Hi , let's think about this problem :

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

Is the answer?Spoiler:

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- Aug 12th 2010, 10:20 PMsimplependulumPutnam Problem of the Day
Hi , let's think about this problem :

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

Is the answer__Spoiler__: - Aug 13th 2010, 03:26 AMUnbeatable0
Let the number be .

Let .

Then .

Therefore .

By backwards argument it's easy to see that all the cycles in are solutions.

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

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

Just write out the multiplcation,

. . and the digits will appear sequentially.

Quote:

Find the smallest natural number with as the last digit,

such that if the final is moved to the front of the number it is multiplied by .

We have: .

In column-6: .

. .

In column-5: .

. .

In column-4: .

. .

In column-3: .

. .

In column-2: .

Therefore: .

- Aug 13th 2010, 05:19 PMWilmer
Soroban, the fact that it is a 6digit number is NOT a given.

- Aug 13th 2010, 06:16 PMSorobanQuote:

Originally Posted by**Wilmer**

I originally set up the problem like this:

. .

And found that "it came out even" after six digits

. . which gave me the smallest number:

The next is the 12-digit number:

- Aug 13th 2010, 07:10 PMundefined
Fyi, there is a name for these, n-parasitic number, where here n=4 and k=6.

- Aug 13th 2010, 09:32 PMWilmer
Let n = number of digits

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

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