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Problem Of The Week 35

with 8 comments

How many 3-d igit numbers are such that one of the digits is the average of the other two?

(A)   96   (B) 112   (C) 120   (D) 104   (E) 256


Written by Implex

September 28, 2008 at 4:44 pm

8 Responses

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  1. 96

    (1,2,3),(2,3,4)….(7,8,9) 7 triplets
    (1,3,5),(2,4,6)…(5,7,9) 5 triplets
    (1,4,7),(2,5,8),(3,6,9) 3 triplets
    (1,5,8) 1 triplet

    Total possible triplets =16

    Total possible no.s =16* 3! =96


    September 29, 2008 at 8:01 am

  2. Zero has not been considered above. So when you consider zero, then one of the other number should be twice of another. there are 4 possibilities and each has 4 ways..so adding 16 the answer is 96+16=112. Answer choice B)

    Celebrating Life

    September 29, 2008 at 10:16 am

  3. nice way…
    consider the terms in ap and arrange 🙂


    September 29, 2008 at 11:09 am

  4. @Celebrating Life

    the case for numbers like 111,222,333………999
    has not been considered…
    9 cases more are added thus my answer appears to be(96+16+9)=121
    would anyone like to correct me?


    October 1, 2008 at 3:44 am

  5. Yeah @wolverine
    your solution counts for those aps when terms are equal
    but this is an old amc problem
    and they did not give option for counting with that approach

    so the option 121 is not there
    answer is right

    so as the term distinct is not mentioned
    we should also count for there 9!

    Good Job


    October 1, 2008 at 11:17 am

  6. then i must say it was a flawed question…!!!


    October 1, 2008 at 5:19 pm

  7. yupp, it was ambiguity is generally avoided by amc


    October 1, 2008 at 6:22 pm

  8. 16*3!+9+4*(3!-2!)=121


    October 2, 2008 at 6:28 pm

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