What does A B C D E \overline{ABCDE} represents?

In the following addition, different letters represent different non-zero digits. What is the 5-digit number A B C D E ? \overline{ABCDE}?

A B C D E D B + B C D E D B + C D E D B + D E D B + E D B + D B + B A A A A A A A \begin{array} { l l l l l } & & A & B & C & D & E & D & B \\+ &&&B & C & D & E & D & B \\ + &&&& C & D & E & D & B \\ + &&&&& D & E & D & B \\ + &&&&&& E & D & B\\ + &&&&&&&D & B \\ + &&&&&&&& B \\ \hline & \\ &&\ A & A & A & A & A & A & A \end{array}

What is A B C D E ? \overline{ABCDE}?

Problem Courtesy Math Meets.


The answer is 84269.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Amos Vagg
Oct 10, 2018

I am only a 13-year-old student so I only have a limited mathematical understanding. Therefore, I solved this by a process of logical elimination, rather than sound mathematical principle. I found that if A=1, B must =3 because there are 7 B's and the number must end in 1 so 7x3=21. In this first round of elimination, all I did was eliminate A=5 as a possibility because all the letters represent different numbers, so the only possible solution of B=5 was actually invalid.

I then proceeded to D, by writing what number would be carried over from B for each question, and discovered that the only possible answers were A=2 B=6 D=8, A=8 B=4 D=1 or A=8 B=4 D=6. My next task was to carry over the remaining numbers to figure what value E could be. I then realised that the only possible answer was A=8 B=4 D=6 because neither 5 could be added to 5 or 0 to result in 2 or 0 be added to 5 or 0 to result in 8.

I then figured out what values of E would be possible, based on the fact I had previously discovered D=6. I knew, therefore, that the total of D in the fourth column had to be 24 and because A=8, we had to carry 4 over from E, meaning the value of E would have to be 9 (9x5=45).

I then used the same elimination to find that the value of C would be 2 and checked my answers by substituting them into the equation.

Therefore A=8 B=4 C=2 D=6 E=9, resulting in 8426964+426964+26964+6964+964+64+4=8888888, giving the final answer of 84269.

Thank you for sharing your solution.

Hana Wehbi - 2 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...