Question

The sum of "z" consecutive integers is 90.

Which of the following cannot be the value of "z".

3, 4, 5, 6, 9.

Answer 1

it seems that the answer given by saurav is very correct

Answer 2

i still don't get it. lol.

Answer 3

maybe first put it like this:
\[n+(n+1)+(n+2)+...+(n+(z-1))=90\]
and find it from here

Answer 4

How? Why? @myko

Answer 5

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...

Answer 6

the expretion would be equal to:
n*z+ sum of first (z-1) integers = 90

Answer 7

@mukushla your sum is of z+1 consecutive integers

Answer 8

no its z

n+1
n+2
...
n+z

Answer 9

should be
nz+z(z-1)/2=90

Answer 10

So..?

Answer 11

2nz+z^2-z=180
z^2+(2n-1)z-180=0
and I think just check for possible natural solutions of this

Answer 12

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

Answer 13

@mukushla : Your's and @myko 's methods are same right?

Answer 14

yes they are

Answer 15

for example \(z=3\) gives

\[n=\frac{90}{3}-\frac{3+1}{2}=30-2=28\]

Answer 16

yes they are same

Answer 17

only difference is that i start fomr n and @mukushla from n+1

Answer 18

So answer is 3?

Answer 19

it asks: Which of the following cannot be the value of "z".
and 3 aparently , is.

Answer 20

But it's incorrect. :S

Answer 21

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

Answer 22

4and 6 would be the answer

Answer 23

i think

Answer 24

Myko's thinking is going the correct way!

Answer 25

i got it by mine first formula using n like a start integer.

Answer 26

but aparently there should be no difference in starting from n+1 or n+2 or n+whatever

Answer 27

hmm

Answer 28

for \(z=4\)

\[21+22+23+24=90\]

Answer 29

4

90/4 - 4+1/2
90/4 - 10/4

80/4


is an integer

Answer 30

4 is incorrect too!

Answer 31

its impossible ... see my last reply

Answer 32

The answer is 6. :S

Answer 33

I'm going to type the solution.. just a sec.

Answer 34

@saifoo.khan

thats right answer is 6...

Answer 35

How? Why? :D

Answer 36

man... we already solved it...

Answer 37

But you were getting 11.5 when we inserted 6.

Answer 38

well that means there is not 6 consecutive integers such that their sum is 90

Answer 39

(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

Answer 40

@mukushla , got it right!

Answer 41

I totally get it. Thanks a ton!
@mukushla & @myko

Answer 42

yw :)

Answer 43

yw