ABCDE + BCDE + CDE + DE + E = AAAAA

Question

Today I came through this question when one of my friends asked. I don't know if this math.stackexchange community is the right place. I am new but active on other SO communities. Take me easy if this is not the right place to ask the question.

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

What is the solution?

You might be interested in this SO Question about verbal arithmetic. What are your own thoughts about the problem? — hardmath, Aug 27 '14 at 20:49
@hardmath My approach for this problem was trial and error method — Raghavendra, Aug 28 '14 at 04:50
@ThomasAndrews yes those have to be distinct which is why i gave five variables A,B,C,D and E — Raghavendra, Aug 28 '14 at 04:52
$x+y+z=2, x+2y+3z=3, 2x+3y+z=5$ has three variables, but $x=y=1,z=0$ is a solution. Having different letters does not mean that they are not equal. @Raghav — Thomas Andrews, Aug 28 '14 at 04:54
@ThomasAndrews Yes, in algebra, variables can have same values. This comes more under verbal arithmetic where each letter denotes each distinct number — Raghavendra, Aug 28 '14 at 04:57

score 6 · Accepted Answer · answered Aug 27 '14 at 18:01

The given equation can be written as $10000A + 2000B + 300C + 40D + 5E = 11111A$.

Since the left side is a multiple of $5$, so is the right side. Thus, $A = 5$ (since $A = 0$ isn't possible).

Plugging in $A = 5$ yields $2000B + 300C + 40D + 5E = 5555$, i.e. $400B + 60C + 8D + E = 1111$.

If $B \le 1$, we have $400B + 60C + 8D + E \le 400 \cdot 1 + 60 \cdot 9 + 8 \cdot 9 + 9 = 1021 < 1111$, a contradiction. If $B \ge 3$, then $400B + 60C + 8D + E > 400 \cdot 3 = 1200 > 1111$, a contradiction. Thus $B = 2$.

Plugging in $B = 2$ yields $60C + 8D + E = 311$. Using similar logic as above, we can easily find the solutions $(C,D,E) = (5,1,3)$ and $(4,8,7)$.

Thus, $(A,B,C,D,E) = (5,2,5,1,3)$ or $(5,2,4,8,7)$.

It is typically required for verbal arithmetic puzzles that different letters correspond to different digits, which would narrow down the solution to the last one found. — hardmath, Aug 27 '14 at 20:56

score 0 · Answer 2 · answered Sep 02 '14 at 04:06

0

Please check those scratch paper below.

Scratch paper 1 enter image description here Scratch paper 2 Scratch paper 3

answered Sep 02 '14 at 04:06

Crane

1

2

Is this an answer, or a question? If it's an answer, what does it say that the other one doesn't? Apart from that... people don't like reading handwriting on the Internet. Also, search engine bots don't. Also, people reliant on screen readers don't. – John Dvorak Sep 02 '14 at 04:12

ABCDE + BCDE + CDE + DE + E = AAAAA

2 Answers2