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13 Dec 2022, 02:30
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C
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Difficulty:
45%
(medium)
Question Stats:
71% (02:07) correct
29%
(02:25)
wrong
based on 240
sessions
History
Date
Time
Result
Not Attempted Yet
How many integers between 100 and 400, inclusive, contain the digit 2?
A. 100
B. 125
C. 138
D. 145
E. 150
A. 100
B. 125
C. 138
D. 145
E. 150
13 Dec 2022, 05:14
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Total numbers: \((400-100)+1 = 301\)
Total numbers which do not contain 2: Imagining the numbers as three blocks xyz = [x][y][z]
let x be \(1\) or \(3\)
y can be any 9 integers excluding 2
z can also be any 9 integers excluding 2
\(2*9*9 = 162\) and then lastly xyz can be \(400\). That means \(163\) numbers don't contain \(2\).
Numbers between 100 and 400, inclusive, which do contain 2: \(301-163 = 138\)
Answer C
Total numbers which do not contain 2: Imagining the numbers as three blocks xyz = [x][y][z]
let x be \(1\) or \(3\)
y can be any 9 integers excluding 2
z can also be any 9 integers excluding 2
\(2*9*9 = 162\) and then lastly xyz can be \(400\). That means \(163\) numbers don't contain \(2\).
Numbers between 100 and 400, inclusive, which do contain 2: \(301-163 = 138\)
Answer C
13 Dec 2022, 06:49
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Bunuel
An alternative approach to solve the question
Let's assume that the number is represented by xyz
Case 1: When the unit digit is 2
x y 2
- x can be filled in three ways - it can be 1 , 2 and 3
- y can be filled in nine ways - all numbers from 0 to 9 except 2. We will handle 2 separately.
Total number of ways = 3 * 9 = 27
Case 2: When the tens digit is 2
x 2 z
- x can be filled in three ways - it can be 1 , 2 and 3
- z can be filled in nine ways - all numbers from 0 to 9 except 2. We will handle 2 separately.
Total number of ways = 3 * 9 = 27
Case 3: When the hundreds digit is 2
2 y z
- y can filled in nine ways - all numbers from 0 to 9 except 2. We will handle 2 separately.
- z can be filled in nine ways - all numbers from 0 to 9 except 2. We will handle 2 separately.
Total number of ways = 9 * 9 = 81
Total = 81 + 27 + 27 = 135
Case 4: Handling multiple 2s
So far we have handled scenarios wherein xyz has only one two in either tens, or units or hundreds place. However, we can have multiple twos in xyz. We can have three numbers wherein two can appear in multiple places.
1 2 2
2 2 2
3 2 2
Total = 135 + 3 = 138
Option C
