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Number of 7s

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Source: — Problem Solving |

by GmatKiss » Sat Aug 06, 2011 10:33 pm
IMO: A
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by ajaykpat » Sat Aug 06, 2011 10:52 pm
hi,

From :1-100 = 20 ( 7,17,27,37,47,57,67,70-79,87,97 =20 , PS=77 has two 7 digits)
From :101-200 = 20
From :201-300 = 20
From :301-400 = 20
From :401-500 = 20
From :501-600 = 20
From :601-699 = 20
From :700-800 = 20 + 100
From :801-900 = 20
From :901-1000 = 20

Total = 20x10 + 100= 300


Br,

ajp
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by Anurag@Gurome » Sat Aug 06, 2011 10:54 pm
MBA.Aspirant wrote:How many number of times will the digit '7' be written when listing the integers from 1 to 1000?
Effectively we have to find out the number of times the digit '7' appears when listing all the integers from 000 to 999. Hence, we have three places to be filled by any digit. Consider the following cases,
  • 1. --7 : First two places can be filled by any of the 10 digits (0 to 9)
    2. -7- : First and last places can be filled by any of the 10 digits (0 to 9)
    1. 7-- : Last two places can be filled by any of the 10 digits (0 to 9)
Hence, a total of (10*10 + 10*10 + 10*10) = 300 times 7 will appear.

The correct answer is B.
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by GMATGuruNY » Sat Aug 06, 2011 11:28 pm
MBA.Aspirant wrote:How many number of times will the digit '7' be written when listing the integers from 1 to 1000?

(a) 271
(b) 300
(c) 252
(d) 304
(e) 512
If we use 0 as a placeholder -- so that 007 represents 7, 072 represents 72, etc. -- then we need to count the number of times that 7 will appear among the 3-digit integers from 000 to 999, inclusive.

Total number of 3-digit integers from 000 to 999, inclusive = biggest - smallest + 1 = 999-000+1 = 1000.

Each of these 1000 integers includes 3 digits.
Thus, the total number of digit appearances = 3*1000 = 3000.

Among these 3000 digit appearances, each of the 10 digits will appear the same number of times.
Thus, the number of times that 7 will appear = 3000/10 = 300.

The correct answer is B.
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