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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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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??
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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??

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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Hello bb
Can I get more question of these type?
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Hello bb
Can I get more question of these type?

Check Similar Questions block at the bottom of the page.


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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


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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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.
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RRINSEAD2025
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?

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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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= \(1000-729=271\)
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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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