[seqfan] Re: Counting problem

David Wilson dwilson at gambitcomm.com
Wed Jun 17 23:13:49 CEST 2009

BTW, I found this while working on Neil's old problem about the maximum 
persistence of a number under the product of digits map (A003001). This 
problem devolves to the persistence of numbers of the form 2^i*3^j*7^k 
and 5^i*3^j*7*k.

To eliminate some of the 5^i*3^j*7*k numbers, I was hoping to bound i by 
showing that a sufficiently large 5^i factor would force an even digit 
in the last i digits of 5^i*3^j*7*k. This would mean 5^i*3^j*7*k has an 
even digit and the digit 5, at the end. The product of its digits would 
then end in 0, and the second product of its digits would be 0, so that 
5^i*3^j*7*k would have persistence 2 and could be eliminated from 

Alas, I found that for any factor 5^i, no matter how large i was, there 
was some value of 3^j*7*k that made the last i digits of 5^i*3^j*7*k all 
odd. Oddly (sic), these last i digits seems to be unique.

I did some experiments which indicated there was exactly one multiple of 
5^i ending in i odd digits for 1 <= i <= 100, so I supposed it was true. 
Tanya confirms this.

So let's take this as our hypothesis: For any k, there is a multiple of 
5^k that ends in a unique string v_k of k odd digits.

Letting k = i, we see that some multiple of 5^i ends in the unique 
string v_i of i odd digits. In other words

	c*5^i mod 10^i = v_i

for some c.

Now take w to be v_i with it first digit lopped of, that is

	w = v_i mod 10^(i-1)

So w has i-1 odd digits. But

	w = (c*5^i mod 10^i) mod 10^(i-1)
	  = c*5^i mod 10^(i-1) (since 10^(i-1) divides 10^i)
	  = 5c*5^(i-1) mod 10^(i-1).

so that w is the last i-1 odd digits of a multiple of 5^(i-1).

But by setting k = i-1 in our hypothesis, we see that the last i-1 odd 
digits of a multiple of 5^(i-1) must be v_(i-1). Since w fills this 
description, w = v_(i-1).

But this means that v_(i-1) = w is the last i-1 digits of v_i, by 
definition of w. This shows that v_(i-1) is a suffix of v_i, or in other 
words, v_i is formed by prepending an odd digit to v_(i-1), as you observed.

I can't help but think that there is some simply-defined 10-adic number 
which includes these suffixes.

Harvey P. Dale wrote:
> David:
> 	Is the answer always 1?  The numbers satisfying your conditions,
> starting with n=1, are 5, 75, 375, 9375, 59375, 359375, and 3359375.
> Note that once any such number is found, the number for the next higher
> n has the same ending digits with a prepended different odd digit.
> 	Best,
> 	Harvey

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