The number abcde has five distinct digits, none of which are zero, each represented by one of the letters $a, b, c, d, e$. Multiplying this number by 4 results in a five-digit number edcba. What is the value of $a+b+c+d+e$?
A) 22
B) 23
C) 24
D) 25
E) 27
Solution

## ALTERNATIVE E

The multiplication can be schematized as

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-18.jpg?height=114&width=246&top_left_y=1942&top_left_x=1005)

The solution is based on the following observations:

- The digit $a$ can only be 1 or 2, because if $a \geq 3$, then $4a$ would be a two-digit number and therefore the number edcba would have 6 digits. But $a$ cannot be 1 because $edcba$, being a multiple of 4, is even, so its last digit is even. Therefore, $a=2$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-18.jpg?height=117&width=254&top_left_y=2414&top_left_x=1004)

- The digit $e$ can only be 8 or 9, because $2 \times 4=8$ and $edcba$ has only 5 digits. However, $e$ cannot be 9 because $9 \times 4$ ends in 6 and not in 2. Therefore, $e=8$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-19.jpg?height=126&width=242&top_left_y=234&top_left_x=813)

- The digit $b$ can only be 1 or 2, because $4 \times b$ must be a single-digit number. Since $a=2$ and the five digits of abcde are distinct, we can only have $b=1$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-19.jpg?height=120&width=243&top_left_y=548&top_left_x=815)

- The digit $d$ can only be 2 or 7, because $4 \times d+3$ is a number ending in 1. Since $a=2$ and the five digits of abcde are distinct, we can only have $d=7$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-19.jpg?height=120&width=243&top_left_y=868&top_left_x=815)

- The digit $c$ can only be 9, because $4c+3$ is a number ending in $c$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-19.jpg?height=122&width=243&top_left_y=1144&top_left_x=815)

Therefore, the answer is $8+7+9+1+2=27$.

3 The disks make turns - Solution

## ALTERNATIVE D

The figure shows that disks $A$ and $B$ rotate in the same direction, disks $B$ and $C$ in opposite directions, and disks $C$ and $D$ in the same direction.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-19.jpg?height=400&width=488&top_left_y=1722&top_left_x=681)

Thus, $D$ rotates counterclockwise. We recall that the perimeter $p$ of a circle with radius $r$ is given by $p=2 \pi r$. Since the radius of disk $A$ is four times that of $D$, it follows that the perimeter of $A$ is also four times the perimeter of $D$. Therefore, $D$ makes four turns for each turn of $A$.

Note: In the argument above, we used the fact that the radii of disks $B$ and $C$ are irrelevant to the solution of this question; it is interesting to show this rigorously. Denoting by $a, b, c$ and $d$ the radii of $A, B, C$ and $D$ and by $n_{a}, n_{b}, n_{c}$ and $n_{d}$ the number of turns made by disks $A, B, C$ and $D$, respectively, then:

$$
n_{a} 2 \pi a=n_{b} 2 \pi b, \quad n_{b} 2 \pi b=n_{c} 2 \pi c, \quad n_{c} 2 \pi c=n_{d} 2 \pi d
$$

which implies that:

$$
n_{a} 2 \pi a=n_{d} 2 \pi d,
$$

thus,

$$
\frac{n_{d}}{n_{a}}=\frac{a}{d}
$$

Therefore, if $n_{a}=1$ then, using that $a=8$ and $d=2$, we get that $n_{d}=\frac{8}{2}=4$.