Putnam Problem of the Day

**simplependulum** · August 12th 2010, 10:20 PM

Hi , let's think about this problem :

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 $4$ .

Is the answer

Spoiler:

?

**Unbeatable0** · August 13th 2010, 03:26 AM

Let the number be $a_1a_2...a_n6$ .

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

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

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

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

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

This technique can be applied to this kind of problems in many cases.

**Soroban** · August 13th 2010, 02:03 PM

Hello, simplependulum!

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

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 $4$ .

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

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

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

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

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

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

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

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

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

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

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

**Wilmer** · August 13th 2010, 05:19 PM

Soroban, the fact that it is a 6digit number is NOT a given.

**Soroban** · August 13th 2010, 06:16 PM

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:

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

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

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

**undefined** · August 13th 2010, 07:10 PM

Fyi, there is a name for these, n-parasitic number, where here n=4 and k=6.

**Wilmer** · August 13th 2010, 09:32 PM

Let n = number of digits

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

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

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