# Problem

• Nov 11th 2011, 08:33 AM
emilhp
Problem
Find all 6-digit numbers x with the following property:

If the first digit in x is deleted and added after the 5 other digits, you get a number that is 3 times as large as x.

For instance, the number x=158249 gets turned into 582491.

The number x=158249 does not meet the condition, because 3*158249 ≠ 582491

I hope that someone can help me with this problem - I have no idea how to solve it..(Speechless)
• Nov 11th 2011, 09:18 AM
veileen
Re: Problem
$x=\overline{abcdef}$, a, b, c, d, e, f are digits, a > 0

$3\cdot \overline{abcdef}=\overline{bcdefa} \Leftrightarrow 3\cdot a \cdot 10^5+3\cdot \overline{bcd} \cdot 10^2+3\cdot \overline{ef}=\overline{bcd}\cdot 10^3+\overline{ef}\cdot 10+a$ $\Leftrightarrow (3\cdot 10^5-1)a=700 \overline{bcd}+7\overline{ef}\Leftrightarrow 299.999\cdot a=7 \overline{bcdef} \Leftrightarrow 42.857 \cdot a=\overline{bcdef}$

a is a digit and $42.857 \cdot a = \overline{bcdef} < 99.999$, so a can be 1 or 2. If a is 1, then x is 142857, if a is 2, x is 242857.
• Nov 11th 2011, 04:17 PM
Soroban
Re: Problem
Hello, emilhp!

This can be solved as an Alphametic.

Quote:

Find all 6-digit numbers x with the following property:

If the first digit in x is deleted and added after the 5 other digits,
you get a number that is 3 times as large as x.

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

We see that: $A \:=\:1,\,2$
If $A \ge 3$, the product will be a seven-digit number.

Suppose $A = 1.$

We have: . $\begin{array}{cccccc} 1&B&C&D&E&F \\ \times &&&&& 3 \\ \hline B&C&D&E&F&1 \end{array}\qquad\text{We see that: }\,F = 7.$

We have: . $\begin{array}{cccccc} 1&B&C&D&E&7 \\ \times &&&&& 3 \\ \hline B&C&D&E&7&1 \end{array} \qquad\text{We see that: }E = 5.$

We have: . $\begin{array}{cccccc} 1&B&C&D&5&7 \\ \times &&&&& 3 \\ \hline B&C&D&5&7&1 \end{array} \qquad\text{We see that: }D = 8.$

We have: . $\begin{array}{cccccc} 1&B&C&8&5&7 \\ \times &&&&& 3 \\ \hline B&C&8&5&7&1 \end{array} \qquad\text{We see that: }C = 2.$

We have: . $\begin{array}{cccccc} 1&B&2&8&5&7 \\ \times &&&&& 3 \\ \hline B&2&8&5&7&1 \end{array} \qquad\text{We see that: }B = 4.$

We have: . $\begin{array}{cccccc} 1&4&2&8&5&7 \\ \times &&&&& 3 \\ \hline 4&2&8&5&7&1 \end{array}$

Assume $A = 2$ .and using similar reasoning,

. . we have: . $\begin{array}{cccccc} 2&8&5&7&1&4 \\ \times &&&&& 3 \\ \hline 8&5&7&1&4&2 \end{array}$