   SEARCH HOME Math Central Quandaries & Queries  Question from Elmer, a parent: I have part of a phone number. the first 3 digits are 212, the last digit is four, and the three missing digits are all odd, how many numbers will I have to call to get the right number? 212- _ _ _ 4? We have two responses for you

Elmer,

If you are lucky only one, but I think you want the number of possible phone numbers that meet your requirements.

The remaining three digits are odd so they are 1, 3, 5, 7 or 9, that is there are 5 choices for each missing digit. Suppose you start with the first digit after the -. You can fill in this blank in 5 ways

212- 1 _ _ 4
212- 3 _ _ 4
212- 5 _ _ 4
212- 7 _ _ 4
212- 9 _ _ 4

Each of the five possibilities listed above can be extended in 5 ways by inserting an odd digit in the next place.

212- 1 1 _ 4
212- 1 3 _ 4
212- 1 5 _ 4
212- 1 7 _ 4
212- 1 9 _ 4

212- 3 1 _ 4
212- 3 3 _ 4
and so on.

This gives a list of 5 × 5 = 25 possibilities with two of the missing three digits filled in.

How many possibilities are there when filling in all three missing digits?

Penny

Hi Elmer. The digits you know have nothing to do with the digits you don't know. So you simply need to think of how many three digit numbers can you make with only odd digits.

There are 5 odd digits: 1, 3, 5, 7, 9. If you select one for the first digit, there are the same five to choose for the second (so for every choice of the first digit, there are five choices for the second). Likewise, there are 5 choices for the third digit for every possible two-digit prefix.

So you simply need to multiply the number of choices for each digit together !

Hope this helps,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.