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mathmath333

  • one year ago

Sam has forgotten his friend’s seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674, the number is odd, and the number 9 appears once. If Sam were to use a trial and error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{ Sam has forgotten his friend’s seven-digit telephone number.}\hspace{.33em}\\~\\ & \normalsize \text{ He remembers the following: the first three digits are either}\hspace{.33em}\\~\\ & \normalsize \text{ 635 or 674, the number is odd, and the number 9 appears }\hspace{.33em}\\~\\ & \normalsize \text{ once. If Sam were to use a trial and error process to reach }\hspace{.33em}\\~\\ & \normalsize \text{ his friend, what is the minimum number of trials he has to }\hspace{.33em}\\~\\ & \normalsize \text{ make before he can be certain to succeed?}\hspace{.33em}\\~\\ & a.)\ 1000 \hspace{.33em}\\~\\ & b.)\ 2430 \hspace{.33em}\\~\\ & c.)\ 3402 \hspace{.33em}\\~\\ & d.)\ 3006 \hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    r ut yping

  3. imqwerty
    • one year ago
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    he had to make combinations like - 1)number starting with635 |dw:1441713928711:dw| 1st we take the case where the number has 9 at the end and calculate possibilities nd then u take cases with 1,3,5,7 and fix 9 at one of these points-|dw:1441714010300:dw| and calculate the possibilities nd then multiply it by 3 cause there are 3 such places :) so this way u get possibilities in this case :P similarly do it with 674 :)

  4. mathmath333
    • one year ago
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    10*9*9 for number ending with 1,3,5,7

  5. mathmath333
    • one year ago
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    and 9*9*9 for number ending with 9

  6. imqwerty
    • one year ago
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    nope :) 1) u have 4 numbers fixed like this-|dw:1441714329392:dw| so u have only 3 vacant places to be filled so u have total possibilities =9*9*4 and u can place the 9 at two other places too so u get total combinations=9*9*4*3

  7. imqwerty
    • one year ago
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    yea for 9 at the end it is 9*9*9 :)

  8. imqwerty
    • one year ago
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    but the difficulty is

  9. imqwerty
    • one year ago
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    the options :)

  10. mathmath333
    • one year ago
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    |dw:1441714551371:dw|

  11. imqwerty
    • one year ago
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    no wait we haven't yet counted the case with674

  12. mathmath333
    • one year ago
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    see my photu

  13. imqwerty
    • one year ago
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    yea LOL :) fat monk

  14. mathmath333
    • one year ago
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    i m talkinh about this

  15. mathmath333
    • one year ago
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    |dw:1441714695772:dw|

  16. imqwerty
    • one year ago
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    :O yes i didn't think abt that so the number of combinations wuld be=9*9*10*3

  17. mathmath333
    • one year ago
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    but the answer in book is 3402

  18. imqwerty
    • one year ago
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    so where did we go wrng :)

  19. mathmath333
    • one year ago
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    did the book allow repetitions

  20. imqwerty
    • one year ago
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    hey wait

  21. imqwerty
    • one year ago
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    no nvm

  22. imqwerty
    • one year ago
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    yes repetitions must be allowed if they were not then why wuld they specially say that 9 comes only once :)

  23. mathmath333
    • one year ago
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    ok

  24. imqwerty
    • one year ago
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    hey wait

  25. imqwerty
    • one year ago
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    |dw:1441715221286:dw| hehe u haven't fixed the 9 so it can be like the 9 doen't even appears so this is wrong :)

  26. imqwerty
    • one year ago
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    what i did earlier ws correct u knw that 9 comes once nd lets say its not at the end so u have 3 places for it nd then u calculate like this- |dw:1441715357131:dw| so u get 9*9*4 cases similarly u can move it to 3 alt places so u have a total of 9*9*4*3 combinations :)

  27. mathmath333
    • one year ago
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    omg imquerty is a genius again . O: :O

  28. imqwerty
    • one year ago
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    nd then if 9 is at the end u have 9*9*9 combinations total combinations =9*9*9 + 9*9*4*3 =1701 and u multiply it by 2 cause u also get same number of combinations for 674 so answer is 1701x2 =3402

  29. imqwerty
    • one year ago
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    lol

  30. mathmath333
    • one year ago
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    do u have querty keypad

  31. imqwerty
    • one year ago
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    on my laptop :) yea

  32. mathmath333
    • one year ago
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    i m angry on myself , i cant ssolve counting questions

  33. imqwerty
    • one year ago
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    lol i've learnt so much permutation and combination questions haha we haven't started it yet :)

  34. imqwerty
    • one year ago
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    u jst don need to do anything except multiply numbers nd think ways ways to adjust :)

  35. imqwerty
    • one year ago
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    ahahah looks like hes goin to tumble :)

  36. mathmath333
    • one year ago
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    monks dont tumble, sumos tumble

  37. imqwerty
    • one year ago
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    :O

  38. imqwerty
    • one year ago
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    ༼•͟ ͜ •༽

  39. mathmate
    • one year ago
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    Hi, This is what I have.

  40. imqwerty
    • one year ago
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    hi, i like it what u have :)

  41. mathmate
    • one year ago
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    Considering only the last four digits Pattern(XXX9)=9^3=729 ways Patterns(9XX, X9X, XX9 {1,3,5,7})=3*4*9^2=972 Total for last 4 digits = 729+972=1701 Multiply by two prefixes, 2*1701=3402

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