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How many odd three-digit integers greater than 800 are there

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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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New post 20 Jun 2012, 00:40
Hi,

My solution is as follows:
You have to find the odd (3digit) numbers greater than 800, all distict digits. Available digits would be (0 to 9)
Odd numbers starting with 8 = 1*8*5 = 40 (Hundredth digit is 8 - so, only 1 choice. Unit digit can be (1, 3, 5, 7, 9). Now tens digit will not be 8 & a digit chosen at units place - 8 possibilities)
Odd numbers starting with 9 = 1*8*4 = 40 (Hundredth digit is 9 - so, only 1 choice. Unit digit can be (1, 3, 5, 7). Now tens digit will not be 9 & a digit chosen at units place - 8 possibilities)
Total numbers = 32

Answer (C)

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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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New post 20 Jun 2012, 11:11
1 8 4 = 32
- - -


1 means that we can only 9 at that place, 4 is for odd numbers, 1,3,5,7 we cant take 9 bacause we have already taken 9 , and 8 is for putting (10-2=8 numbers which we already taken so question asked differnt number so we can 8 digits over there)

1 8 5=40
- - -
so 1 for 8, 5 for using 1,3,5,7,9 five odd digits and again (10-2=8) ..

40+32=72...
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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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New post 23 Jun 2012, 10:12
2*9*8=144/2 = 72

1st digit: 2 (can be either 8 or 9)
2nd digit: 9 (can be any number out of 10 but must be different from the 1st digit, hence 9 options)
3rd digit: 8 (8 options after filling first two spots)
/2 (because we need only odd numbers)
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New post Updated on: 01 Aug 2012, 06:39
Here is my solution, using reasoning based on symmetry:

The total number of three digit numbers, greater than 800 with all three digits distinct, is 2 * 9 * 8 = 144.
We have two choices for the first digit (8 or 9), 9 choices for the second digit (it must be different from the first digit) and 8 choices for the third digit, as it should be different from the two previous digits).

Now, here is where, I think, symmetry can help:

Among those starting with 8, there are less even numbers, as the last digit cannot be 8 (so, only 4 choices), while odd choices for the last digit are 5.
For the numbers starting with 9, the situation is reversed, as there are only 4 choices for the third digit for odd numbers and 5 choices for the even numbers.
If we put all the numbers together, at the end, we have a balanced outcome, there must be the same number of each type.
It means that among the above 144 numbers, there are as many even as odd numbers, so 72 of each type.

Therefore, Answer C.

Did you meet questions where some type of reasoning based on symmetry can be used?
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Originally posted by EvaJager on 01 Aug 2012, 03:07.
Last edited by EvaJager on 01 Aug 2012, 06:39, edited 1 time in total.
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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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New post 01 Aug 2012, 06:18
first we consider 1*8*5=40, but doesn't this mean, that among 8 numbers there are 5 odd ones? The restriction in the question is
Quote:
such that all their digits are different

the same is for 1*8*4=32

i solved this problem this way: 1*8*5+1*8*4-2=70, since there is no such an answer among choices went with 72, but could you please help me to understand where i was wrong?
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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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New post 01 Aug 2012, 06:34
Galiya wrote:
first we consider 1*8*5=40, but doesn't this mean, that among 8 numbers there are 5 odd ones? The restriction in the question is
Quote:
such that all their digits are different

the same is for 1*8*4=32

i solved this problem this way: 1*8*5+1*8*4-2=70, since there is no such an answer among choices went with 72, but could you please help me to understand where i was wrong?


The second digit can be odd or even, the sole restriction is just to have all three digits distinct. You are not supposed to subtract the 2.
You have the 8 because the first and the last digits are eliminated, so you can choose from 10 - 2 = 8 possibilities for the second digit.
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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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New post 01 Aug 2012, 06:59
EvaJager
afraid we are talking about different things:
8 includes 5 odd numbers (4 in the second case).
for instance,lets assume the unit digit is 3, we can't consider 3 in tenths - so have to eliminate 1 choice from 8
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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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New post 08 Jul 2013, 00:10
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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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New post 13 Aug 2015, 07:36
dimitri92 wrote:
How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40
B. 60
C. 72
D. 81
E. 104

From 801 to 899, there are 9x8=72 numbers have all their digits different
From 900 to 999, there are 9x8=72 numbers have all their digits different
so, we have 144 number in all and half of them odd.
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Re: How many odd three-digit integers greater than 800 are there  [#permalink]

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