Numbers swap

Number Theory Level 2

$5\times \overline{{\color{#D61F06}ABC}{\color{#3D99F6}DE}} =8\times \overline{{\color{#3D99F6}DE}{\color{#D61F06}ABC}}$

If the last 2 digits ( $\overline{{\color{#3D99F6}DE}}$ ) of a 5-digit number $\overline{{\color{#D61F06}ABC}{\color{#3D99F6}DE}}$ are swapped to the front, then a new 5-digit number $\overline{{\color{#3D99F6}DE}{\color{#D61F06}ABC}}$ is formed. The value of the new number decreases and the above equation holds.

Find the minimum value of the 5-digit number $\overline{{\color{#D61F06}ABC}{\color{#3D99F6}DE}}$ .

Note: $A, D\neq 0$

The answer is 19512.

1 solution

Chan Lye Lee
Jan 5, 2019

Let $x=\overline{{\color{#D61F06}ABC}}$ and $y=\overline{{\color{#3D99F6}DE}}$ .

$5\times \overline{{\color{#D61F06}ABC}{\color{#3D99F6}DE}} =8\times \overline{{\color{#3D99F6}DE}{\color{#D61F06}ABC}}$

implies that $5\left(100x+y \right) = 8\left(1000y+x\right)$

Hence, $492x=7995y$ or equivalently, $\frac{x}{y}=\frac{65}{4}= \frac{130}{8}=\frac{195}{12}$ .

This means that the corresponding minimum value of the 3-digit number $x=195$ and 2-digit number $y=12$ . So $\overline{{\color{#D61F06}ABC}{\color{#3D99F6}DE}}=\boxed{19512}$ .

Numbers swap

The answer is 19512.

1 solution

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