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

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Bunuel

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16 Dec 2024, 01:27

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GGGMAT2

Bunuel I took this approach to solve the Q.

Please let me know what's wrong with the approach. Apologies, I am unable to figure out the problem with my approach by going through the solutions provided.

The number of 1 digit integers that can have 7 is 1

The number of 2 digit integers where 7 occupies the one's place 9X1 = 9
The number of 2 digit integers where 7 occupies the tens's place 1X10 = 10
However, with this we have counted 77 twice, so total numbers would be 9 + 10 - 1 = 18.

The number of 3 digit integers where 7 occupies the one's place 9X10X1 = 90
The number of 3 digit integers where 7 occupies the tens's place 9X1X10 = 90
The number of 3 digit integers where 7 occupies the hundred's place 1X10X10 = 100
However, with this we have counted 777 thrice, so total numbers would be 90 + 90 + 100 - 2 = 278.

Total number of integers are 1 + 18 + 278 = 297

What about other overlaps? Such as 177, 277, 377, ..., 977. Also, 707, 717, 727, 737, 747, 757, 767, 787, 797.

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14 Mar 2025, 11:09

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HouseStark

Bunuel

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

A. 143
B. 152
C. 171
D. 190
E. 271

PS20404

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= \(1000-729=271\)

COULD YOU SHARE THE FORMULA HOW DID YOU HAVE NO OF INTEGERS IN WHICH 7 DOESN'T APPEAR

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15 Apr 2025, 02:56

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Set theory can also be used here =>

P(A) + P(B) + P(C) - P(A∩B) - P(A∩C) - P(B∩C) + P(A∩B∩C)

P(A) = 7 can occur in units digit in 10*10 ways = 100
P(B) = 7 can occur in tenths digit in 10*10 ways = 100
P(C) = 7 can occur in hundredths digit in 10*10 ways = 100

P(A∩B) = 7 can occur in units and tenths digit in 10 ways = 10
P(A∩C) = 7 can occur in units and hundredths digit in 10 ways = 10
P(B∩C) = 7 can occur in tenths and hundredths digit in 10 ways = 10

P(A∩B∩C) = 7 can occur in units, tenths and hundredths digit in 1 way = 1

Total = 100 + 100 + 100 - 10 - 10 - 10 + 1 = 271

IMO: E

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15 Apr 2025, 03:12

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Bunuel

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

A. 143
B. 152
C. 171
D. 190
E. 271

PS20404

from 1 to 99 digit 7 appears in 19 numbers

from 100 to 199 digit 7 appears in 19 numbers
...
...
...

There are 9 such centuries of 100 numbers (except 100 numbers from 700-799) so total occurences on unit and tens place = 19*9 = 171

7 appears in 100 numbers = 100

So total such numbers = 171 + 100 = 271

Answer: Option E

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Solved it as such, use negation, as in subtract from 1000 the number of integers where 7 does not appear

1] how many times 7 comes in a single digit integer series
2] how many times 7 comes in a double digit integer series
3] how many times 7 comes in a three digit integer series
4] then we are left with 1000 (because the question asks for the first 1000 integers)

negation, single digits without 7, double without 7, three digits without 7 and then 1000

8
8 9
8 9 9
1 0 0 0

using permutation method, multiply and sum all

so 8+72+648+1 = 729

hence 1000 - 729 = 271
E

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03 Aug 2025, 06:17

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HouseStark

Bunuel

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

A. 143
B. 152
C. 171
D. 190
E. 271

PS20404

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= \(1000-729=271\)

why 9 * 9 * 9 it does not appear??

Bunuel

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03 Aug 2025, 08:44

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arjunaaho

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Bunuel

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

A. 143
B. 152
C. 171
D. 190
E. 271

PS20404

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= \(1000-729=271\)

why 9 * 9 * 9 it does not appear??

Because for each digit (hundreds, tens, units), there are 9 options that are not 7. So:

9 (choices for hundreds) * 9 (tens) * 9 (units) = 729 numbers without a 7.

1000 total - 729 = 271 numbers with at least one 7.

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04 Aug 2025, 01:43

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Hello bb
Can I get more question of these type?

Bunuel

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quibusdamofficiis

Hello bb
Can I get more question of these type?

Check Similar Questions block at the bottom of the page.

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14 Aug 2025, 01:18

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at the thousands place 1 is possible
at the 100's place 9 digits are possible (excluding 7 and including 0)
and so on.

Hence we considered 1*9*9*9

Hope my understanding is correct.

HouseStark

Bunuel

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

A. 143
B. 152
C. 171
D. 190
E. 271

PS20404

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= \(1000-729=271\)

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Hey Bunuel,

Could you explain why you're considering 9 possible digits for the hundreds place? Shouldn't it be 8 given that 0 cannot be in the hundreds place? What am I missing?

My reasoning is this: In the hundreds place you can have 8 digits, excluding 7 and 0. In the tens place you can have 9 digits, including 0 but excluding 7. In the units place you can have again 9 digits.

You then subtract 8 x 9 x 9 from the 1000 positive integers.

However this does not yield the correct answer. What am I missing here?

Bunuel

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numer 7 will appear in the following digits (we don't count how many times the single number 7 is present in a number, just in how many numbers 7 is present: ex--> in 77 we are counting as 1 the number 77)
7,17,27,37,47,57,67,70,71,72,73,74,,75,76,77,78,79,87,97 -->
19 times in the first 99 numbers, so also: (19 times)
19 times in the range 100-199, (28 times)
19 times in range 200-299, (57 times)
19 times in range 300-399, (76 times)
19 times in range 400-499, (95 times)
19 times in range 500-599, (114 times)
19 times in range 600-699, (133 times)
100 times in range 700-799, (233 times)
witout finishing we can already exclude all answers, except for one.

Bunuel

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23 Aug 2025, 11:51

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RRINSEAD2025

You’re missing that the “9 options” for the hundreds place already works if we treat every integer from 000 to 999 as a 3-digit string. That way, we get 10 * 10 * 10 = 1000 numbers in total. Then the count of numbers without any 7 is 9 * 9 * 9 = 729. Subtracting gives 1000 - 729 = 271 numbers that contain at least one 7.

Including “000” is just a counting device to make the calculation consistent, it still corresponds exactly to the 1000 positive integers from 1 to 1000.

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05 Oct 2025, 07:16

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How did you get 9 x 9 x 9 = total where 7 does not appear?

HouseStark

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= \(1000-729=271\)

Bunuel

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05 Oct 2025, 07:44

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Othompson18

How did you get 9 x 9 x 9 = total where 7 does not appear?

This is addressed in the thread:

Because for each digit (hundreds, tens, units), there are 9 options that are not 7. So:

9 (choices for hundreds) * 9 (tens) * 9 (units) = 729 numbers without a 7.

1000 total - 729 = 271 numbers with at least one 7.

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15 Feb 2026, 22:43

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Since 1000 does not contain a digit, instead we can think of 0-999 and how many times is 7 written. That would be 9*9*9 and hence we have 729 digits where 7 is not written as a digit. Hence we have 271 integers with a written 7 on it.

Bunuel

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

A. 143
B. 152
C. 171
D. 190
E. 271

PS20404