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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} S=\{1,2,3,4,\cdots , 1000\} \hspace{.33em}\\~\\ \end{align}}\) How many arithmatic progressions can be formed from the elements of \(S\) that start with \(1\) and end with \(1000\) and have at least \(3\) elements.

  2. anonymous
    • one year ago
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    may u pls help me with my math?

  3. anonymous
    • one year ago
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    ok, let's work on this problem

  4. anonymous
    • one year ago
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    \(a_1=1\), \(a_n=1000\) let difference of terms be \(d\)

  5. mathmath333
    • one year ago
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    u mean common difference

  6. anonymous
    • one year ago
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    yeah

  7. anonymous
    • one year ago
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    Now we can calculate \(a_n\)

  8. mathmath333
    • one year ago
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    \(a_n=1+(n-1)d\)

  9. anonymous
    • one year ago
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    very good

  10. anonymous
    • one year ago
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    then, we have\[999=(n-1)d\]so \(d\) must be a divisor of \(999\)

  11. mathmath333
    • one year ago
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    yeas

  12. anonymous
    • one year ago
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    ok, how many values \(d\) can take?

  13. anonymous
    • one year ago
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    considering\[999=3^3 \times 37\]

  14. anonymous
    • one year ago
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    @mukushla can u help me pls

  15. anonymous
    • one year ago
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    @mathmath333 can u pls help me with my question

  16. mathmath333
    • one year ago
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    i didnt undestand about the value of d

  17. anonymous
    • one year ago
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    well\[n-1=\frac{999}{d}\]right?

  18. mathmath333
    • one year ago
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    yes

  19. anonymous
    • one year ago
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    since \(n-1\) is an integer so \(\frac{999}{d}\) must be an integer, right?

  20. mathmath333
    • one year ago
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    yes

  21. anonymous
    • one year ago
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    so d must be a divisor of 999

  22. mathmath333
    • one year ago
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    yes

  23. anonymous
    • one year ago
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    ok, how many positive divisors does 999 have?

  24. mathmath333
    • one year ago
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    i forgot how to do it ,do we add the powers of its prime divisors

  25. anonymous
    • one year ago
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    yeah, that right,\[(3+1)(2+1)=8\]which are\[1, 3, 9, 27, 37, 111, 333, 999\]

  26. anonymous
    • one year ago
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    how many of them are acceptable considering conditions of problem?

  27. mathmath333
    • one year ago
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    i m still in doubt how u got this (3+1)(2+1)

  28. anonymous
    • one year ago
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    sry, It's (3+1)(1+1)

  29. mathmath333
    • one year ago
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    ok np

  30. anonymous
    • one year ago
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    final answer will be 7, see if you can get this number as your final answer

  31. mathmath333
    • one year ago
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    ans given is 7 , but how u got thaat

  32. anonymous
    • one year ago
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    well all of values for \(d\) are acceptable, except \(d=1\) which gives a value of \(n=2\), and our arithmatic progressions have at least 3 elements

  33. anonymous
    • one year ago
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    for example when you let \(d=333\) then \(n\) becomes:\[n-1=\frac{999}{333}=3\]\[n=4\]

  34. anonymous
    • one year ago
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    which is acceptable

  35. mathmath333
    • one year ago
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    u checked all 8 options

  36. anonymous
    • one year ago
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    no need to check all of them because it's obvious that all of values give a \(n\) that is acceptable

  37. mathmath333
    • one year ago
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    ohk thnx briliant/bright

  38. anonymous
    • one year ago
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    np, anytime math

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