nazii wrote:
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
Bunuelcan 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?
HouseStark wrote:
Bunuel wrote:
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= \(1000-729=271\)
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.