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)