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# M03-12

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Math Expert
Joined: 02 Sep 2009
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GMAT 1: 650 Q47 V34
GMAT 2: 710 Q48 V39
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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

Thanks
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GMAT 1: 760 Q48 V46
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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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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Brilliant!The problem is just solved using counting method.
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I think this is a high-quality question and I agree with explanation.
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HI Bunuel

Thanks!
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JIAA wrote:
HI Bunuel

Thanks!

Check Constructing Numbers, Codes and Passwords Questions topic from our Special Questions Directory.
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GMAT 1: 690 Q48 V36

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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578vishnu wrote:

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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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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Bunuel your method is too good. I spent forever on this question analyzing every condition.
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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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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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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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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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­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
­