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# How many 3-digit numbers greater than 500 contain the digit 9 appearin

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Math Expert
Joined: 02 Sep 2009
Posts: 64880
How many 3-digit numbers greater than 500 contain the digit 9 appearin  [#permalink]

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12 May 2020, 06:15
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66% (02:06) correct 34% (02:13) wrong based on 94 sessions

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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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Joined: 19 Oct 2018
Posts: 1968
Location: India
Re: How many 3-digit numbers greater than 500 contain the digit 9 appearin  [#permalink]

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12 May 2020, 07:30
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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 wrote:
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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Joined: 24 Jan 2019
Posts: 5
Re: How many 3-digit numbers greater than 500 contain the digit 9 appearin  [#permalink]

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12 May 2020, 07:40
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Total nos - Nos that contain no 9

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

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Joined: 13 Jun 2019
Posts: 20
Re: How many 3-digit numbers greater than 500 contain the digit 9 appearin  [#permalink]

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13 May 2020, 05:04
1
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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Shrekey

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Joined: 19 Oct 2018
Posts: 1968
Location: India
Re: How many 3-digit numbers greater than 500 contain the digit 9 appearin  [#permalink]

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13 May 2020, 05:15
1
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 wrote:
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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Re: How many 3-digit numbers greater than 500 contain the digit 9 appearin  [#permalink]

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19 May 2020, 06:37
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1
Bunuel wrote:
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

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.

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Joined: 23 Feb 2020
Posts: 43
Re: How many 3-digit numbers greater than 500 contain the digit 9 appearin  [#permalink]

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30 May 2020, 05:11
rsrighosh wrote:
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 = $$499 - 324 = 175$$

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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Joined: 11 Feb 2019
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Re: How many 3-digit numbers greater than 500 contain the digit 9 appearin  [#permalink]

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30 May 2020, 05:29
1
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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NJ
Re: How many 3-digit numbers greater than 500 contain the digit 9 appearin   [#permalink] 30 May 2020, 05:29