Looking for exactly two of the 6-digits

First Hundred   Second Hundred     Etc.,       Etc., ........
6                 106
16                116
26                126
36                  '
46                  '
56                  ', 166, 160
66                  '
76                  '
86                  '
96                196
======           =========

1 of them        2 of them           ..         ...

Consider all integer numbers from 0 to 999, inclusive.


In all, there are exactly 1000 such numbers.


Will consider one-digit numbers, like 3, 7 as three digit numbers with leading zeroes  003, 007.

Will consider two digit numbers, like 37 as three digit numbers with leading zero 037.

It will change NOTHING in the solution.



All such three-digit numbers, that have only two digits "6", are in the following three disjoint categories:

    - having "6" in the first  and the second positions, only;

    - having "6" in the first  and the third  positions, only;

    - having "6" in the second and the third positions, only.



Now, let's consider all three-digit numbers with the digits 6 in "hundreds" and "tens" position.

    The amount of such numbers is 9, obviously, because there are 9 such thre-digit numbers that have 9 possible digits 
    (all excepting 6)  in the "ones" position.



Next, consider all three-digit numbers with the digits 6 in "hundreds" and "ones" position.

    The amount of such numbers is 9, obviously, because there are 9 such thre-digit numbers that have 9 possible digits 
    (all excepting 6) in the "tens" position.



Finally, consider all three-digit numbers with the digits 6 in "tens" and "ones" position.

    The amount of such numbers is 9, obviously, because there are 9 such thre-digit numbers that have 9 possible digits 
    (all excepting 6) in the "ones" position.


ANSWER.  In all, there are 9+9+9 = 27 numbers from 0 to 1000, having only two digits "6" in their records.

I got 28 numbers with EXACTLY 2 "6s."

Number of integers BETWEEN 0 and 1,000, non-inclusive (this is what BETEWEEN means) is: 999 

Therefore this should be  

Answer by greenestamps(13216)   (Show Source): You can put this solution on YOUR website!
 


(1) True, there is no algebraic method for solving this problem.


(2) But just counting them is not a very efficient method.  A formal mathematical solution does not require an algebraic solution; logical analysis is perfectly good mathematics, and it can make this problem easy.


(3) There is nothing particularly brilliant or elegant, as one tutor called it, about the method for solving the problem; only a very little logical analysis is required.


(4) A response simply saying "I got 28 for the answer", without any justification, is of no use to the student -- especially since that answer is wrong.


(5) Finally, asking for the "fraction of the integers between 0 and 1000" makes it unclear what the answer should be, because of the ambiguity of the meaning of "between 0 and 1000"; we can't be sure whether the denominator is 999 or 1000, or even 1001.  To make the problem (and expected answer) clear, the problem should simply ask how many numbers between 0 and 1000 have exactly two 6's.


The relatively simple solution:


(1) integers of the form 66A, where A is not 6: 9 choices for A (any digit except 6)

(2) integers of the form 6A6, where A is not 6: 9 choices for A (any digit except 6)

(3) integers of the form A66, where A is not 6: 9 choices for A (any digit except 6)


Note for the last case we can have 0 as the leading digit, representing the integer 66, which is between 0 and 1000.


ANSWER: There are 27 integers between 0 and 1000 that contain exactly two 6's.




 

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