## B x DE = ABC

$A,B, \textup{and } C$ are consecutive digits. $D \textup{ and }E$ are also consecutive digits. $A,B,C,D,\textup{and }E$ are all different digits. If $B\times DE=ABC$, find the sum of $A + B + C + D + E$. (Note: $C \textup { and }D$ are not necessarily consecutive digits.
Source: mathcontest.olemiss.edu 10/29/2007

SOLUTION
Let $A=a$. Then, $B=a+1 \textup { and } C=a+2$ . Let $D=d$, then $E=d+1$.

Since $D$ is in the tens place and $E$ in the ones place, the value of $DE$ is represented by
$DE=10d+1\left (d+1\right )$
$=11d+1$

Since $A$ is in the hundreds place, $B$ in the tens place, and $C$ in the ones place, we represent the value of $ABC$ by
$ABC=100a+10\left (a+1\right )+1\left (a+2\right )$
$=100a+10a+10+a+2$
$=111a+12$

If $B\times DE=ABC$, then we have the following equation:
$\left (a+1\right )\left (11d+1\right )=111a+12$

Let’s try a few values for $a$.
a=0
$\left (0+1\right )\left (11d+1\right )=111\times 0+12$
$11d+1=12$
$11d=11$
$d=1$
Not a solution because it is conflicting with $b=1$.

a=1
$\left (1+1\right )\left (11d+1\right )=111\times 1+12$
$2\left (11d+1\right )=123$
Not a solution because the left side is even but the right side is odd.

a=2
$\left (2+1\right )\left (11d+1\right )=111\times 2+12$
$3\left (11d+1\right )=234$
$33d+3=234$
$33d=231$
$d=7$

Check
$A=2,\;B=3,\;C=4,\;D=7,\;E=8$
$B\times DE=ABC$
$3\times 78=234$?
$234=234$
YES

Thus, $A+B+C+D+E=2+3+4+7+8=24$.