My answer for this problem is far from that given by the answer sheet.

In the multiplication $abcde \times e=edade$, each letter represents a different digit. what digit does the letter C represent?

From the multiplication, I got: $$ee=e \tag1$$ $$ed=d \tag2$$ $$ec=a \tag3$$ $$eb=d \tag4$$ $$ea=e \tag5$$

from 1, I got $e=1$, and from 5 , I got $a=1$. Then I use 3 to get $c=1$.

  • 2
    $\begingroup$ From 1, you could also have gotten $e=6$ or $e=5$. This would've been mighty boring if the question was to examine $abcde\cdot1$. $\endgroup$
    – Arthur
    Oct 17 '16 at 13:01
  • $\begingroup$ Note that $5\times 5=25$ and $6\times 6=36$, so you can't conclude that $e=1$, in fact, since you're told that the letters represent different digits, $e$ can't be $1$. $\endgroup$ Oct 17 '16 at 13:01
  • $\begingroup$ When you multiply you can carry. So $ee$ is equal to $e$ plus some multiple of $10$. $\endgroup$
    – GEdgar
    Oct 17 '16 at 13:01
  • $\begingroup$ $e$ cannot be $1$, since it must be bigger than $a$: look at the leftmost digit. $\endgroup$ Oct 17 '16 at 13:01
  • $\begingroup$ @Arthur, I do not understand why from 1, I can get e=6 or 5 ? Could you explain it a little more? $\endgroup$
    – math
    Oct 17 '16 at 13:06

Since $e^2$ ends in $e$, $e$ must be $1,5$, or $6$. Clearly $e\ne 1$, since $abcde\cdot 1=abcde$, not $edade$. In fact, $a$ must be $1$ in order for $e$ to be the first digit of the product. Thus, we now have

$$1bcde\cdot e=ed1de\;,$$

where $e$ is $5$ or $6$. Next, $be$ does not produce a carry. If $b$ were $2$ or more, $be$ would have to produce a carry, since $e\ge 5$, and $b$ cannot be $1$, so $b=0$, and we have either

$$10cd5\cdot 5=5d1d5$$


$$10cd6\cdot 6=6d1d6\;.$$

In the first case $5\cdot d+2$ ends in $d$; $5\cdot d$ ends in either $0$ or $5$, so $5\cdot d+2$ ends in $2$ or $7$, and you can check that both $d=2$ and $d=7$ satisfy the condition that $5\cdot d+2$ end in $d$.

In the second case $6\cdot d+3$ ends in $d$, and you can check that there is no value of $d$ for which this is true. Thus, $e=5$, and $d$ is $2$ or $7$. The possibilities at this point are therefore

$$10c25\cdot 5=52125$$


$$10c75\cdot 5=57175\;.$$

From here you should be able to finish it pretty easily.

Added: I wrote this up exactly as I worked out the problem myself, as an illustration of the thought processes involved and, to be honest, because I didn’t feel like trying to polish it; as a result, it’s not as efficient as it could have been. As Henry notes in the comments, $$10999\cdot5=54995<55000\;;$$

had I noticed this, I could immediately have ruled out the possibility $d=7$.


Here is the explanation why $a$ has to be 1: $a$ has to be 1 because 1) $e$ equals 5 or 6, and 2) the product is also a 5-digit integer. If $a$ is larger than 1, the product should be six-digit integer.


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