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

TOOK 30 SEC TO SOLVE THIS

TOTAL NO. = 1000
NO. OF INTEGERS IN WHICH 7 DOESN'T APPEAR = 9*9*9=729

NO. OF INTEGER IN WHICH 7 APPEAR= \(1000-729=271\)

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*7 (from zero to 699) + 19 in 800-899 + 19 from 900-999 = 19*6+19+19= 171 times.
Add: 100 times of when 7 appeared as hundreds digit (from 700 to 799)

171+100 =271

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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.
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I think the question means something else. It does not say "how many times..." It says "...in how many ... integers?"
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Got It, the digits in 700 category will create redundancy. Thank You sir.
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answer is E
19*9+100 = 271

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
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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
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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 !!

yashikaaggarwal Could you please help me out here? Really struggling with my approach and can't figure out the flaw.

From 0-99 -> 19 digits
Now, fixing positions :
1. Fixed 7 at ones position : 9*10*1 = 90
2. Fixed 7 at tens position : 9*1*10 =90
3. Fixed 7 at hundreds position: 1*10*10 = 100

Buddy tell me a thing!
Why are you considering 3 places for numbers till 100?
we just have 2 places for a set of hundred no. except the ones from 700-799

answer will be like:
1. Fixed 7 at ones position will be : 9*1(7 is fixed here) = 9
2. Fixed 7 at tens position : (7 is fixed here)1*9 = 9
3. Fixed 7 at both ones and tens place = 1*1 = 1
Total = 19

we have that for numbers 0-100
for numbers 0-699
the 7 digit appearance = 19*7 = 133

for numbers 800-900
the 7 digit appearance = 19*1 = 19

for numbers 900-1000
the 7 digit appearance = 19*1 = 19

Total= 133+19+19 = 171
Add 100 numbers from 700-799

171+100 =271
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Apologies let me re-phrase. I am not considering 3 places for numbers till 100. That count remains 19.
My approach for counting the number of times 7 appears in the 3 digit numbers is slightly different to what you've explained and I want to understand my mistake there.

So,instead of going to regular 100-699, then 700-799 way, I am fixing 7 at one's, ten's and so on.
Which leads me to the below :
1. Fixed 7 at ones position : 9*10*1 = 90
2. Fixed 7 at tens position : 9*1*10 =90
3. Fixed 7 at hundreds position: 1*10*10 = 100

I know point 3 is correct, from 700-799, we will have 100 numbers. What is wrong in Point 1 and 2?
Appreciate your guidance yashikaaggarwal

I got the mistake,
See the highlighted part.
There will be 9 cases in there when tens will have 7 digit as well. (077,177,277,377,477,577,677,877,977)
But we already covered them in 2nd point.
(71,72,73.....77,78,......171,172,....177,178.... And so on)

So first point will have 7 at ones and 9*9 at hundred and tens place.
Making total no. = 9*9*1 = 81
Second = 90
Third = 100

Total is 271

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