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by tohellandback » Wed Aug 12, 2009 7:26 pm
numbers from 001 to 999
7 can appear as XX7,X7X,7XX
where X is any value from 0-9
XX7- 10*10 ways=100
same for X7X and 7XX.
total 300 ways

answer D
The powers of two are bloody impolite!!
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by ashis979 » Wed Aug 12, 2009 7:40 pm
I'd go with D - 300 times.

- For the numbers in which the digit 7 occurs only once such as 7, 17, 27...
With the exception of the number 7, for the other numbers, the remaining digits can be any of the digits from 0-9, except 7. So there are 81 (1*9*9) such numbers. But since 7 can appear as the first, second, or third digit, there will be 81*3=243 numbers in which the digit 7 occurs only once.

- For the numbers in which the digit 7 occurs twice such as 77, 771, 773...
For these numbers, there is one digit that is not 7, and the digits can be any of digits from 0-9, except 7. So there are 9 (1*9) such numbers. But since 7 can appear as the first, second, or third digit, there will be 9*3=27 numbers in which the digit seven occurs twice. Therefore 27*2=54 occurences of the digit 7.

- There is only one number between 1-1,000 that has three 7's, the number 777. So, that's 3 occurences of the digit 7.

Therefore, total number of occurence of the digit 7=> 243+54+3=300.
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by Sher1 » Thu Aug 13, 2009 2:13 pm
Ashish can you clarify.
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by ashis979 » Thu Aug 13, 2009 4:40 pm
Sher 1,

Apologies the first explanation wasn't clear. I guess the best way to conceptualize/visualize this problem is to think of all the numbers as being three digits. There are no 7's in 1,000 so we can ignore it.

7 can the represented as 007, 70 as 070, 17 as 017, and so on...

So, for the numbers that will have one occurence of of the digit 7:
7XX, X7X, or XX7. If X was 0 the numbers presented are 700, 070 (70), and 007 (7). So let's peg 7 in the first position, that means there are 9 ways to pick the second digit (only 9 ways because you cannot pick 7=> remember, we only want numbers with one occurence of the digit 7. So your choices are 0,1,2,3,4,5,6,8,9). Similarly, there are 9 ways to pick the third digit. Therefore, with 7 pegged in the first position, you can pick the remaining two digits in 9*9=81 ways. That is a total of 81 numbers where the digit 7 will occur once. BUT, since that 7 we pegged can be in the first position, the second position, and the third position, the total number of possibilities is in fact 3*9*9=243. I hope this is making sense.

For the numbers that will have two occurences of of the digit 7:
77X, X77, 7X7. If X was 0 the numbers presented are 770, 077 (77), and 707. As I did above, if you peg the positions of the two 7's you will have 9 ways to pick the third digit (again, here also just the numbers 0,1,2,3,4,5,6,8,9 because we only want numbers where the digit 7 occurs twice). So that means 9 different numbers. BUT since the third digit can be in the first position, the second position, and the third position, the total number of possibilities is in fact 3*9=27. Now, don't forget the last step. All of these 27 numbers have 2 7's in them. Therefore, the total number of occurences is actually 2*27=54.

And for the last case where we need a number between 1-1,000 where there are three 7's, there is only one number: 777. So, that is a total of 3 occurences of the digit 7.

Therefore, putting it all together, the total number of occurences (that's what the question is asking for, occurences NOT numbers) = 243+54+3=300.

I hope this makes sense now. Let me know if still unclear.
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by ghacker » Thu Aug 13, 2009 7:05 pm
1 to 1000 How many 7s

7 once : 9*9*3 =243

7 Twice : 3*9*2 =54

7 Trice : 1 *3


Hence total = 300
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by conquistador » Tue Sep 29, 2015 9:46 am
ashis979 wrote:Sher 1,

Apologies the first explanation wasn't clear. I guess the best way to conceptualize/visualize this problem is to think of all the numbers as being three digits. There are no 7's in 1,000 so we can ignore it.

7 can the represented as 007, 70 as 070, 17 as 017, and so on...

So, for the numbers that will have one occurence of of the digit 7:
7XX, X7X, or XX7. If X was 0 the numbers presented are 700, 070 (70), and 007 (7). So let's peg 7 in the first position, that means there are 9 ways to pick the second digit (only 9 ways because you cannot pick 7=> remember, we only want numbers with one occurence of the digit 7. So your choices are 0,1,2,3,4,5,6,8,9). Similarly, there are 9 ways to pick the third digit. Therefore, with 7 pegged in the first position, you can pick the remaining two digits in 9*9=81 ways. That is a total of 81 numbers where the digit 7 will occur once. BUT, since that 7 we pegged can be in the first position, the second position, and the third position, the total number of possibilities is in fact 3*9*9=243. I hope this is making sense.

For the numbers that will have two occurences of of the digit 7:
77X, X77, 7X7. If X was 0 the numbers presented are 770, 077 (77), and 707. As I did above, if you peg the positions of the two 7's you will have 9 ways to pick the third digit (again, here also just the numbers 0,1,2,3,4,5,6,8,9 because we only want numbers where the digit 7 occurs twice). So that means 9 different numbers. BUT since the third digit can be in the first position, the second position, and the third position, the total number of possibilities is in fact 3*9=27. Now, don't forget the last step. All of these 27 numbers have 2 7's in them. Therefore, the total number of occurences is actually 2*27=54.

And for the last case where we need a number between 1-1,000 where there are three 7's, there is only one number: 777. So, that is a total of 3 occurences of the digit 7.

Therefore, putting it all together, the total number of occurences (that's what the question is asking for, occurences NOT numbers) = 243+54+3=300.

I hope this makes sense now. Let me know if still unclear.
I don't understand the boldened part.
Even if there are two sevens do we get two different numbers by inter changing them.
If not then what is the point of multiplying them with number 2.

can someone explain?
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by DavidG@VeritasPrep » Tue Sep 29, 2015 9:59 am
I don't understand the boldened part.
Even if there are two sevens do we get two different numbers by inter changing them.
If not then what is the point of multiplying them with number 2.

can someone explain?
The question is asking how many 7's would appear if you'd written out all the integers from 1 to 1000. (It's not asking for the number of integers that contain a 7.)

Take a simpler example. Say we just wanted to count the number of 7's contained in the range from 770 to 774 inclusive. Clearly there are 5 integers in the range - 770, 771, 772, 773, 774 - but each of these integers contains two 7's, so if we wanted the number of 7's contained in this range, we'd have to multiply the number of integers, 5, by 2.
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by conquistador » Tue Sep 29, 2015 10:42 am
DavidG@VeritasPrep wrote:
I don't understand the boldened part.
Even if there are two sevens do we get two different numbers by inter changing them.
If not then what is the point of multiplying them with number 2.

can someone explain?
The question is asking how many 7's would appear if you'd written out all the integers from 1 to 1000. (It's not asking for the number of integers that contain a 7.)

Take a simpler example. Say we just wanted to count the number of 7's contained in the range from 770 to 774 inclusive. Clearly there are 5 integers in the range - 770, 771, 772, 773, 774 - but each of these integers contains two 7's, so if we wanted the number of 7's contained in this range, we'd have to multiply the number of integers, 5, by 2.
thanks David.
It is very clear now after re-reading the question.
I missed the very essence of the question while solving it..............Ironic :wink:
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by GMATGuruNY » Tue Sep 29, 2015 11:21 am
Sher1 wrote:How many times will the digit 7 be written when listing the integers from 1 to 1000?


A 110
B 111
C 271
D 300
E 304
To make the math easier, consider the following set:
000, 001, 002...997, 998, 999.

Total number of values in this set = 1000.
Since each value is composed of 3 digits, the total number of digits = 3*1000 = 3000.

Of the 10 digits 0-9, each will appear the SAME NUMBER OF TIMES.
Thus, 7 will constitute 1/10 of the 3000 digits:
(1/10)(3000) = 300.

The correct answer is D.
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by conquistador » Tue Sep 29, 2015 12:18 pm
Of the 10 digits 0-9, each will appear the SAME NUMBER OF TIMES.
Why did u assume that each will appear the same number of times within 999.
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by GMATGuruNY » Wed Sep 30, 2015 4:02 am
Mechmeera wrote:
Of the 10 digits 0-9, each will appear the SAME NUMBER OF TIMES.
Why did u assume that each will appear the same number of times within 999.
Consider "1".
In every set of 10 integers (0-9, 10-19, 20-29), "1" will appear in the units place exactly once.
In every set of 100 integers (0-99, 100-199, 200-299), "1" will appear in the tens place exactly 10 times.
In every set of 1000 integers (0-999, 1000-1999, 2000-2999), "1" will appear in the hundreds place exactly 100 times.

The same reasoning can be applied to EVERY DIGIT.
Implication:
If we list the values from 000 to 999, inclusive, every digit will appear the SAME NUMBER OF TIMES.
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by Brent@GMATPrepNow » Sat Oct 19, 2019 1:45 pm
Sher1 wrote:How many times will the digit 7 be written when listing the integers from 1 to 1000?
A 110
B 111
C 271
D 300
E 304
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: B

Cheers,
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