I have difficulty counting

Question

Take this example:

$$84 + 87 + 90 + 93 .. + 180 + 183$$

If we want to use Gauss' way of finding this sum, we have to find the number of elements in this. What I do is just take the difference of the highest and lowest sum (99) and divide by 3, but this yields 33 and that can't be correct because it needs to be an even number in order to make sense. Why does it count 34 elements instead of the 33 my (faulty) arithmetic gives me?

How many natural numbers are there from $0$ to $33$ (inclusive)? — tomasz, May 12 '13 at 20:26
consider 1,2,3,4: $4-1 = 3$ yet there are 4 elements. You need to consider the difference $(183 - 81)/3 = 34$ — Myself, May 12 '13 at 20:26
@Kaster Otherwise the sum you obtain through Gauss' method has a fraction in it which is nonsense of course. — Gallus, May 12 '13 at 20:29
@Kaster the numbers in the odd places (first, third, fifth ...) are all even. The final number is odd, so must come in an even place. — Mark Bennet, May 12 '13 at 20:30

score 4 · Answer 1 · answered May 12 '13 at 20:30

It's one of those typical "off by one" errors. The best way to see how to count the elements in such a sequence is to start with few elements:

If you have $84+87$ you have $(87-84)/3=1$ and there are $2$ elements in the sum
If you have $84+87+90$ you have $(90-84)/3=2$ and there are $3$ elements in the sum
If you have $84+87+90+93$ you have $(93-84)/3=3$ and there are $4$ elements in the sum

I guess by now you see the pattern ...

If you have $84+87+90+\cdots183$ you have $(183-84)/3=33$ and there are $34$ elements in the sum

score 3 · Answer 2 · answered May 12 '13 at 20:23

3

Try finding $3+6+\cdots+183$, and then $3+6+\cdots+81$, and subtracting.

answered May 12 '13 at 20:23

vadim123

82,796

score 2 · Accepted Answer · answered May 12 '13 at 20:26

2

There are 34 terms starting with $84+0 \times 3=84$ and ending with $84+33\times 3=183$

answered May 12 '13 at 20:26

Mark Bennet

100,194

score 0 · Answer 4 · answered May 12 '13 at 20:28

We want \begin{align} \color{red}{S = 84 + 87 + 90 + 93 + \cdots + 180 + 183} \end{align} We also have \begin{align} \color{blue}{S = 183 + 180 + 177 + 174 + \cdots + 87 + 84} \end{align} This gives us \begin{align} \color{green}{2S} & \color{green}{= (84+183) + (87+180) + (90+177) + \cdots + (180+87) + (183+84)}\\ & \color{green}{= (267) + (267) + (267) + \cdots + (267) + (267)}\\ \end{align} Let us count how many such $267$'s are there: We have $183 = 84 + 3k$. Note that $k+1$ gives the number of terms in the summation. We get $k = \dfrac{183-84}3 = 33$. Hence, there are $34$ terms. We hence get that $$\color{green}{2S = 34 \times 267 \implies S = 17 \times 267 = 4539}$$

score 0 · Answer 5 · answered May 12 '13 at 20:28

0

When you count $1$ through $10$ you have $10$ elements, but $10-1=9$. The extra one comes because you are counting both ends, while taking the difference counts the gaps between them. You have to add $1$ to account for that.

answered May 12 '13 at 20:28

Ross Millikan

374,822

Stefan4024 · Answer 6 · 2013-05-12T20:50:37.880

Every number in your series is divisible by 3, so we can write it as:

$$84+87+90+93..+180+183 = 3(28+29+...+60+61)$$

Know using Gauss formula we can get:

$$3(28+29+...+60+61) = 3(\frac{61*62}{2} - \frac{27*28}{2})$$

We can simplify this to:

$$3(\frac{34*62 + 27*62 - 27*28}{2}) = $$ $$3(\frac{34*62 + 27(62-28)}{2}) = $$ $$3(\frac{34*62 + 27*34}{2}) = $$ $$3(\frac{34(62+27)}{2}) = $$ $$3(\frac{34*89)}{2}) = $$ $$\frac{102*89}{2} = 4539$$

I have difficulty counting

6 Answers6