# Putnam Problem of the Day

• Aug 12th 2010, 10:20 PM
simplependulum
Putnam Problem of the Day

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

Spoiler:
153846
?
• Aug 13th 2010, 03:26 AM
Unbeatable0
Let the number be $\displaystyle a_1a_2...a_n6$.

Let $\displaystyle x=0.\overline{a_1a_2...a_n6}$.

Then $\displaystyle 4x = 0.\overline{6a_1a_2...a_n} \Rightarrow 40x = 6.\overline{a_1a_2...a_n6} = 6+x$.

Therefore $\displaystyle x = \frac{6}{39} = \frac{2}{13} = 0.\overline{153846}$.

By backwards argument it's easy to see that all the cycles in $\displaystyle \frac{2}{13}$ are solutions.

Thus we have found all the solutions, the smallest of which is $\displaystyle 153846$.

This technique can be applied to this kind of problems in many cases.
• Aug 13th 2010, 02:03 PM
Soroban
Hello, simplependulum!

Just write out the multiplcation,
. . and the digits will appear sequentially.

Quote:

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

We have: .$\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ A&B&C&D&E&6 \\ \times &&&&& 4\\ \hline 6&A&B&C&D&E \end{array}$

In column-6: .$\displaystyle 6 \!\times\! 4 \,=\,24 \quad\Rightarrow\quad E = 4$

. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ A&B&C&D&4&6 \\ \times &&&&& 4\\ \hline 6&A&B&C&D&4 \end{array}$

In column-5: .$\displaystyle 4 \!\times\! 4 + 2 \:=\:18 \quad\Rightarrow\quad D = 8$

. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ A&B&C&8&4&6 \\ \times &&&&& 4\\ \hline 6&A&B&C&8&4 \end{array}$

In column-4: .$\displaystyle 8 \!\times\! 4 + 1 \:=\:33 \quad\Rightarrow\quad C = 3$

. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ A&B&3&8&4&6 \\ \times &&&&& 4\\ \hline 6&A&B&3&8&4 \end{array}$

In column-3: .$\displaystyle 3\!\times\!4 + 3 \:=\:15 \quad\Rightarrow\quad B = 5$

. . $\displaystyle \begin{array}{cccccc} ^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\ A&5&3&8&4&6 \\ \times &&&&& 4\\ \hline 6&A&5&3&8&4 \end{array}$

In column-2: .$\displaystyle 5\!\times\!4 + 1 \:=\:21 \quad\Rightarrow\quad A = 1$

Therefore: . $\displaystyle \begin{array}{cccccc} 1&5&3&8&4&6 \\ \times &&&&& 4\\ \hline 6&1&5&3&8&4 \end{array}$

• Aug 13th 2010, 05:19 PM
Wilmer
Soroban, the fact that it is a 6digit number is NOT a given.
• Aug 13th 2010, 06:16 PM
Soroban
Quote:

Originally Posted by Wilmer
Soroban, the fact that it is a 6-digit number is NOT a given. . Right!

I originally set up the problem like this:

. . $\displaystyle \begin{array}{ccccccccccc} A & B & C & D & E & F & G & \hdots & M & N & 6 \\ \times &&&&&&&&&& 4 \\ \hline 6 & A & B & C & D & E & F & G & \hdots & M & N \end{array}$

And found that "it came out even" after six digits
. . which gave me the smallest number: $\displaystyle 153,\!846.$

The next is the 12-digit number: $\displaystyle 153,\!846,\!153,\!846.$

• Aug 13th 2010, 07:10 PM
undefined
Fyi, there is a name for these, n-parasitic number, where here n=4 and k=6.
• Aug 13th 2010, 09:32 PM
Wilmer
Let n = number of digits

{20[10^(n-1) - 4] + 78} / 13

Gives integer solution for n=6k where k = any integer > 0