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

1. mathslover

it seems that the answer given by saurav is very correct

2. saifoo.khan

i still don't get it. lol.

3. myko

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

4. saifoo.khan

How? Why? @myko

5. mukushla

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...

6. myko

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

7. myko

@mukushla your sum is of z+1 consecutive integers

8. mukushla

no its z n+1 n+2 ... n+z

9. myko

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

10. saifoo.khan

So..?

11. myko

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

12. mukushla

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

13. saifoo.khan

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

14. myko

yes they are

15. mukushla

for example $$z=3$$ gives $n=\frac{90}{3}-\frac{3+1}{2}=30-2=28$

16. mukushla

yes they are same

17. myko

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

18. saifoo.khan

19. myko

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

20. saifoo.khan

But it's incorrect. :S

21. mukushla

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

22. myko

4and 6 would be the answer

23. myko

i think

24. saifoo.khan

Myko's thinking is going the correct way!

25. myko

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

26. myko

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

27. myko

hmm

28. mukushla

for $$z=4$$ $21+22+23+24=90$

4 90/4 - 4+1/2 90/4 - 10/4 80/4 is an integer

30. saifoo.khan

4 is incorrect too!

31. mukushla

its impossible ... see my last reply

32. saifoo.khan

33. saifoo.khan

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

34. mukushla

@saifoo.khan thats right answer is 6...

35. saifoo.khan

How? Why? :D

36. mukushla

37. saifoo.khan

But you were getting 11.5 when we inserted 6.

38. mukushla

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

39. 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

40. saifoo.khan

@mukushla , got it right!

41. saifoo.khan

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

42. mukushla

yw :)

43. myko

yw