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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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A. 191
B. 189
C. 176
D. 175
E. 153
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Expert reply
Posts: 22,738
Bunuel
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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General Discussion
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
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
Total nos - Nos that contain no 9
5*10*10 - 4*9*9 =176
Posted from my mobile device
5*10*10 - 4*9*9 =176
Posted from my mobile device
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 = \(499 - 324 = 175\)
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 = \(499 - 324 = 175\)
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
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
rsrighosh
Hi rsrighosh
Can you please further elabourate step 2. Or first solution jn step 1. I coukdnot understand that method.
Thank you!
Posted from my mobile device
Posts: 266
C.
500-599 => 19 times
600-699= > 19 times
700-799 => 19 times
800-899= > 19 times
900-999 => 100 times
Counting: 176
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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