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