Welcome to our community

Be a part of something great, join today!

Find The Smallest Natural Number

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

After some simplification, we obtain:

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

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:

\(\displaystyle \frac{615384}{153846}=4\)
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Bravo, MarkFL!

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

soroban

Well-known member
Feb 2, 2012
409
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:- [tex]\text{. . . }A\,B\,C\,D\,E\,6[/tex]

[tex]\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}[/tex]


[tex]\text{In column-6, }E = 4[/tex]
[tex]\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}[/tex]


[tex]\text{In column-5, }D = 8[/tex]
[tex]\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}[/tex]


[tex]\text{In column-4, }C = 3[/tex]
[tex]\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}[/tex]


[tex]\text{In column-3, }B = 5[/tex]
[tex]\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}[/tex]


[tex]\text{In column-2, }A = 1[/tex]
[tex]\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}[/tex]

[tex]\text{ta-}DAA![/tex]
 
  • Thread starter
  • Admin
  • #5

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Hello, anemone!

This can be solved as an alphametic.



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

[tex]\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}[/tex]


[tex]\text{In column-6, }E = 4[/tex]
[tex]\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}[/tex]


[tex]\text{In column-5, }D = 8[/tex]
[tex]\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}[/tex]


[tex]\text{In column-4, }C = 3[/tex]
[tex]\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}[/tex]


[tex]\text{In column-3, }B = 5[/tex]
[tex]\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}[/tex]


[tex]\text{In column-2, }A = 1[/tex]
[tex]\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}[/tex]

[tex]\text{ta-}DAA![/tex]
Hi soroban,:) thank you for showing us another great method to solve this problem and you too deserve a pat on the back!:cool:(Clapping)