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Bunuel
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How many 3-digit numbers greater than 500 contain the digit 9 appearin 12 May 2020, 07:15
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C
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How many 3-digit numbers greater than 500 contain the digit 9 appearing at least once?

A. 191
B. 189
C. 176
D. 175
E. 153


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C
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How many 3-digit numbers greater than 500 contain the digit 9 appearin 12 May 2020, 08:30
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Number that contains only one 9= 1*9*9+4*1*9+4*9*1= 153

Number that contains two 9's= 1*1*9+4*1*1+1*9*1= 22

Number that contains three 9's= 1*1*1=1

Total numbers = 153+22+1=176





Bunuel
How many 3-digit numbers greater than 500 contain the digit 9 appearing at least once?

A. 191
B. 189
C. 176
D. 175
E. 153


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How many 3-digit numbers greater than 500 contain the digit 9 appearin 12 May 2020, 08:40
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Total nos - Nos that contain no 9

5*10*10 - 4*9*9 =176

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How many 3-digit numbers greater than 500 contain the digit 9 appearin 13 May 2020, 06:04
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I think the answer is 175
As statement says 3-digit numbers greater than 500 . This means \(501<=x<=999\)
Therefore

Step 1:- total number of numbers between 501 and 999 (all inclusive) = \(5 * 10 * 10 = 500 - 1 = 499\)
We can also find this by \(999-501 = 498+1 = 499\) (Adding 1 to include 501)

Step 2:- total number of numbers which does not contain 9 =\( 4 * 9 * 9 = 324\)

Step 3:- total number of numbers containing at least one 9 = \(499 - 324 = 175\)
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How many 3-digit numbers greater than 500 contain the digit 9 appearin 13 May 2020, 06:15
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You count 500 in Step 2....Deduct there too. You'll get 176. Point is it does not matter whether you count from 500 or 501, Numbers that contain 9 remain the same

Another way is-

Numbers from 501-599 that contain 9 = 10+10-1=19 (Deduct 1, as we counted 599 twice)

Numbers from 600-699 that contain 9 = 10+10-1=19

Numbers from 700-799 that contain 9 = 10+10-1=19

Numbers from 800-899 that contain 9 = 10+10-1=19

Numbers from 900-999 that contain 9 = 100

Total numbers = 19*4+100=176


rsrighosh
I think the answer is 175
As statement says 3-digit numbers greater than 500 . This means \(501<=x<=999\)
Therefore

Step 1:- total number of numbers between 501 and 999 (all inclusive) = \(5 * 10 * 10 = 500 - 1 = 499\)
We can also find this by \(999-501 = 498+1 = 499\) (Adding 1 to include 501)

Step 2:- total number of numbers which does not contain 9 =\( 4 * 9 * 9 = 324\)

Step 3:- total number of numbers containing at least one 9 = \(499 - 324 = 175\)
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How many 3-digit numbers greater than 500 contain the digit 9 appearin 19 May 2020, 07:37
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Bunuel
How many 3-digit numbers greater than 500 contain the digit 9 appearing at least once?

A. 191
B. 189
C. 176
D. 175
E. 153


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All 100 numbers in the 900s have at least one digit of 9, The number of numbers in 500s, 600s, 700s, and 800s that have at least one digit of 9 must be equal. That is, there is the same amount of numbers in 500s that have at least one digit of 9 as in 600s, 700s, or 800s. If this amount is x, then the total number of numbers is 4x + 100. Notice this must be an even number and since 176 is the only even number in the given choices, 176 must be the correct answer.

Answer: C
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How many 3-digit numbers greater than 500 contain the digit 9 appearin 30 May 2020, 06:11
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rsrighosh
I think the answer is 175
As statement says 3-digit numbers greater than 500 . This means \(501<=x<=999\)
Therefore

Step 1:- total number of numbers between 501 and 999 (all inclusive) = \(5 * 10 * 10 = 500 - 1 = 499\)
We can also find this by \(999-501 = 498+1 = 499\) (Adding 1 to include 501)

Step 2:- total number of numbers which does not contain 9 =\( 4 * 9 * 9 = 324\)

Step 3:- total number of numbers containing at least one 9 = \(499 - 324 = 175\)
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Hi rsrighosh
Can you please further elabourate step 2. Or first solution jn step 1. I coukdnot understand that method.

Thank you!

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How many 3-digit numbers greater than 500 contain the digit 9 appearin 30 May 2020, 06:29
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C.

500-599 => 19 times
600-699= > 19 times
700-799 => 19 times
800-899= > 19 times
900-999 => 100 times

Counting: 176
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How many 3-digit numbers greater than 500 contain the digit 9 appearin 21 Nov 2025, 16:42
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