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GMATinsight
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My solution.

there are 19 numbers containing atleast one seven from 100 to 200. The same will be true for the eight following hundreds sequences except the 700-800 sequence. The 700 - 800 sequence contains 100 numbers that has atleast one seven in it .
So the total no of digits can be calculated as 8*19 + 100 = 252. Ans D

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GMATinsight
In how many no. between 100 and 1000 at least one of their digits is 7?

(A) 900
(B) 648
(C) 720
(D) 252
(E) 729

Source: https://www.GMATinsight.com

Method 1:

From 100-199 there are 19 numbers with atleast one of their digits 7 {107, 117, 127, 137, 147, 157, 167, 177, 187, 197}

Likewise every 100 numbers except series 700-799 will have 19 numbers with one of their digits 7

so total such numbers = 19*8

from 700-799, all 100 numbers will have one of their digits 7 hence total required numbers = 19*8+100 = 252

Method 2:

Total Three digit numbers from 100-1000 = 900

Total three digit numbers without use of digit 7 = 8*9*9 = 648

Total Numbers with one of the digits 7 = 900 - 648 = 252

Answer: Option D

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