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Re M0312
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15 Sep 2014, 23:20
Official Solution:In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"? A. 180 B. 196 C. 286 D. 288 E. 324 Examine what digits these set members can contain:  First digit (hundreds): 8 choices (1, 2, 3, 4, 6, 7, 8, 9  cannot be 0 or 5)
 Second digit (tens): 9 choices (0, 1, 2, 3, 4, 6, 7, 8, 9  cannot be 5)
 Last digit (units): 4 choices (1, 3, 7, 9  cannot be 0, 2, 4, 5, 6, 8)
The answer is \(8 * 9 * 4 = 32 * 9 = 288.\) Answer: D
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Re: M0312
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13 Jan 2015, 09:18
Hi Bunnel ,
Could you elaborate on this one ?
Thanks and Regards , Shelrod007



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Re: M0312
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19 May 2015, 11:10
Hey Bunell,
I would love some clarification too.



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Re: M0312
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19 May 2015, 12:20
101,103,107, 109... total + 4 111,113,117,119.... total + 4 ....... 151, 153....... don't count ....... ....... 191,193,197,199..... total +4 ===================== Total = 9*4 =36 Repeat same for numbers with 2 at hundred,3 at hundred,4 at hundred, 6 at hundred..... 9 at hundred position.
Therefore, total odd numbers that don't contain digit '5' are 36 * 8 = 288
Thanks



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20 May 2015, 01:36



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Re: M0312
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30 Oct 2015, 09:16
Last digit (units): 4 choices (1, 3, 7, 9  cannot be 0, 2, 4, 5, 6, 8)
Why can't the last digits be anything even?



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31 Oct 2015, 02:58
HunterJ wrote: Last digit (units): 4 choices (1, 3, 7, 9  cannot be 0, 2, 4, 5, 6, 8)
Why can't the last digits be anything even? The question asks about the number of odd integers. An odd integer cannot have even units digit.
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10 Jul 2016, 04:53
Brilliant!The problem is just solved using counting method.



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20 Aug 2016, 18:04
I think this is a highquality question and I agree with explanation.



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Re: M0312
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05 Jan 2017, 10:22
Hi Bunuel,
I followed an approcach considering the intersection of 2 sets: (1) odd, (2) numbers not containing 5. Please culd you tell in what I am wrong? Thanks.
What is asked is the intersection of 1&2. So,
Total numbers = 1000100+1= 901 numbers Set (1): 450 odd numbers Set (2): Total of 649 numbers 3 digits: 8*9*9 = 648 4 digits: 1 (note that there is a similar problem in the forum to find set (2): search "How many three digit numbers contain the digit 5 at least once")
Now, according to the sets theory: 649 + 450  (int 1 & 2) = 901 numbers. Hence, the answer must be 649 + 450  901 = 198, not 288.
What am I doing wrong? Thank you!



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06 Jan 2017, 01:37
ricardobs006 wrote: Hi Bunuel,
I followed an approcach considering the intersection of 2 sets: (1) odd, (2) numbers not containing 5. Please culd you tell in what I am wrong? Thanks.
What is asked is the intersection of 1&2. So,
Total numbers = 1000100+1= 901 numbers Set (1): 450 odd numbers Set (2): Total of 649 numbers 3 digits: 8*9*9 = 648 4 digits: 1 (note that there is a similar problem in the forum to find set (2): search "How many three digit numbers contain the digit 5 at least once")
Now, according to the sets theory: 649 + 450  (int 1 & 2) = 901 numbers. Hence, the answer must be 649 + 450  901 = 198, not 288.
What am I doing wrong? Thank you! Why should (odd) + (numbers not containing 5)  (both) give total of 901? Where is the group of even numbers containing 5?
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Re: M0312
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17 Apr 2017, 05:47
I did an alternative approach, although I admit, that the solution is far more easier.
\(901\) numbers \(450\) odd numbers \( 50\) numbers (remaining odd numbers from 501 to 599) \( 8*5\) numbers (8x times (100range, 200range till 900range; excluding 500range) the remaining odd numbers from 51 to 59 (5 numbers)) \( 8*9\) numbers (8x times (100range, 200range till 900range; excluding 500range) the remaining 9 odd numbers out of every 10range; excluding 50range) \(=288\) numbers



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Re M0312
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01 Aug 2017, 22:00
I think this is a highquality question and I agree with explanation. amazing clarification



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Joined: 09 Aug 2014
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Hi Bunuel,
A small confusion with my methodology:
Answer = Total ODD numbers (1001000) Less: numbers where the digit 5 appears
Odd numbers between 1001000 = 450 [1000100/2 +1]
Cases where 5 appears:
_ _ 5 = 8*9 = 72 cases _ 5 _ = 8*4 = 32 cases 5 _ _ = 9*4 = 36 cases i.e. total 140 cases
Hence, Answer = 450 140 = 310!
Thanks in advance
Regards Srinath



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Re: M0312
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22 Dec 2017, 20:37
This could be solved using this approach
Nos starting with 1: _ _ _ (First one has to be 1, second can be any number  0,1,2,3,4,6,7,8,9; third can only be 1,3,7,9) Therefore = 1*9*4 = there are 36 numbers that start with 1, and are odd, and do not have a 5 Nos starting with 2: 36 numbers following same method nos starting with 3: 36 nos nos starting with 4: 36 nos nos starting with 6: 36 nos nos starting with 7: 36 nos nos starting with 8: 36 nos nos starting with 9:36 nos
36*8 = 288



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Re: M0312
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30 May 2018, 18:25
Followed a similar approach to Bhawna If we count such numbers between 100200 which can be replicated 8 times, we get 36. (odd numbers that don't contain the digit 5) Final answer: 36*8=288
It is multiplied by 8 because the range we are considering is as follows: 100200, 200300, 300400, 400500, 600700, 700800, 800900, 9001000



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Re: M0312
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24 Nov 2018, 01:19
Have anyone collected a similar type of these exercise under 1 link?










