Hi All,

There are a couple of different way to conceive of this question, and each has its own 'organization' to it, so you should try to think in whatever terms are easiest for you.

To start, it shouldn't be hard to figure out how many 3-DIGIT numbers will START with 7... 700 to 799 inclusive... so that's 100 appearances of a 7 right there.

Next, you might find it easiest to think about the UNITs DIGIT. Consider the following pattern...
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
etc

Notice how 1 out of every 10 numbers has a unit's digit of 7? That pattern repeats over-and-over. Since we have 1000 total consecutive integers in our list, 1/10 of them will have a unit's digit of 7... (1000)(1/10) = 100 appears of a 7...

Now, think about what you've seen so far... a group of 100 and another group of 100. I wonder what will happen when we deal with the TENS DIGITS...
10 20 30 40 50 60 70 80 90 100
__ 71 72 73 74 75 76 7_ 78 79 80

In the first 100 integers, there are 10 additional 7s in the TENS 'spot' (I removed 70 and the units digit 7 from 77 since I already counted those). There are 10 sets of 100 to consider, so there are (10)(10) = 100 additional 7s...

Total = 100 + 100 + 100 = 300

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

Here's one way to look at it.
Write all of the numbers as 3-digit numbers.
That is, 000, 001, 002, 003, .... 998, 999

NOTE: Yes, I have started at 000 and ended at 999, even though though the question asks us to look at the numbers from 1 to 1000. HOWEVER, notice that 000 and 1000 do not have any 7's so the outcome will be the same.

First, there are 1000 integers from 000 to 999
There are 3 digits in each integer.
So, there is a TOTAL of 3000 individual digit. (since 1000 x 3 = 3000)

Each of the 10 digits is equally represented, so the 7 will account for 1/10 of all digits.

1/10 of 3000 = 300

So, there are 300 0's, 300 1's, 300 2's, 300 3's, . . ., and 300 9's in the integers from 000 to 999

Answer: D

Cheers,
Bren
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Solution using permutation combination:

Digits needed to form numbers: 0,1,2,3,4,5,6,7,8,9 (10 digits)

Case 1: 7 comes in exactly one position:

7 _ _ , _ 7 _ and _ _ 7
For the above three sub cases, the number of possibilities are (1 x 9 x 9, 9 x 1 x 9 and 9 x 9 x 1) which is 81 for each sub case.
Total number of 7s : 3 x 81 = 243
Note: (a) 9 possibilities (0-9, excluding 7)
(b) This covers 2-digit 7-numbers as well since we have included 0.

Case 2: 7 comes in exactly 2 positions:

77_ , 7_7 and _77
For the above sub cases the number of possibilities are: (1x1x9 , 1x9x1 , 9x1x1)
Each sub case has 9 possibilities. But the question asks about the number of 7s so each sub case will have 2x9 = 18 possibilities.
Total number of 7s = 3x18 = 54

Case 3: 7 comes in all places:

777 : 1 possibility ; Total number of 7s=3

Adding 243 +54 +3 = 300

If you like the explanation, then don't forget to give kudos.

Dear Bunuel

According to your solution, 7 as tens digit - 10 time (71, 72, 73, ..., 79);

Should the above starts from 70 not 71? (there are only 9 numbers between 71-79 inclusive)

Thank you