Find The Smallest Natural Number

[anemone]

Find the smallest natural number with 6 as the last digit, such that if the final 6 is moved to the front of the number it is multiplied by 4.

 

[MarkFL]

Spoiler

Let's choose to let $x$ represent the string of $n$ digits to the left of 6 initially. Using the given information, we may state:

$$4(10x+6)=6\cdot10^n+x$$

After some simplification, we obtain:

$$x=\frac{2\left(10^n-4 \right)}{13}$$

By trial and error, I find the smallest value for $n$ which gives an integral value for $x$ is:

$n=5$

and thus:

$x=15384$

which means the original number is $153846$. And we find:

$$\frac{615384}{153846}=4$$

 

[anemone]

Bravo, MarkFL!

You deserve a round of applause for this solution!(Clapping)

 

[soroban]

Hello, anemone!

This can be solved as an alphametic.

Find the smallest natural number ending with 6,
such that if the final 6 is moved to the front of the number ,
it is multiplied by 4.

Suppose the number has the form:- \text{. . . }A\,B\,C\,D\,E\,6

\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & B & C & D & E & 6 \\ \times &&&&& 4 \\ \hline 6 & A & B & C & D & E \end{array}\text{In column-6, }E = 4
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & B & C & D & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & A & B & C & D & 4 \end{array}\text{In column-5, }D = 8
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & B & C & 8 & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & A & B & C & 8 & 4 \end{array}\text{In column-4, }C = 3
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & B & 3 & 8 & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & A & B & 3 & 8 & 4 \end{array}\text{In column-3, }B = 5
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & 5 & 3 & 8 & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & A & 5 & 3 & 8 & 4 \end{array}\text{In column-2, }A = 1
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ 1 & 5 & 3 & 8 & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & 1 & 5 & 3 & 8 & 4 \end{array}

\text{ta-}DAA!

 

[anemone]

Hello, anemone!

This can be solved as an alphametic.


Suppose the number has the form:- \text{. . . }A\,B\,C\,D\,E\,6

\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & B & C & D & E & 6 \\ \times &&&&& 4 \\ \hline 6 & A & B & C & D & E \end{array}\text{In column-6, }E = 4
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & B & C & D & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & A & B & C & D & 4 \end{array}\text{In column-5, }D = 8
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & B & C & 8 & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & A & B & C & 8 & 4 \end{array}\text{In column-4, }C = 3
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & B & 3 & 8 & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & A & B & 3 & 8 & 4 \end{array}\text{In column-3, }B = 5
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ A & 5 & 3 & 8 & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & A & 5 & 3 & 8 & 4 \end{array}\text{In column-2, }A = 1
\text{We have: }\;\begin{array}{cccccc} _1&_2&_3&_4&_5&_6 \\ 1 & 5 & 3 & 8 & 4 & 6 \\ \times &&&&& 4 \\ \hline 6 & 1 & 5 & 3 & 8 & 4 \end{array}

\text{ta-}DAA!

Hi soroban,:) thank you for showing us another great method to solve this problem and you too deserve a pat on the back!(Clapping)