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M03-12

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M03-12  [#permalink]

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New post 15 Sep 2014, 23:19
1
23
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

50% (01:33) correct 50% (01:54) wrong based on 219 sessions

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Re M03-12  [#permalink]

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New post 15 Sep 2014, 23:20
5
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 \(8 * 9 * 4 = 32 * 9 = 288.\)


Answer: D
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Re: M03-12  [#permalink]

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New post 13 Jan 2015, 09:18
Hi Bunnel ,

Could you elaborate on this one ?

Thanks and Regards ,
Shelrod007
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Re: M03-12  [#permalink]

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New post 19 May 2015, 11:10
Hey Bunell,

I would love some clarification too.
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Re: M03-12  [#permalink]

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New post 19 May 2015, 12:20
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


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Re: M03-12  [#permalink]

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New post 20 May 2015, 01:36
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Re: M03-12  [#permalink]

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New post 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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Re: M03-12  [#permalink]

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New post 31 Oct 2015, 02:58
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Re: M03-12  [#permalink]

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New post 10 Jul 2016, 04:53
Brilliant!The problem is just solved using counting method.
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Re M03-12  [#permalink]

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New post 20 Aug 2016, 18:04
I think this is a high-quality question and I agree with explanation.
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Re: M03-12  [#permalink]

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New post 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 = 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!
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Re: M03-12  [#permalink]

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New post 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 = 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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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M03-12  [#permalink]

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New post 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 (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
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Re M03-12  [#permalink]

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New post 01 Aug 2017, 22:00
I think this is a high-quality question and I agree with explanation. amazing clarification
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M03-12  [#permalink]

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New post 09 Sep 2017, 00:18
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!

Thanks in advance

Regards
Srinath
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Re: M03-12  [#permalink]

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New post 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: M03-12  [#permalink]

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New post 30 May 2018, 18:25
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)
Final answer: 36*8=288

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
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Re: M03-12  [#permalink]

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New post 24 Nov 2018, 01:19
Have anyone collected a similar type of these exercise under 1 link?
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Re: M03-12 &nbs [#permalink] 24 Nov 2018, 01:19
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