, $2178AB6$ must be divisible by $8$ and $9$. To be divisible by $8$ the last three digits $AB6$ must be divisible by $8$. To be divisible by $9$ the sum of the digits must be divisible by $9$.
$2+1+7+8+A+B+6=24+A+B$
$27$ and $36$ are divisible by $9$
$27=24+A+B$
$A+B=3$
$36=24+A+B$
$A+B=12$
Case 1: $A+B=3$
$03, 036$ not divisible by $8$
$12, 126$ No
$21. 216$ Yes
$30, 306$ No
Case 2: $A+B=12$
$39, 396$ not divisible by 8
$48, 486$ No
$57, 576$ Yes
$66, 666$ No
$75, 756$ No
$84, 846$ No
$93, 936$ Yes