How many times will the digit 7 be written when listing the integers

Thanks Bunuel. This is logical answer by you. When I had seen such a question, I counted all the numbers!! It was funny
Anyhow, thanks for giving reasons for the correct answer.
You are the ROCKSTAR!

excellent approach Bunuel. You are indeed a rockstar.

Wow, I really like Brunel's method.
But I tried to do it in a different way. Not a very intelligent method, but it sure will not take much time.
First do it from 1-100, then look at the bigger picture.

But I still am facing a problem.

In one to hundred,
7
17
27
37
47
57
67
-
87
97
70
:
:
79

Thats 19 * 10 = 190 7s in the tens place and units places in teh first thousand numbers.

Hundreds place:
700
701
:
:
799


Older 190 + hundreds place 100 7s = 290

Where am I missing the 10 others?


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Kudos me if you like this method or post!

in case of counting from 0 to 100 you are missing 77--the count is 2 here for 7, you considered just one 7.
So for all 0 to 1000 you are missing total of 10 such counts.

How many times will the digit 7 be written when listing the integers from 1 to 1000?

110
111
271
300
304

any easy way to do this question?
here is how i did it...
let xyz be a 3-digit#, first, let Z=7, we have 10 options ( 0-9)for x and 10 options for y, then we get 10 x 10 =100, same logic is applied when y =7 and x = 7, we get 300 times.
is it correct?

Thanks Bunuel... But please explain how can i proceed using permutation to solve this one for my understanding...I am able to get a count =280 but not 300 using permutation method.

I made simple thing complex by using permutations here. I got answer but kicked myself after reading simple concept presented by bunuel. I fixed 7 number on specific place and counted how many numbers can be generated with other digits

Single digit = 1
Double digit = 9C1+8C1 + 2 = 19 (7 _ and _7 and 77)
Three digit = 9C1*9C1 + 8C1*9C1 + 8C1*9C1 ( 7 _ _, _ 7 _; _ _ 7)
2( 9C1 + 9C1 + 8C1) (77_; 7_7; _77)
+ 3 (777)
1 + 19 + (81+72+72) + 52 + 3 = 300

First note that we only need consider the numbers 0 to 999. (In other words, we
consider 3-digit numbers which may start with a zero; e.g., 11 = 011.) There are
3 possible places for a 7 to appear. We can proceed by counting the number of times
a 7 appears in each place. So fix a place for a 7 to appear; then for the remaining 2
digits, there are 10^2 possibilities. So the total is 3 · 10^2.


To generalize the case, if you see:
"How many times is the digit X written when listing all numbers from 1 to Y"
The total is (#of digits in Y-1)*10^(#of digits in Y-2)

Not sure what can I add to this... It just means that out of 3000 digits we used to write down first 1000 numbers (000, 001, 002, 003, ..., 999), each digit from 0 to 9 is used equal number of times, how else? Thus we used each digit 3000/10=300 times.

Another approach:
7 in unit's place:
In every 10 consecutive numbers, 7 will appear once. Since we have 1000 numbers, we have 100 sequences of 10 consecutive numbers (1-10, 11-20 etc) so 7 will appear in unit's place 100 times.
7 in ten's place:
In every 100 consecutive numbers, 7 appears in ten's place 10 times (from 70 to 79). We have 10 sequences of 100 consecutive numbers (1-100, 101-200 etc) so we get that 7 will appear in ten's place 10*10 = 100 times.
In every 1000 numbers, 7 will appear in hundred's place 100 times (from 700 to 799).
Total = 100 + 100 + 100 = 300

My method is not as simple, but I think it is easier to understand than the previous methods.

First off for single and double digit numbers, lets assume everyone can figure out that there are 20 7's.

Once we get into the hundreds, we know that the 700s will be tricky, but for all the other hundreds, we should be seeing 20 7's in each set of hundred. Other than the 700s, there are only 8 other sets of hundreds (100s, 200s, 300s, etc.). So we multiply 20 * 8 = 160...bringing the total to 180.

So far we have accounted for all the 7's for single and double digit numbers and all hundreds not including the 700s.

For the 700s, we already know that there is at least one 7 from the numbers 700-799. So there's 100 right there. At this point the only remaining numbers with 7 are the same ones we identified for single and double digit numbers. (20 of em)

So in the end we have...
20+160+100+20 = 300

compared to the above methods, this is a very crude way of doing this problem

7 does not occur in 1000. So we have to count the number of times it appears between 1
and 999. Any number between 1 and 999 can be expressed in the form of xyz where
0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other
9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second
or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and
3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with
the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or
m second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
243 + 54 + 3 = 300

MOST EFFICIENT METHOD USING COMBINATORIX

As Bunuel said, think of each number as XXX ie 001 to 999

Numbers with one 7.

Either 7XX, X7X or XX7, so 3 x 1C9 x 1C9 = 3x9x9 = 243

Numbers with two 7's (nb question asks how many sevens are displayed, so this answer must be doubled)

Either 77X, 7X7 or X77, so 3x 1C9 = 27 (or 54 once you double it)

Numbers with three 7's

Only 1 (777). Triple it as there are three 7,s = 3

Add all together: 243+54+3 = 300

Answer => D

Alternate solution:

Using probability of occurrence = desired outcomes / total outcomes, desired outcomes = probability x total outcomes.

total outcomes = 1000.

probability of occurrence - to simplify this, the question statement could be interpreted as "what is the probability that the digit 7 is picked at least once when 3 digits are chosen at random?"

probability of at least one 7 = 7 in first digit OR 7 in second digit OR 7 in third digit = 1/10 + 1/10 + 1/10 = 3/10

desired outcomes = probability x total outcomes = 3/10 x 1000 = 300.

The first time I did the question, my answer was C : 271 since I thought the question is "how many numbers from 0 - 1000, that contain the 7 digit". So I counted 77, 707,717,....,777,...797, 771, 772, 773,..., or 779 as one.
That may be the reason why 23% of people who answered the question decided to pick answer C.
That is really interesting question

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