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anonymous
 one year ago
The sum of K consecutive integers is 41. If the least integer is 40, then k?
anonymous
 one year ago
The sum of K consecutive integers is 41. If the least integer is 40, then k?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here'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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The cutoff part says \(+...=41\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, if the first numebr is 40, just omit the 41 in my equations.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2does that equation work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmmmm so it's 41 then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhhhh crap .... it's 81

anonymous
 one year ago
Best ResponseYou've already chosen the best response.040 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{i=1}^k(40+i)=41 \\ 40k+\frac{k(k+1)}{2}=41\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0it is a quadratic equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm you know what i do is i add 41 to 40 and then multiply by 81

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I think I should have started it at i=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The key here is that they are "consecutive" integers. That's why just multiplying like that doesn't work.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[40(k+1)+\frac{k(k+1)}{2}=41\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles I got 82 as well. Might've had an error though.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0you said 81 earlier that is right

freckles
 one year ago
Best ResponseYou've already chosen the best response.0that 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 \[40(k+1)+\frac{k(k+1)}{2} \text{ which we want to this output } 41 \\ \text{ so we need \to solve } \\ 40(k+1)+\frac{k(k+1)}{2}=41\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{80(k+1)}{2}+\frac{k(k+1)}{2}=41 \\ \frac{(k+1)}{2}(80+k)=41 \\ (k+1)(80+k)=82 \\ k^279k80=82 \\ k^279k8082=0 \\ k^279k162=0 \\ (k+2)(k81)=0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0only one of these k's make sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@freckles You can use the sum of an arithmetic series.\[a_k=40, 39, ...\]\[S_k=\frac{k}{2}(2a_1+(k1)d)\]\[41=\frac{k}{2}(k81)\]\[82=k^281k\]\[0=k^281k82\]\[0=(k82)(k+1)\]\[k=1 ~or ~82\]And 1 is extraneous.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you prefer the algebraic method.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I gave an algebraic method above 81 works

freckles
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=sum%2840%2Bi%2Ci%3D0..k%29%3D41

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By starting at i = 0, the number of terms in that series is actually k + 1. I think that's the error.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0then why is wolfram saying 81 that is weird

dan815
 one year ago
Best ResponseYou've already chosen the best response.2freckles the indice starts at 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The number of terms can easily be calculated by ba+1 given\[\sum_{a}^{b}\]What you used equates to k+1 terms.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0so I should have used this: \[\sum_{i=1}^{k}(40+(i1)) =41 \\ 40k+\frac{k(k+1)}{2}k =41 \] i think this should arrive to the correct answer and yes this is k terms now my bad

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[80k+k(k+1)2k=82 \\ k^2+k82k=82 \\ k^281k82=0 \\ (k82)(k+1)=0\]
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