Question

The sum of K consecutive integers is 41. If the least integer is -40, then k?

Answer 1

Here's a way of thinking about this. If you're adding consecutive integers and the first is -41, then you're adding:\[-41+(-40)+(-39)+(-38)+...=41\]However, at some point, you're going to get into the positive numbers when you're adding.\[-41+(-40)+(-39)+...+(-3)+(-2)+(-1)+(0)+(1)+(2)+(3)+...=41\]At the point where you get to the positive numbers, the negatives and positives will begin to cancel out. This is because -3 + 3 = 0, -2 + 2 = 0, -1 + 1 = 0. Does this help you with the rest of the problem?

Answer 2

The cutoff part says \(+...=41\)

Answer 3

the least number is -40

Answer 4

oops

Answer 5

Ok, if the first numebr is -40, just omit the -41 in my equations.

Answer 6

does that equation work

Answer 7

hmmmmm so it's 41 then?

Answer 8

Not quite. 41 would be the last term that you would be adding to get the sum to be 41. However, k is the number of numbers you're adding.

Answer 9

ohhhhh crap .... it's 81

Answer 10

-40 to 0 is 41 numbers. 1 to 41 is 41 numbers. So, 41 + 41 should be what k is. Is this making sense to you?

Answer 11

\[\sum_{i=1}^k(-40+i)=41 \\ -40k+\frac{k(k+1)}{2}=41\]

Answer 12

it is a quadratic equation

Answer 13

hmmm you know what i do is i add 41 to -40 and then multiply by 81

Answer 14

I think I should have started it at i=0

Answer 15

The key here is that they are "consecutive" integers. That's why just multiplying like that doesn't work.

Answer 16

\[-40(k+1)+\frac{k(k+1)}{2}=41\]

Answer 17

crap so it's 82

Answer 18

almost

Answer 19

huh?

Answer 20

@freckles I got 82 as well. Might've had an error though.

Answer 21

you said 81 earlier that is right

Answer 22

that is what you get when you solve that quadratic equation above
\[\sum_{i=0}^k (-40+i)\]
this does mean
-40+(-39)+(-38)+...+(-40+k)

or if you used those formula things
you can write that sum thingy as
\[-40(k+1)+\frac{k(k+1)}{2} \text{ which we want to this output } 41 \\ \text{ so we need \to solve } \\ -40(k+1)+\frac{k(k+1)}{2}=41\]

Answer 23

\[\frac{-80(k+1)}{2}+\frac{k(k+1)}{2}=41 \\ \frac{(k+1)}{2}(-80+k)=41 \\ (k+1)(-80+k)=82 \\ k^2-79k-80=82 \\ k^2-79k-80-82=0 \\ k^2-79k-162=0 \\ (k+2)(k-81)=0\]

Answer 24

only one of these k's make sense

Answer 25

82 is right i believe