The digit 7 appears in how many of the first thousand positive integer

Hi Yashika, the question asks 7 appears in how many positive integers. Which means in 777, you have 1 occurrence not 3. Does this make sense?

** Just speaking from personal experience, Bunuel is right about 99.99% of the time!

[quote="AnirudhaS"][quote="yashikaaggarwal"]7 appearing only in unit digit till 100 = 9 times (7,17,27,37,47,57,67,87,97)

7 appearing only in tens digit is = 9 times (70,71,72,73,74,75,76,78,79)

7 appearing in both units as well as tens digit is = 1 (77)

Total 1+9+9=19 times till 100
19*10 till 1000 = 190 times.
Add: 100 times of when 7 appeared as hundreds digit (from 700 to 799)

190+100 =290

Bunuel Sir kindly check the options. Thank You.

[size=80][b][i]Posted from my mobile device[/i][/b][/size][/quote]
Hi Yashika, the question asks 7 appears in how many positive integers. Which means in 777, you have 1 occurrence not 3. Does this make sense?

** Just speaking from personal experience, [url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=Bunuel][b]Bunuel[/b][/url] is right about 99.99% of the time![/quote]

I personally love to solve Questions post by Sir, I am not saying he is wrong. Just put forward my doubt. As the question says how many times. So 777 will be considered as 3 times 7. As per mine knowledge.

I think the question means something else. It does not say "how many times..." It says "...in how many ... integers?"

Got It, the digits in 700 category will create redundancy. Thank You sir.

7 will appear in 19 integers from 0-99 and similarly will appear in 100-199 ... 900-999 except for the range 700-799 (all will include digit 7)

Thus total integers = 19*9 + 100 = 171+100 = 271

Answer - E

how many of the first thousand positive integers = 1- 1000
Q: The digit 7 appears in how many of the first thousand positive integers?
to find : numbers of integers from 1-1000, that have atleast 1 digit as 7.

lets take 1-100
7, 17, 27, 37, 47, 57, 67, 70-79, 87, 97 = 19

100-200
107, 117, 127, 137, 147, 157, 167, 170-179, 187, 197 = 19
similarly, 200-300, 300-400, 400-500, 500-600, 600-699, 800-900, 900-1000 will each have 19 integers.


700-799
701, 702, 703......, 798, 799 =100

total
1-100 = 19
101-699= 6*19
800-1000= 2*19
700-799= 100
====> 19+ 6*19+ 2*19+100= 19*9 +100= 171+100=271.

solve using combinatrics

target is find digit 7 from 1- 1000
so from 1-10 ; 1
from 10-99; ( 8*9) ; 72 ; 90 is total 2 digits ; so 90-72 ; 18
and from 100-999; ( 8*9*9) ; 648 ; total 3 digits are ( 999-100+1= 900) so with 7 digit ; 900-648 ; 252

total 7 digit appears ; 252+18+1 ; 271
OPTION E

Note : This question is basically asking us to find the total number of integers between 1-1000 which has atleast one '7' as a digit. This question is NOT asking us to count the total number of 7s in the integers between 1 -1000.


There are 19 integers between 1-99 which has '7' as a digit in it :

7, 17,27,37,47,57,67,70,71,72,73,74,75,76,77,78,79,87,97

Consequently, 100-199,200-299,300-399,400-499,500-599,600-699 has 19*6 integers in each case which has the digit 7 in it.

700-799 has 100 integers with 7s in it.

Subsequently, numbers 800-899 and numbers 900-999 has a total of 19*2 integers which has a 7 in it.

Therefore, total number of integers between 1-1000 which has at least one 7 in it is 19*9 + 100 = 271

Answer : E

A more easy approach would be to fix the position of 7

1) 7 fixed at hundreds place would give : 7_ _ : 100 possible numbers

2) 7 Fixed at tens place would give : _ 7 _ : 10(numbers that can take ones place) x 9 (numbers that can take hundreds place excluding 7) = 90 possible numbers

3) 7 fixed at ones place would give : _ _ 7 : 9(numbers that can take hundred place excluding 7) x 9(numbers that can take tens place excluding 7) = 81 possible numbers

= 1 +2 +3

= 100 + 90 + 81

= 271

Solution:

In each of the hundreds (i.e., 1-99, 100s, 200s, etc.) of the first 1000 positive integers, except for the 700s, the digit 7 appears in 19 integers, either as the units digit or the tens digit or both. In the 700s, 7 appears in every integer; therefore, the total number of integers that has (at least) one digit of 7 is:

9 x 19 + 100 = 171 + 100 = 271

Answer: E

Could you please explain how you got the 9

Thanks

The unit place can have any 9 values ( 0 to 9 except 7 i.e 0,1,2,3,4,5,6,8,9)
the same goes for tens and hundreds places(0 to 9 except 7)

hope you understand !!

HI GMATGuruNY , AndrewN , GMATCoachBen, TestPrepUnlimited

Can you help me with this question?

So from 0-1 = 7
From 10-99 = 8*1 except 70-79 = 10

Total 1+8+10 =19
So similarly 100-999?

Almost, NandishSS. Your rationale would lead to (D), 190. You cannot forget that every integer between 700 and 799, inclusive, will include a 7. Thus, you can take your 19 and multiply by 9 (accounting for the ranges 1-699 and 800-1000), then add in the extra 100 integers that include a 7 within the other range. The answer will be (E).

I hope that helps. Thank you for thinking to ask me about the question.

- Andrew

Total number of time 7 appears in units digit
=>7+n-1 * 10 =97
=>n=10

Total number of 7 occurs in tens place exluding that included in units digit =9

Therefore total number of times 7 occurs in units and tens digit =19*10 =190

In adddition we need to take into account the the appearence in hundreds digit which makes it more than 190
Therefore IMO E

Hello my friend
But in the unit place 0 can not be !
so I think 8 digit are verified to place at unit . (1,2,3,4,5,6,8,9)! can you help on your approcach?
Dear Bunuel
can we solve this problem by this approach?
I didn't see your response on this post, would you pls your approach on this question pls?

While this approach leads to the correct answer, it's not entirely accurate.

Consider the sequence of numbers written as follows:
000, 001, 002, ..., 999.

In this case, the total count of numbers from 0 to 999 is represented by 10x10x10, as each digit can take 10 values from 0 to 9. To find the count of numbers that do not contain the digit 7, we subtract the count of such numbers (9x9x9) from this total, giving us the count of numbers from 0 to 999 that include the digit 7.

However, there's a subtle detail to consider: the total count of 1,000 numbers includes 0, which shouldn't be counted as it's not a positive integer, and excludes 1,000, which should be included. Fortunately, these two omissions balance each other out, so we still arrive at the correct count of total numbers, which is 1,000.

Similarly, in calculating the count of numbers without the digit 7, we include 0 (which we shouldn't) and exclude 1,000 (which we should include). Yet, these two factors again balance each other out, ensuring that our final count of numbers without the digit 7 remains accurate.

Therefore, the final calculation of 10^3 - 9^3 is accurate but with the caveat mentioned above.

Hope it's clear.