# Thread: Smallest number when you change digits' positions

1. ## Smallest number when you change digits' positions

What is the smallest number with the property that
when the leftmost digit is moved to the rightmost position, the new number
is three times the original?

2. Hello, Intsecxtanx!

What is the smallest number with the property that
when the leftmost digit is moved to the rightmost position,
the new number is three times the original?

We want a number such that: .$\displaystyle (ABCDE\hdots F) \times 3 \:=\:(BCD{E}F\hdots A)$

We have: . $\displaystyle \begin{array}{cccccc}A & B & C & D & E & F \\ \times &&&&& 3 \\ \hline B&C&D&E&F&A \end{array}$

Since we want the smallest number, let $\displaystyle A = 1$.

. . $\displaystyle \begin{array}{cccccc}{\color{red}1} & B & C & D & E & F \\ \times &&&&& 3 \\ \hline B&C&D&E&F&{\color{red}1} \end{array}$

In the rightmost column, $\displaystyle 3\!\times\!F$ ends in 1.
. . $\displaystyle 3F \to 1 \quad\Rightarrow\quad F = 7$

. . $\displaystyle \begin{array}{cccccc}1 & B & C & D & E & {\color{red}7} \\ \times &&&&& 3 \\ \hline B&C&D&E&{\color{red}7}&1 \end{array}$

In the second column, $\displaystyle 3\!\times\!E + 2 \to 7 \quad\Rightarrow\quad E = 5$

. . $\displaystyle \begin{array}{cccccc}1 & B & C & D & {\color{red}5} & 7 \\ \times &&&&& 3 \\ \hline B&C&D&{\color{red}5}&7&1 \end{array}$

In the next column, $\displaystyle 3\!\times\!D+1 \to 5 \quad\Rightarrow\quad D = 8$

. . $\displaystyle \begin{array}{cccccc}1 & B & C & {\color{red}8} & 5 & 7 \\ \times &&&&& 3 \\ \hline B&C&{\color{red}8}&5&7&1 \end{array}$

In the next column, $\displaystyle 3\!\times\!C + 2 \to 8 \quad\Rightarrow\quad C = 2$

. . $\displaystyle \begin{array}{cccccc}1 & B & {\color{red}2} & D & 5 & 7 \\ \times &&&&& 3 \\ \hline B&{\color{red}2}&8&5&7&1 \end{array}$

In the next column, $\displaystyle 3\!\times\!B \to 2 \quad\Rightarrow\quad B = 4$

. . $\displaystyle \begin{array}{cccccc}1 & {\color{red}4} & 2 & 8 & 5 & 7 \\ \times &&&&& 3 \\ \hline {\color{red}4}&2&8&5&7&1 \end{array}$

In the last column, $\displaystyle 3\!\times\!1 + 1$ must equal 4 . . . Yes!

Solution: .$\displaystyle \boxed{1\,4\,2\,8\,5\,7}$