Challenge Yourself: Count and Add 1-100 Without a Calculator!

AI Thread Summary
The discussion centers on the ability to sum the integers from 1 to 100 without a calculator, referencing a famous anecdote about mathematician Carl Friedrich Gauss. Participants highlight that Gauss reportedly calculated the sum quickly as a child by using a formula: sum = 0.5n(n+1), which simplifies to 5050 for the numbers 1 through 100. The conversation explains that this formula derives from finding the average of the first and last terms, which is 50.5, and multiplying it by the total number of terms (100). Clarifications are made regarding the average value, emphasizing that 50.5 represents the average of all integers in the sequence from 1 to 100. The discussion concludes with a suggestion for further learning through a linked lecture on the topic.
konartist
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Can you count and add all the numbers 1-100 without using a calculator?
 
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konartist said:
Can you count and add all the numbers 1-100 without using a calculator?

You got me!
 
As a metter of fact, I think I can :wink:
If I am not mistaken then Gauss did it centuries ago...
 
konartist said:
Can you count and add all the numbers 1-100 without using a calculator?

Let's hope not! ... Maybe all the *integers*, tho :)
(And yeah, you can use the old ((S+F)/2)*(F-S+1) trick to add all the integers between S and F)[/color]

DaveE
 
konartist said:
Can you count and add all the numbers 1-100 without using a calculator?
Well it wasn't easy, but there are 100[/color] of them.
 
VietDao29 said:
As a metter of fact, I think I can :wink:
If I am not mistaken then Gauss did it centuries ago...

Yes sir, an old wive's tale is that he one day was punished in 3rd grade class and the teacher asked him to add all the numbers 1-100 thinking it would keep him busy, but he was able to reply the answer back to her in a matter of seconds.
 
yea u just use the formula...sum = 0.5n(n+1)

so 0.5 x 100 x 101 =50 x 101 = 5050
 
Yes, 1 + ... + 100 = (1 + 100) + (2 + 99) + ... (50 + 51) = (101) + (101) + ... + (101) [50 times] = 50 x 101 = 5050. I heard that this is how Gauss did it.
 
So what you're all saying is, find the average value (50.5) and mutiply by the number of terms.
 
  • #10
Cybersteve said:
So what you're all saying is, find the average value (50.5) and mutiply by the number of terms.
What is 50.5 the average value of?
 
  • #11
mattmns said:
What is 50.5 the average value of?
I think he meant the average value of the first term and the last term, ie:
(1 + 100) / 2 = 50.5
 
  • #12
mattmns said:
What is 50.5 the average value of?

50.5 is the average value of all the numbers 1 to 100 inclusive.

If there were an odd number of terms in an evenly spaced sequence the average would be the middle term.
As this sequence is even the average value is the average of the two middle terms - 50 & 51.
 
  • #13
Ahh, I must have misread something.
 
  • #14
mattmns said:
Ahh, I must have misread something.
Not so much misread as partially read. 50.5 is not only the average of 1 and 100. It is also the average of 2 and 99, 3 and 98, etc. That is, it is the average of the set of integers from 1 to 100.
 
  • #15
im lost...
 
  • #16
This is easy.

Here is a lecture that shows you how to do it.
http://www.jimloy.com/algebra/gauss.htm
 
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