20060411, 16:49  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
one to a billion
: ":smile
Once a bright young lady called Lillian, Summed the numbers from one to a billion, But it gave her the fidgets To add up the digits; If you can help her, she'll thank you a million." Mally 
20060411, 17:25  #2 
Jul 2005
2·193 Posts 
5E17 or 5E23?

20060411, 17:51  #3 
Sep 2005
Detroit, MI
2^{3} Posts 
Easy one. Just have to use the formula for adding 1n :)
500,000,000,500,000,000 or ~5^17 Last fiddled with by Jamiaz on 20060411 at 17:51 
20060411, 19:53  #4 
Jun 2003
2·3·5·173 Posts 
Is the question about sum of 1..n OR sum of _digits_ of 1..n ?!

20060411, 19:57  #5 
Aug 2002
61×137 Posts 
$ cat a.pl
#!/usr/bin/perl w use strict; my $sum; for my $x ( 1 .. 1_000_000_000 ) { $sum += $x; } print "$sum\n"; $ ./a.pl 500000000500000000 
20060411, 20:03  #6  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
11,027 Posts 
Quote:
Paul 

20060411, 23:03  #7 
Jul 2005
2·193 Posts 
Indeed, in any reasonable language[1] that would produce possibly random/incorrect answers.
bash# cat a.c #include <stdio.h> #include <stdint.h> int main(void) { uint64_t sum; /* you think this is equal to 0? */ printf( "%llu\n", sum ); return(0); } bash# gcc a.c o a bash# ./a 18807377184 bash# gcc O2 a.c o a bash# ./a 6994688905387704324 Anyway, my original answers were wrong (yeah, yeah) but why did I give two (wrong) answers to this question? [1] I'm joking. I love perl, but I've programmed in C for too long to let uninitialised variables pass me by (and it even had 'use strict' in there!). P.S. axn1, you raise a good point. Last fiddled with by Greenbank on 20060411 at 23:09 
20060411, 23:55  #8  
Aug 2002
North San Diego County
5×139 Posts 
Quote:


20060412, 08:45  #9  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Sum of digits
Quote:
You are on the right track axn1. The poem is ,as it is, self explanatory. It is the sum of all the digits thats required. Take a billion as having 9 zeros None of the answers given so far are correct. Mally 

20060412, 09:26  #10 
Dec 2005
2^{2}·7^{2} Posts 
do not forget the last one
Leaving alone 1.000.000.000 which adds 1 to the final sum (see subject title) we
are summing all 9 digit numbers, not beginning with 0. Fixing a position (say the last one) we can have 0,1,2,3,4,5,6,7,8 or 9 on it and for every number in this choice we have 9*10^7 possible 9digit numbers. Summing over the last 8 positions gives 8*9*10^7*45. If we fix the first digit, we have 10^8 possible 9digit numbers, giving 10^8*45. Adding these two numbers and adding 1 should give the required result: 36900000001 
20060412, 12:33  #11 
Jun 2003
5190_{10} Posts 
000 000 000 000 000 001 ........... 999 999 998 999 999 999 1 billion numbers * 9 digits per number = 9 billion digits. Number of occurrences of any one digit = 9 billion / 10 = 900 million Sum = 900 million * Sum(0..9) = 45 * 900 million = 40.5 billion And adding 1 (for 1 billion), we get "40.5 billion and 1" Last fiddled with by axn on 20060412 at 12:34 
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