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Old 2006-04-11, 16:49   #1
mfgoode
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Cool 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
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Old 2006-04-11, 17:25   #2
Greenbank
 
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5E17 or 5E23?
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Old 2006-04-11, 17:51   #3
Jamiaz
 
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Easy one. Just have to use the formula for adding 1-n :)

500,000,000,500,000,000 or ~5^17

Last fiddled with by Jamiaz on 2006-04-11 at 17:51
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Old 2006-04-11, 19:53   #4
axn
 
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Is the question about sum of 1..n OR sum of _digits_ of 1..n ?!
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Old 2006-04-11, 19:57   #5
Xyzzy
 
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$ 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
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Old 2006-04-11, 20:03   #6
xilman
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Down not across

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Quote:
Originally Posted by Xyzzy
$ 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
That is a truly disgusting piece of code!



Paul
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Old 2006-04-11, 23:03   #7
Greenbank
 
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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 2006-04-11 at 23:09
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Old 2006-04-11, 23:55   #8
sdbardwick
 
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Quote:
Originally Posted by Greenbank
Anyway, my original answers were wrong (yeah, yeah) but why did I give two (wrong) answers to this question?
Because the term billion is ambiguous; 10^12 or 10^9?
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Old 2006-04-12, 08:45   #9
mfgoode
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Smile Sum of digits

Quote:
Originally Posted by axn1
Is the question about sum of 1..n OR sum of _digits_ of 1..n ?!

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
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Old 2006-04-12, 09:26   #10
Kees
 
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Dec 2005

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Default 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 9-digit numbers.
Summing over the last 8 positions gives 8*9*10^7*45.
If we fix the first digit, we have 10^8 possible 9-digit numbers, giving 10^8*45.
Adding these two numbers and adding 1 should give the required result:

36900000001
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Old 2006-04-12, 12:33   #11
axn
 
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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 2006-04-12 at 12:34
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