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

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More answers

saifoo.khan
  • saifoo.khan
How? Why? @myko
anonymous
  • anonymous
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...
anonymous
  • anonymous
the expretion would be equal to: n*z+ sum of first (z-1) integers = 90
anonymous
  • anonymous
@mukushla your sum is of z+1 consecutive integers
anonymous
  • anonymous
no its z n+1 n+2 ... n+z
anonymous
  • anonymous
should be nz+z(z-1)/2=90
saifoo.khan
  • saifoo.khan
So..?
anonymous
  • anonymous
2nz+z^2-z=180 z^2+(2n-1)z-180=0 and I think just check for possible natural solutions of this
anonymous
  • anonymous
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
saifoo.khan
  • saifoo.khan
@mukushla : Your's and @myko 's methods are same right?
anonymous
  • anonymous
yes they are
anonymous
  • anonymous
for example \(z=3\) gives \[n=\frac{90}{3}-\frac{3+1}{2}=30-2=28\]
anonymous
  • anonymous
yes they are same
anonymous
  • anonymous
only difference is that i start fomr n and @mukushla from n+1
saifoo.khan
  • saifoo.khan
So answer is 3?
anonymous
  • anonymous
it asks: Which of the following cannot be the value of "z". and 3 aparently , is.
saifoo.khan
  • saifoo.khan
But it's incorrect. :S
anonymous
  • anonymous
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
anonymous
  • anonymous
4and 6 would be the answer
anonymous
  • anonymous
i think
saifoo.khan
  • saifoo.khan
Myko's thinking is going the correct way!
anonymous
  • anonymous
i got it by mine first formula using n like a start integer.
anonymous
  • anonymous
but aparently there should be no difference in starting from n+1 or n+2 or n+whatever
anonymous
  • anonymous
hmm
anonymous
  • anonymous
for \(z=4\) \[21+22+23+24=90\]
rsadhvika
  • rsadhvika
4 90/4 - 4+1/2 90/4 - 10/4 80/4 is an integer
saifoo.khan
  • saifoo.khan
4 is incorrect too!
anonymous
  • anonymous
its impossible ... see my last reply
saifoo.khan
  • saifoo.khan
The answer is 6. :S
saifoo.khan
  • saifoo.khan
I'm going to type the solution.. just a sec.
anonymous
  • anonymous
@saifoo.khan thats right answer is 6...
saifoo.khan
  • saifoo.khan
How? Why? :D
anonymous
  • anonymous
man... we already solved it...
saifoo.khan
  • saifoo.khan
But you were getting 11.5 when we inserted 6.
anonymous
  • anonymous
well that means there is not 6 consecutive integers such that their sum is 90
saifoo.khan
  • saifoo.khan
(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
saifoo.khan
  • saifoo.khan
@mukushla , got it right!
saifoo.khan
  • saifoo.khan
I totally get it. Thanks a ton! @mukushla & @myko
anonymous
  • anonymous
yw :)
anonymous
  • anonymous
yw

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