Determine whether there exists a number that begins with 2 having the property that, when the 2 is moved to the end, the number is

(a) doubled;

(b) tripled.
There is no such number. Suppose that it existed; represent it by $2 \cdots a b$ where $a$ and $b$ are the last two digits of the number. Let us write the digit-by-digit multiplication:

![](https://cdn.mathpix.com/cropped/2024_04_17_45b306ab1b407a5798d6g-1.jpg?height=130&width=371&top_left_y=1122&top_left_x=866)

From the last column it is clear that there are only two choices for $b: 1$ and 6 . We now look at the next-to-last column. If $b$ is 1 , there is no way that the multiplication of $a$ by 2 , without an added carry, can give the odd number 1 . On the other hand, if $b$ is 6 , there is no way that the multiplication of $b$ by 2 , with the added carry of 1 resulting from $6 \times 2=12$, can give the even number 6 .

The answer is yes; such a number is 285714 , which satisfies $285714 \times 3=857142$. There is no need to say more; we have rigorously proved that such a number exists. However, we will mention two of many methods for finding this number. The first is the end-digit analysis used in part (a):

$$
\begin{gathered}
\cdots x \times 3=\cdots 2 \text { gives } x=4 \text { (no other digit works) } \\
\cdots x 4 \times 3=\cdots 42 \text { gives } x=1, \text { etc. }
\end{gathered}
$$

This process continues until we reach a number that begins with 2 .

The second method is an algebraic one. Suppose that the number is $x$ and has $n$ digits. Then moving a 2 from the beginning to the end of $x$ can be achieved by first multiplying $x$ by 10 (thus adding a 0 at the end), adding 2 (thus changing this 0 to a 2 ), and subtracting $2 \cdot 10^{n}$ (deleting the 2 at the beginning). We have the condition

$$
10 x+2-2 \cdot 10^{n}=3 x,
$$

which we can solve for $x$ :

$$
\begin{aligned}
10 x-3 x & =2 \cdot 10^{n}-2 \\
7 x & =2 \cdot 10^{n}-2 \\
x & =\frac{2 \cdot 10^{n}-2}{7} .
\end{aligned}
$$

Thus for $x$ to exist, $2 \cdot 10^{n}-2$ must be divisible by 7 . As is easily verified, this does not hold for $n=1,2,3,4$, or 5 , but it works for $n=6$, yielding

$$
x=\frac{2000000-2}{7}=285714 .
$$