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  • KMC 2015 Cadet No. 24

    Anonymous Christian
    Apr 19, 2016
    KMC, MATH, Math Olympiad
    Comments Off on KMC 2015 Cadet No. 24

     

    Question:

    Yesterday I wrote down my friend Ekin’s telephone number. The telephone number on my note has six digits, but I remember that Ekin said that the number had seven digits. I have no idea what digit I forgot to write down, or its position in the number. How many different telephone numbers do I have to try to be sure that I use the correct one? (note that a telephone number may start with any digit, including 0).

    Solution

    Please see picture. Note that a common “wrong” answer is “70”. The “correct” answer is “64” as explained in the picture.

    The idea to solve it is like this:

    Let the six digit telephone number on your notes be ABCDEF (in that order).

    Now, the “seventh missing digit” may be placed as follows:

    (i) To the “left of A”, any one of the 10 digits (0 to 9) may be placed.

    Or

    (ii) In a space in between “A and B”, any digit “except digit A” may be placed, meaning 9 digits may be placed here.

    This is because placing “digit A” on the left of A as in case (i) or placing it in between A and B both produces the “same” 7-digit number AABCDEF.

    Hence, to avoid “double counting or over counting”, it is equivalent to require “another digit A” to “not” be placed “between the original digits A and B”.

    (iii) Likewise we have “9 possible digits” for placement “between the original digit B and C” (excluding “another” digit B here from 0 to 9).

    Or

    (iv) Next, we have “9 possible digits” for placement “between the original digit C and D” (excluding “another” digit C here from 0 to 9).

    Or

    (v) In like manner, we have “9 possible digits” for placement “between the original digit D and E” (excluding “another” digit D here from 0 to 9).

    Or

    (vi) Also we have “9 possible digits” for placement “between the original digit E and F” (excluding “another” digit E here from 0 to 9).

    Or

    (vii) Lastly, we have “9 possible digits” for placement “after the original digit F” (excluding “another” digit F here from 0 to 9).

    So, by the “addition principle” (since the cases above are all separately considered by an “or”), we count:

    The total of such numbers
    = 10 + 9 + 9 + 9 + 9 + 9 + 9
    = 10 + (9 x 6)
    = 10 + 54
    = 64 (Answer)

    DSC_0001

anonymous-christian

Jonathan Ramachandran is founder at AnonymousChristian.org which serves as a Christian Blog discussing the Greater Hope of Biblical Christian in Christ Alone. He is also an IMO (International Math Coach) and a featuring Guitarist for many bands.

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