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anonymous

  • 5 years ago

can anyone solve these problems??? M,A,M,J,J,A,S,O,..........What comes next? a.N b.M c.A d.P

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  1. anonymous
    • 5 years ago
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    sathi, you posted a wrong problem, whats wrong with you,

  2. anonymous
    • 5 years ago
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    the sequence is 3, 7, 31, 211

  3. anonymous
    • 5 years ago
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    NOT 3, 7, 32, 211,

  4. anonymous
    • 5 years ago
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    so how did u prove that 3 , 7 31, 211

  5. anonymous
    • 5 years ago
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    but the question was same in the exam paper. 31 not 32

  6. anonymous
    • 5 years ago
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    but anyway how did u solve that ..

  7. anonymous
    • 5 years ago
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    you posted 32 , scroll to your question

  8. anonymous
    • 5 years ago
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    look at the problem you posted earlier you will find 32

  9. anonymous
    • 5 years ago
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    oh... i mean 32 not 31

  10. anonymous
    • 5 years ago
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    whats ur process

  11. anonymous
    • 5 years ago
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    no, wait, admit that you posted an incorrect problem

  12. anonymous
    • 5 years ago
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    ok...

  13. anonymous
    • 5 years ago
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    look at the post , on the left, scroll down

  14. anonymous
    • 5 years ago
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    follow me , one sec

  15. anonymous
    • 5 years ago
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    wow i got answer

  16. anonymous
    • 5 years ago
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    ok so the sequence 3, 7, 31, 211, is 1 + 2, 1 + 2 * 3 , 1 + 2 * 3 * 5 , 1 + 2*3*5*7

  17. anonymous
    • 5 years ago
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    its option no.A...N

  18. anonymous
    • 5 years ago
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    wait, lets do the sequence

  19. anonymous
    • 5 years ago
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    what is special about 2,3,5,7

  20. anonymous
    • 5 years ago
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    they are primes

  21. anonymous
    • 5 years ago
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    so the next number will be 1 + 2*3*5*7*11 = 2311

  22. anonymous
    • 5 years ago
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    2311, 30031, 510511

  23. anonymous
    • 5 years ago
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    yea i agree with ur idea...thanks

  24. anonymous
    • 5 years ago
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    how did you solve your MAMJAN problem?

  25. anonymous
    • 5 years ago
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    its the series of month starting from M(march)

  26. anonymous
    • 5 years ago
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    is that a code, letters mean numbers

  27. anonymous
    • 5 years ago
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    haha

  28. anonymous
    • 5 years ago
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    nice one

  29. anonymous
    • 5 years ago
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    X % of Y is Y % of : a.x b.y/100 c.x/100 d.100x

  30. anonymous
    • 5 years ago
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    wait, this is an open question, the sequence before are called euclid numbers

  31. anonymous
    • 5 years ago
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    are there an an infinite number of prime euclid numbers, E6 is not prime for example

  32. anonymous
    • 5 years ago
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    hey man m the student of medicine so i dont know the mathmatical calculation much more......so would u plz describe in simple way.

  33. anonymous
    • 5 years ago
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    my guess was a x

  34. anonymous
    • 5 years ago
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    yes a

  35. anonymous
    • 5 years ago
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    x is correct

  36. anonymous
    • 5 years ago
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    but its c in book

  37. anonymous
    • 5 years ago
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    x% of y = x/ 100 * y = x * y/100 = y% of x

  38. anonymous
    • 5 years ago
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    oh , hmmmm

  39. anonymous
    • 5 years ago
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    well another question.

  40. anonymous
    • 5 years ago
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    in a class with a certain number of students , if one student weighing 50kg is added then average weight of class incereases by 1 kg ,if one more student weighin 50kg is added then the average weight of class increasses by 1.5kg over the original average? what isthe original average weight (in kg) of the class?

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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