Re[2]: Math problem

["Joseph Nguyen" <Joseph_Nguyen@bmc.boehringer-mannheim.com>]

Hi Hong,

You overcounted the number of numbers that divisible of 7 in the group of 
numbers divisible of 49.

there are at least 100 7's in 700!
>From 1 to 700 there are 1 7^2 (49) and 1 7^3 (343).  So, we heve 103 7's in 
700! we need only 100 7's.  
100 - 3 = 97. and so, the least number should be 7*97 = 679


______________________________ Reply Separator _________________________________
Subject: Re: Math problem 
Author:  vnsa-l@csd.uwm.edu at SMTP
Date:    3/19/97 9:23 AM


  
  
Hi Tuan,
  
On Wed, 19 Mar 1997, Tuan V Nguyen wrote:
  
> Hello Thu Huong and Thanh Hong,
> 
>         You are very smart, that is the correct number (679). But, you have 
> not told us how did you arrive at the number! Please do.
> 
>         Cheers,
> 
>         Tuan V Nguyen
> 
  
As my algorithm, count from 1 to  609, there are 87 numbers that is 
indivisable for 7, such as 602, 595, 588, and so on. Among these numbers, 
there are 12 numbers that is divisable for 49 ( = 7*7) such as 588, 539,.. 
and the number of 343 is equal 7*7*7. Thus, 609! will divisable for 
7^(87+12+1 = 100).
  
Could you test whether this algorithm is true or not?
  
Cheers,
  
Hong.