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The sum of "z" consecutive integers is 90. Which of the following cannot be the value of "z". 3, 4, 5, 6, 9.

Mathematics
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it seems that the answer given by saurav is very correct
i still don't get it. lol.
maybe first put it like this: \[n+(n+1)+(n+2)+...+(n+(z-1))=90\] and find it from here

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Other answers:

How? Why? @myko
guys lets work it step by step \[n+1+n+2+...+n+z=90\]\[nz+\frac{z(z+1)}{2}=90\]\[z^2+(2n+1)z=180\]\[z(z+2n+1)=180\] we can find all integers \(z\) that satisfies the last equality...but here we have options and we can plug them into equation and see what happens...
the expretion would be equal to: n*z+ sum of first (z-1) integers = 90
@mukushla your sum is of z+1 consecutive integers
no its z n+1 n+2 ... n+z
should be nz+z(z-1)/2=90
So..?
2nz+z^2-z=180 z^2+(2n-1)z-180=0 and I think just check for possible natural solutions of this
now u have \[n=\frac{90}{z}-\frac{z+1}{2}\] now plug the values of \(z\) and if the value of \(n\) becomes an integer then that \(z\) is your answer
@mukushla : Your's and @myko 's methods are same right?
yes they are
for example \(z=3\) gives \[n=\frac{90}{3}-\frac{3+1}{2}=30-2=28\]
yes they are same
only difference is that i start fomr n and @mukushla from n+1
So answer is 3?
it asks: Which of the following cannot be the value of "z". and 3 aparently , is.
But it's incorrect. :S
let me complete the solution \[z=3 ---------> n=\frac{90}{3}-\frac{3+1}{2}=28\] \[z=4 ---------> n=\frac{90}{4}-\frac{4+1}{2}=20\] \[z=5 ---------> n=\frac{90}{5}-\frac{5+1}{2}=15\] \[z=6 ---------> n=\frac{90}{6}-\frac{6+1}{2}=11.5\] \[z=9 ---------> n=\frac{90}{9}-\frac{9+1}{2}=5\] so the answer is only z=6
4and 6 would be the answer
i think
Myko's thinking is going the correct way!
i got it by mine first formula using n like a start integer.
but aparently there should be no difference in starting from n+1 or n+2 or n+whatever
hmm
for \(z=4\) \[21+22+23+24=90\]
4 90/4 - 4+1/2 90/4 - 10/4 80/4 is an integer
4 is incorrect too!
its impossible ... see my last reply
The answer is 6. :S
I'm going to type the solution.. just a sec.
@saifoo.khan thats right answer is 6...
How? Why? :D
man... we already solved it...
But you were getting 11.5 when we inserted 6.
well that means there is not 6 consecutive integers such that their sum is 90
(behind the book) Solution: Plug in numbers! This may take some time, but we will eventually find the answer. Three numbers 20+30+31=90,eliminate A, four numbers 21+22+23+24=90,eliminate b,five numbers 16+17+18+19+20=90,elimnate c.nine numbers 6+7+8+9+10+11+12+13+14=90 elimate e.no six consecutive numbers add up to 90
@mukushla , got it right!
I totally get it. Thanks a ton! @mukushla & @myko
yw :)
yw

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