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M03-12

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Bunuel
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M03-12 [#permalink] New post   16 Sep 2014, 00:19
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Question Stats:

47% (02:11) correct 53% (02:24) wrong based on 897 sessions
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How many odd integers between 100 and 1000, do not contain the digit "5"?

A. 180
B. 196
C. 286
D. 288
E. 324
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D
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Bunuel
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Re M03-12 [#permalink] New post   16 Sep 2014, 00:20
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Official Solution:

How many odd integers between 100 and 1000, do not contain the digit "5"?

A. 180
B. 196
C. 286
D. 288
E. 324


Let's examine the possible options for each digit in these numbers:

The hundreds digit has 8 options: 1, 2, 3, 4, 6, 7, 8, or 9 (excluding 0 and 5).

The tens digit has 9 options: 0, 1, 2, 3, 4, 6, 7, 8, or 9 (excluding 5).

The units digit has 4 odd options: 1, 3, 7, or 9 (excluding 0, 2, 4, 5, 6, and 8).

Consequently, the total number of odd integers between 100 and 1000 that do not contain the digit "5" is \(8 * 9 * 4 = 32 * 9 = 288.\)


Answer: D
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ChrisL1988
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Re: M03-12 [#permalink] New post   17 Apr 2017, 06:47
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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] New post   19 May 2015, 13:20
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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] New post   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?
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Bunuel
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Re: M03-12 [#permalink] New post   31 Oct 2015, 03:58
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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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mafuz
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Re: M03-12 [#permalink] New post   10 Jul 2016, 05:53
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Brilliant!The problem is just solved using counting method.
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Re: M03-12 [#permalink] New post   20 Aug 2016, 19:04
I think this is a high-quality question and I agree with explanation.
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M03-12 [#permalink] New post   15 Apr 2019, 01:37
HI Bunuel

Can you please provide link to Qs similar to this ??

Thanks!
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Bunuel
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M03-12 [#permalink] New post   15 Apr 2019, 02:50
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JIAA wrote:
HI Bunuel

Can you please provide link to Qs similar to this ??

Thanks!


Check Constructing Numbers, Codes and Passwords Questions topic from our Special Questions Directory.
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Re: M03-12 [#permalink] New post   05 Jul 2019, 03:02
Bunuel chetan2u souvik101990 VeritasKarishma egmat

Hoping somebody would reply.

Though I agree with Bunuel's solution, I wanted to understand what is wrong in my approach.

_ _ _

In 1st dash, any number except for 0 & 5 ---- 8
In 2nd dash, any number except for 5 ---- 9
In 3rd dash, any number except for 5 ---- 9

Now we will get 8 * 9 * 9 = 648. Half of these numbers are even, other half odd. So, 648/2 = 324

It'll be great if somebody replies! :)
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Re: M03-12 [#permalink] New post   05 Jul 2019, 03:22
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578vishnu wrote:
Bunuel chetan2u souvik101990 VeritasKarishma egmat

Hoping somebody would reply.

Though I agree with Bunuel's solution, I wanted to understand what is wrong in my approach.

_ _ _

In 1st dash, any number except for 0 & 5 ---- 8
In 2nd dash, any number except for 5 ---- 9
In 3rd dash, any number except for 5 ---- 9

Now we will get 8 * 9 * 9 = 648. Half of these numbers are even, other half odd. So, 648/2 = 324

It'll be great if somebody replies! :)


Not half of them will be odd. You excluded 5, so 4/9th will be odd. 648*4/9 = 288.
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Bunuel
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Re: M03-12 [#permalink] New post   07 Oct 2019, 07:10
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ghnlrug wrote:
Bunuel could you please tell me what is incorrect in my approach:

(total odd) - (total odd containing 5) = 450 - (1*10*5+9*1*5+9*10*1)

5_ _ –> 1*10*5
_ 5 _ –> 9*1*5
_ _ 5 –> 9*10*1


You are subtracting more odd numbers with 5, then there are. For example, number 515 is counted in the first group as well as in the third one.
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Re: M03-12 [#permalink] New post   02 Jun 2020, 08:19
Bunuel your method is too good. I spent forever on this question analyzing every condition.
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DixitNiket
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Re: M03-12 [#permalink] New post   28 Nov 2020, 11:37
Invested almost 21 minutes on this question to solve through number system. But didn't think through to use permutations and combinations.
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Re: M03-12 [#permalink] New post   19 Dec 2020, 05:00
Kritisood wrote:
Bunuel your method is too good. I spent forever on this question analyzing every condition.


yes exactly. I literally freaked out thinking about all the solutions only to realise that it was so so simple.
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Bunuel
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Re: M03-12 [#permalink] New post   21 Apr 2023, 13:14
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M03-12 [#permalink] New post   04 Sep 2023, 03:05
Another way to do this question


We get 19 unique instances of 5 in every 100 numbers

Out of these 19 numbers, 5 are even namely x50, x52, x54, x56, x58. Therefore, you have a total of 19-5=14 odd numbers containing 5.

So for every 100 excluding 500-600, you get 14*8 = 112 odd numbers containing 5

Remember, there are 50 odd numbers between 500-599

So total number of odd numbers containing 5 is 112+50=162

Total number of odd integers between 100 and 1000 is 450

Subtract 450-162 = 288 - The final answer
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M03-12 [#permalink] New post   23 Apr 2024, 07:53
­Hello,
I did it in a worst way (more time spended).

-Between 100 and 1000 there are 900 numbers.
- all 500´s (501,502...599) = 100
- Numbers containing a 5 : every 100 there are 10 (50,51,52...59) + 9 (105,115,125,135,145,165..) = 19*8 (not counting 500´s) = 152
- All odd numbers (5 every 10 without counting 50´s and 500´s) = 5*9*8 = 360
Total:
900-100-152-360 = 288
 ­
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M03-12 [#permalink]
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