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.

Answer: 24.


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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , . Bookmark the permalink.

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