A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

how many 3-digit whole #s can be formed from the digits 3-9 if: the result number is a multiple of 3?

  • This Question is Closed
  1. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    good one

  2. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what you first need to do is to know how a number is can be a multiple of 3

  3. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if the numbers digits add up to a multiple of 3, then we know that the number itself is divisible by 3

  4. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for example, 222 is divisible by 3 because 2+2+2 =6

  5. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so you are going to use the numbers 3,4,5,6,7,8,9 in order to find out all combinations of the sum of the digits that are a multiple of 3

  6. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I would suggest this strategy

  7. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if the exactly one of the digits is a multiple of 3, namely 3,6,9 the other two has to add up to a multiple of 3, so the only pair that does that is (4,5) (5,4) because you are not allowed to use other multiples of 3s. so you have 3,4,5 and all the permutations of that, 6 of them 6,4,5 and it's permutations 6 of them 9,4,5 and its permutations 6 of them

  8. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so if you have exactly one number with a multiple of 3, you have exactly 18 numbers

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i found 7 combinations 3,4,5/3,4,8/4,5,6/4,5,9/5,6,7/6,7,8/7,8,9 but the order doesn't matter because the sum of the digits will be the same

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so would it be 7*6 as the answer?

  11. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now you will think about having exactly two digits that are multiple of 3, If you see this carefully, the other digit MUST be a multiple of 3 as well, so there are no 3 digit number with exactly two of them being a multiple of 3

  12. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now we think of a way to count how many 3 digit numbers are divisible by 3 when it has exactly 3 digits that are multiple of 3s

  13. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this one is tough to count because there is no easy formula for it

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If the digits you choose total 3, 6 or 9 then they are divisible by 3. Start with 300. You now just add 3 each time... 300, 303, 306, 309, ( but you cannot have 312, 315, etc because digits 1-2 cannot be used so next numbers are...) 330. 333, 336, 339, What have you found so far? 300, 303, 306, 309, 330. 333, 336, 339, It is tempting to look for a pttern already. You will see the numbers end, 0,3,6,9 but this could be a trap! Your next choise of starter digits should systematically be 34... but they must total 3, 6 or 9 so that means 34 2 can be the only one but it is illeagal as 2 is not allowed, so move on to 35... 352 (not leagal) 354, (allowed as 3+5+4+= 12 and 1+2=3) 357 also allowed now we have 300, 303, 306, 309, 330. 333, 336, 339, 354, 357,.. te next numbers using this approach yield 360, 363, 366, 369, 375, 378, 384, 387 390, 393, 396, 399. Put them all together and start patern spotting 300, 303, 306, 309, 330. 333, 336, 339, 354, 357,360, 363, 366, 369, 375, 378, 384, 387 390, 393, 396, 399. End digits are 0, 3, 6, 9, 0, 3, 6, 9, 4, 7, 0, 3, 6, 9, 5, 8, 4, 7, 0, 3, 6, 9 This represents all possible 300 type numbers, do the same for the 400s and 500s and see if you can spot a useful pattern.

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    there are 7x7x7 possible 3 digit numbers from that list 3-9, of those subtract how many dont add up to a muiltiple of 3

  16. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    gian franco, 0 is not allowed to use

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you guys have a final answer

  18. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    anyway, now let's count the exactly 3 digits being divisible by 3

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im going to do it this way 7x7x - ( how many with at least one 7, one 4,,5,8

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    because (3,3,7) and permutations thereof dont add up to multiply of 3, (3,7,7) doent add up, etc

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so this seems quicker

  22. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    333, 666,999 3 ways 3,3,6 can be ordered in 3 ways as well as 339,663,669,993,996 so 18 ways in total 3,6,9 can be ordered in 6 ways, so we have 27 ways in total

  23. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now we need to think if there is any 3 digit number that has no multiple of 3s in its digits.

  24. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is the toughest, you need to count them carefully for example, 555 is a multiple of 3 because the sum is 3 * 5 so using these types of digits will be 444,555,777,888 4 of them

  25. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm trying to see if there is anything else...

  26. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    we have 3,2^2, 5, 7, 2^3, 3^2

  27. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't think there is anything else

  28. Yuki
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so the total is,,, 4 + 27 + 18 = 59 of them

  29. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow that was a ton of math.... i had no idea it was that hard. thanks so much yuki could you explain why you went with the strategy of how many possibilities there were based on the number of digits divisble by 3?

  30. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yuki, theres a mistake

  31. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you said if you have one of 3,6,9, the only other numbers that add up to a multipl of 3 is 4,5. what about 5,7

  32. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    3+5+7 = 15

  33. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yuki?

  34. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh that's true!

  35. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and 6 + 5 + 7 = 18 a multiple of three

  36. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    9 + 5 + 7 = 21, a multiple of 3

  37. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so we'd have to take into account (4,8) and (5,7) solutions?

  38. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  39. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it does seem to be the case that you need at least one multiple of 3 to get a sum of three, like 4,5,7 doesnt work, etc. but 3,5,7 does

  40. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh wait

  41. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well the 444 and etc

  42. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right, im wrong

  43. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but the negation case seems to be easier to count, find the number of cases which dont add up to a multiple of 3, subtract by the total 7x7x7

  44. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how would one do that then?

  45. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so lets look at this way you have the numbers 3,4,5,6,7,8,9 , correct

  46. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you get how i have total cases , 7x7x7 , right ?

  47. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i want to find the number of NON multiples of 3, subtract from the total, will give you multiples of 3

  48. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  49. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    youre cases are 1*3, 2*3, 3*3 1*4, 2*4,3*3 1*5,2*5, 3*5 1*7, 2*7,3*7 1*8,2*8,3*8 1*9,2*9,3*9

  50. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you can have 1*3 + 2*9 , for instance

  51. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the coefficients here have to add to 3 ,

  52. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1*3, 2*3, 3*3 1*4, 2*4,3*4 1*5,2*5, 3*5 1*6, 2*6,3*6 1*7,2*7,3*7 1*8,2*8,3*8 1*9,2*9,3*9

  53. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so far?

  54. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm i'm not understanding the multiplication tables at the top

  55. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so clearly, , well i get 3*9 from (9,9,9)

  56. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and we are adding them

  57. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhhhhhh OK

  58. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so if you randomly picked 1*3, and say 2*7, you get (3,7,7)

  59. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah i get it now XD

  60. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so automatically you rule out all the 3's cases, and you rule out 6's, 9's

  61. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i take that back

  62. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you rule out the last column, of 3*3, 3*4, 3*5,... 3*9

  63. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  64. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    only the last column right? so it becomes 1*3, 2*3, 3*3 1*4, 2*4, 1*5,2*5, 1*7, 2*7 1*8,2*8, 1*9,2*9,

  65. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right those are left

  66. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    alright

  67. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait, you left in 3x3, that goes away

  68. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1*3, 2*3, 1*4, 2*4 1*5,2*5, 1*7, 2*7, 1*8,2*8 1*9,2*9

  69. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so we have left 1*3, 2*3, 1*4, 2*4, 1*5,2*5, 1*6, 2*6 1*7, 2*7 1*8,2*8, 1*9,2*9,

  70. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right :)

  71. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now start with 1*3 in the first column, and how many can it go with in the second column, so that its NOT a multiple of 3

  72. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    clearly 1*3 , 2*3 is rule out, since thats just 3,3,3

  73. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1*4, 2*4 1*5,2*5, 1*7, 2*7, 1*8,2*8 1*9,2*9

  74. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh wait no cause 1*3 can go with others 1*4, 2*4 1*5,2*5, 1*7, 2*7, 1*8,2*8 1*9,2*9

  75. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    not the 9's or the 6's

  76. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh good point

  77. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1*3, 2*3, 1*4, 2*4, 1*5,2*5, 1*6, 2*6 1*7, 2*7 1*8,2*8, 1*9,2*9,

  78. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i see, so there is the case of one 3, one 4 , one 5, etc

  79. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah cause 1*3,2*4 works

  80. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so would we have to find the individual sums for all of the outcomes for each 1*{3,4,5,6,7,8,9}

  81. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right ,

  82. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    theres also a strange case 4+8+9 = 21

  83. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so lets rethink here. we have 1*3, 2*3, 1*4, 2*4, 1*5,2*5, 1*6, 2*6 1*7, 2*7 1*8,2*8, 1*9,2*9, the cases of ones we can do seperately 3,4,5 3,4,6 3 , 4 , 7 3, 4 8 3 , 4 , 9

  84. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    4,5,7 4,5,8, 5,6,8 5,6,9

  85. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    6,7,9

  86. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    3,4,6 3,4,7 3,4,9

  87. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right , so the ones cases you can do by hand

  88. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    those are all the ones with non multiples of 3 so 8 * 6 = 48 ways that aren't possible

  89. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and then you have permutations of them

  90. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  91. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ill double check, but that looks right

  92. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so for the one and 2's cases i have 1*3 - ( 2*4, 2*5, 2*7, 2*8, )

  93. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for the one and twos cases. so there are 3 possibilities. either all of them are the same (3 of em) 2 of them are the same, and 1 different. or all of them are different

  94. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and you systematically eliminate

  95. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    4 + (2*3, 2*7, 2*6, 2*9)

  96. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    5 + (2*3, 2*4, 2*6, 2*7, 2*9)

  97. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    6 + (2*4, 2*5, 2*7, 2*8)

  98. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    7 + (2*3, 2*5, 2*6, 2*8, 2*9)

  99. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    8 + (2*3, 2*4, 2*6, 2*7, 2*9)

  100. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    9 + (2*4, 2*5, 2*7, 2*8)

  101. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    looks good

  102. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so then you'd take the permutations of the singles + the permutations of the doubles and subtract from 7 *7*7?

  103. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  104. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do you have the answer in the book,

  105. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that would be nice to check

  106. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    even # :(

  107. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.