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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.

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  1. mathslover
    • 3 years ago
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    it seems that the answer given by saurav is very correct

  2. saifoo.khan
    • 3 years ago
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    i still don't get it. lol.

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

  4. saifoo.khan
    • 3 years ago
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    How? Why? @myko

  5. mukushla
    • 3 years ago
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    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
    • 3 years ago
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    the expretion would be equal to: n*z+ sum of first (z-1) integers = 90

  7. myko
    • 3 years ago
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    @mukushla your sum is of z+1 consecutive integers

  8. mukushla
    • 3 years ago
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    no its z n+1 n+2 ... n+z

  9. myko
    • 3 years ago
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    should be nz+z(z-1)/2=90

  10. saifoo.khan
    • 3 years ago
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    So..?

  11. myko
    • 3 years ago
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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

  12. mukushla
    • 3 years ago
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    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
    • 3 years ago
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    @mukushla : Your's and @myko 's methods are same right?

  14. myko
    • 3 years ago
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    yes they are

  15. mukushla
    • 3 years ago
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    for example \(z=3\) gives \[n=\frac{90}{3}-\frac{3+1}{2}=30-2=28\]

  16. mukushla
    • 3 years ago
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    yes they are same

  17. myko
    • 3 years ago
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    only difference is that i start fomr n and @mukushla from n+1

  18. saifoo.khan
    • 3 years ago
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    So answer is 3?

  19. myko
    • 3 years ago
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    it asks: Which of the following cannot be the value of "z". and 3 aparently , is.

  20. saifoo.khan
    • 3 years ago
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    But it's incorrect. :S

  21. mukushla
    • 3 years ago
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    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
    • 3 years ago
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    4and 6 would be the answer

  23. myko
    • 3 years ago
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    i think

  24. saifoo.khan
    • 3 years ago
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    Myko's thinking is going the correct way!

  25. myko
    • 3 years ago
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    i got it by mine first formula using n like a start integer.

  26. myko
    • 3 years ago
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    but aparently there should be no difference in starting from n+1 or n+2 or n+whatever

  27. myko
    • 3 years ago
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    hmm

  28. mukushla
    • 3 years ago
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    for \(z=4\) \[21+22+23+24=90\]

  29. rsadhvika
    • 3 years ago
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    4 90/4 - 4+1/2 90/4 - 10/4 80/4 is an integer

  30. saifoo.khan
    • 3 years ago
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    4 is incorrect too!

  31. mukushla
    • 3 years ago
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    its impossible ... see my last reply

  32. saifoo.khan
    • 3 years ago
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    The answer is 6. :S

  33. saifoo.khan
    • 3 years ago
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    I'm going to type the solution.. just a sec.

  34. mukushla
    • 3 years ago
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    @saifoo.khan thats right answer is 6...

  35. saifoo.khan
    • 3 years ago
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    How? Why? :D

  36. mukushla
    • 3 years ago
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    man... we already solved it...

  37. saifoo.khan
    • 3 years ago
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    But you were getting 11.5 when we inserted 6.

  38. mukushla
    • 3 years ago
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    well that means there is not 6 consecutive integers such that their sum is 90

  39. saifoo.khan
    • 3 years ago
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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

  40. saifoo.khan
    • 3 years ago
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    @mukushla , got it right!

  41. saifoo.khan
    • 3 years ago
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    I totally get it. Thanks a ton! @mukushla & @myko

  42. mukushla
    • 3 years ago
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    yw :)

  43. myko
    • 3 years ago
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    yw

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