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# M03-12

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Math Expert
Joined: 02 Sep 2009
Posts: 55277

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16 Sep 2014, 00:19
1
25
00:00

Difficulty:

85% (hard)

Question Stats:

50% (02:17) correct 50% (02:29) wrong based on 149 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 55277

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16 Sep 2014, 00:20
6
9
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:
1. First digit (hundreds): 8 choices (1, 2, 3, 4, 6, 7, 8, 9 - cannot be 0 or 5)
2. Second digit (tens): 9 choices (0, 1, 2, 3, 4, 6, 7, 8, 9 - cannot be 5)
3. Last digit (units): 4 choices (1, 3, 7, 9 - cannot be 0, 2, 4, 5, 6, 8)

The answer is $$8 * 9 * 4 = 32 * 9 = 288.$$

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Manager
Joined: 23 Jan 2013
Posts: 132
Concentration: Technology, Other
Schools: Berkeley Haas
GMAT Date: 01-14-2015
WE: Information Technology (Computer Software)

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13 Jan 2015, 10:18
Hi Bunnel ,

Could you elaborate on this one ?

Thanks and Regards ,
Shelrod007
Intern
Joined: 04 Sep 2014
Posts: 5

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19 May 2015, 12:10
Hey Bunell,

I would love some clarification too.
Intern
Joined: 03 May 2014
Posts: 26

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19 May 2015, 13:20
2
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
Math Expert
Joined: 02 Sep 2009
Posts: 55277

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20 May 2015, 02:36
floody84 wrote:
Hey Bunell,

I would love some clarification too.

Can you please specify what part is unclear in the solution? Thank you.
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Intern
Joined: 06 Oct 2015
Posts: 5
GMAT 1: 760 Q48 V46

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30 Oct 2015, 10:16
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?
Math Expert
Joined: 02 Sep 2009
Posts: 55277

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31 Oct 2015, 03:58
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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Joined: 21 Jan 2015
Posts: 8
GMAT 1: 570 Q41 V28
GPA: 3.47

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10 Jul 2016, 05:53
Brilliant!The problem is just solved using counting method.
Intern
Joined: 21 Aug 2012
Posts: 3

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20 Aug 2016, 19:04
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 29 Sep 2016
Posts: 2

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05 Jan 2017, 11:22
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 = 1000-100+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!
Math Expert
Joined: 02 Sep 2009
Posts: 55277

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06 Jan 2017, 02:37
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 = 1000-100+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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Intern
Joined: 07 Feb 2016
Posts: 20
GMAT 1: 650 Q47 V34
GMAT 2: 710 Q48 V39

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17 Apr 2017, 06:47
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 (100-range, 200-range till 900-range; excluding 500-range) the remaining odd numbers from 51 to 59 (5 numbers))
$$- 8*9$$ numbers (8x times (100-range, 200-range till 900-range; excluding 500-range) the remaining 9 odd numbers out of every 10-range; excluding 50-range)
$$=288$$ numbers
Intern
Joined: 24 Jun 2013
Posts: 3
Schools: CBS '17
GMAT 1: 630 Q47 V27

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01 Aug 2017, 23:00
I think this is a high-quality question and I agree with explanation. amazing clarification
Intern
Joined: 09 Aug 2014
Posts: 8

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09 Sep 2017, 01:18
Hi Bunuel,

A small confusion with my methodology:

Answer = Total ODD numbers (100-1000) Less: numbers where the digit 5 appears

Odd numbers between 100-1000 = 450 [1000-100/2 +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!

Regards
Srinath
Intern
Joined: 10 Feb 2017
Posts: 5

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22 Dec 2017, 21:37
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
Intern
Joined: 23 Mar 2018
Posts: 12
GMAT 1: 750 Q48 V44

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30 May 2018, 19:25
1
Followed a similar approach to Bhawna
If we count such numbers between 100-200 which can be replicated 8 times, we get 36. (odd numbers that don't contain the digit 5)

It is multiplied by 8 because the range we are considering is as follows: 100-200, 200-300, 300-400, 400-500, 600-700, 700-800, 800-900, 900-1000
Intern
Joined: 09 Jul 2018
Posts: 1

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24 Nov 2018, 02:19
Have anyone collected a similar type of these exercise under 1 link?
Intern
Joined: 10 Nov 2018
Posts: 3

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31 Mar 2019, 11:45
Hi Bunuel,
what if the question asked ( 0 to 1000)
would we add 32 ( 8*4), am i right.

Thanks,
Manager
Joined: 18 Jul 2018
Posts: 51
Location: United Arab Emirates

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15 Apr 2019, 01:37
Bunuel wrote:
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:
1. First digit (hundreds): 8 choices (1, 2, 3, 4, 6, 7, 8, 9 - cannot be 0 or 5)
2. Second digit (tens): 9 choices (0, 1, 2, 3, 4, 6, 7, 8, 9 - cannot be 5)
3. Last digit (units): 4 choices (1, 3, 7, 9 - cannot be 0, 2, 4, 5, 6, 8)

The answer is $$8 * 9 * 4 = 32 * 9 = 288.$$

HI Bunuel

Thanks!
Re: M03-12   [#permalink] 15 Apr 2019, 01:37

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# M03-12

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