## 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?

1. mathmath333

\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

r ut yping

3. imqwerty

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

10*9*9 for number ending with 1,3,5,7

5. mathmath333

and 9*9*9 for number ending with 9

6. imqwerty

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

yea for 9 at the end it is 9*9*9 :)

8. imqwerty

but the difficulty is

9. imqwerty

the options :)

10. mathmath333

|dw:1441714551371:dw|

11. imqwerty

no wait we haven't yet counted the case with674

12. mathmath333

see my photu

13. imqwerty

yea LOL :) fat monk

14. mathmath333

15. mathmath333

|dw:1441714695772:dw|

16. imqwerty

:O yes i didn't think abt that so the number of combinations wuld be=9*9*10*3

17. mathmath333

but the answer in book is 3402

18. imqwerty

so where did we go wrng :)

19. mathmath333

did the book allow repetitions

20. imqwerty

hey wait

21. imqwerty

no nvm

22. imqwerty

yes repetitions must be allowed if they were not then why wuld they specially say that 9 comes only once :)

23. mathmath333

ok

24. imqwerty

hey wait

25. imqwerty

|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

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

omg imquerty is a genius again . O: :O

28. imqwerty

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

lol

30. mathmath333

31. imqwerty

on my laptop :) yea

32. mathmath333

i m angry on myself , i cant ssolve counting questions

33. imqwerty

lol i've learnt so much permutation and combination questions haha we haven't started it yet :)

34. imqwerty

u jst don need to do anything except multiply numbers nd think ways ways to adjust :)

35. imqwerty

ahahah looks like hes goin to tumble :)

36. mathmath333

monks dont tumble, sumos tumble

37. imqwerty

:O

38. imqwerty

༼•͟ ͜ •༽

39. mathmate

Hi, This is what I have.

40. imqwerty

hi, i like it what u have :)

41. mathmate

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