one to a billion

[Guilherme]

1 + 1.000.000.000 = 1.000.000.001
2 + 999.999.999 = 1.000.000.001
3 + 999.999.998 = 1.000.000.001
4 + 999.999.997 = 1.000.000.001
and so on.

From 1 to 1.000.000.000, we have 500.000.000 pairs of numbers.

Sum(1 to 1.000.000.000) = 500.000.000 x 1.000.000.001 = 500.000.000.500.000.000

[Kees]

Originally Posted by Kees

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

ugh, why do I always forget the simple cases

[mfgoode]

Originally Posted by axn1

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"

Hey axn1 you are pretty good and excellent reasoning.

Kees you are close but not just quite.

Consider the integers from 0 to 10^n -1; (10^n -1) consists of n nines. Pair zero with this number, pair 1 with 10^n -2, pair 2 with 10^n - 3 etc.
Summing up you get 45 n*(10^n-1) +1.
For one billion, n=9 and so the sum is 40,500,000,001
Note 10^n-1 is actually 10 ^ (n-1)
Mally
P.S Whats all the programming about? Just paper and pencil man!
and a calculator!

[Uncwilly]

[QUOTE=mfgoodeThe poem is ,as it is, self explanatory. It is the sum of all the digits thats required.[/QUOTE]
A co-worker and I have a small wager about the general topic.

Is the sum of the digits in a series of numbers, ever useful in math or science?

[ewmayer]

Hard to believe this topic generated more than a two-line summary response along the lines of:

1) Here we take "billion" to mean 10^9 (or 10^12, as the case may be);
2) Cf. C.F.Gauss, circa age 6.

Mally, I suggest you compare your example calculation to reference (2) - your result for N = 10^9 differs appreciably from (N/2)*(N+1).

[mfgoode]

Originally Posted by Uncwilly

A co-worker and I have a small wager about the general topic.

Is the sum of the digits in a series of numbers, ever useful in math or science?

I have no positive data on this aspect, but as you know, a lot of results in pure maths have no applied use for a long time, until someone finds one, maybe years later, or even centuries.

Multi dimensional geometry beyond 4 dimensions was considered superfluous,
till the physicists applied 11 dimensions in string theory two decades ago.

Further back, pre Einstein, three dimensions solved all the classical problems and Reimann 4 dimensional geometry was just a mathematical curiousity. Then Einstein adopted it and explained away all the loop holes, arising from experimrental results, which conflicted with the classical theories of Newton.

Why, in our own backyard, the search for big primes were considered pointless
until it was used for cryptic codes and random numbers.

Perhaps the summation of digits of huge numbers could be also used for generating random numbers and cryptic codes, heaven knows!
Mally

[ewmayer]

While I make no claims as to the sum of N consecutive digits approaching the profundity of non-Euclidean geometry, here's one small example where it comes in: the number of independently specifiable elements of an N x N symmetric matrix is the sum from 1 to N (inclusive), which equals N*(N+1)/2. Analogously, an N x N antisymmetric matrix has N*(N-1)/2 independent components.

[mfgoode]

Originally Posted by ewmayer

Hard to believe this topic generated more than a two-line summary response along the lines of:

1) Here we take "billion" to mean 10^9 (or 10^12, as the case may be);
2) Cf. C.F.Gauss, circa age 6.

Mally, I suggest you compare your example calculation to reference (2) - your result for N = 10^9 differs appreciably from (N/2)*(N+1).

:surprised
Thank you Ernst for making me take a fresh look at the theory behind the question. It has strengthened my conviction that what was derived is flawless, in my previous post.

Yes, there is a difference between what Gauss, at the age of 6, derived and the one in question.

Gauss arrived at 5050 using the well known formula N ( N + 1)/2 where N is the number

The formula I presented is 9n * (10 ^ n)/2 + 1 where n is the power.
Thus for Sn for sum of digits 1 - 100 where n = 2 the sum works out to be 901 - a vast difference.

Eg: take S 10 for integers 1 - 10 this sum is 55 using Gauss’ formula

For the digits the sum is 46. Please try it out by adding the individual digits + 1 for 1 - 10
Mally

[Xyzzy]

This problem has been bugging me for a while. Now I understand the references to Gauss previously posted.

From: http://www.americanscientist.org/tem...686?&print=yes

In the 1780s a provincial German schoolmaster gave his class the tedious assignment of summing the first 100 integers. The teacher's aim was to keep the kids quiet for half an hour, but one young pupil almost immediately produced an answer: 1 + 2 + 3 + ... + 98 + 99 + 100 = 5,050. The smart aleck was Carl Friedrich Gauss, who would go on to join the short list of candidates for greatest mathematician ever. Gauss was not a calculating prodigy who added up all those numbers in his head. He had a deeper insight: If you "fold" the series of numbers in the middle and add them in pairs—1 + 100, 2 + 99, 3 + 98, and so on—all the pairs sum to 101. There are 50 such pairs, and so the grand total is simply 50×101. The more general formula, for a list of consecutive numbers from 1 through n, is n(n + 1)/2.

[mfgoode]

The other story that went the rounds was popularised by G.H. Hardy the English mathematician.
He took a taxi cab to see Ramanujan, who was ill, at Cambridge and mentioned to the latter that the cab number was a dreary one viz: 1729.

Ramanujan, sick as he was, told Hardy "Not so! It is the sum of two different sets of cubes

i.e 10^3 + 9^3 and also

12^3 + 1^3. Thats number theory for you!

BTW: I have 'Men of mathematics' 2 vols by E.T. Bell. Its a worthy read.
Mally

[nfortino]

Originally Posted by mfgoode

The other story that went the rounds was popularised by G.H. Hardy the English mathematician.
He took a taxi cab to see Ramanujan, who was ill, at Cambridge and mentioned to the latter that the cab number was a dreary one viz: 1729.

Ramanujan, sick as he was, told Hardy "Not so! It is the sum of two different sets of cubes

i.e 10^3 + 9^3 and also

12^3 + 1^3. Thats number theory for you!

BTW: I have 'Men of mathematics' 2 vols by E.T. Bell. Its a worthy read.
Mally

In addition to being the smallest integer which can be written as the sum of a pair of cubes in two different ways, 1729 is also the third Carmichael number. Various tellings of the story have Ramanujan pointing this fact out as well, although I don't know if that actually happened.