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Re: How many odd three-digit integers [#permalink]

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23 Sep 2008, 06:50

fresinha12 wrote:

x2suresh wrote:

elmagnifico wrote:

I will provide the OA, could you please give detailed explanation. Thanks

How many odd three-digit integers greater than 800 are there such that all their digits are different? * 40 * 56 * 72 * 81 * 104

Ans = No.of ways when 100th digit is 8 + no. of ways when 100th digit is 9 = 5*8*1 + 4*8u cant have 8 possibilities??*1 =72

Why not.. for odd three digit integer... only unit digit has to be 'odd' and tenth digit can be any digit from 0...9 ( but we have to exclude two digits.. because repetition not allowed).
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Re: How many odd three-digit integers [#permalink]

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24 Sep 2008, 04:19

x2suresh wrote:

elmagnifico wrote:

I will provide the OA, could you please give detailed explanation. Thanks

How many odd three-digit integers greater than 800 are there such that all their digits are different? * 40 * 56 * 72 * 81 * 104

Ans = No.of ways when 100th digit is 8 + no. of ways when 100th digit is 9 = 5*8*1 + 4*8*1 =72

But why the second time we have 4*8, where the 8 come from and why 4. can you list the numbers you are using for the 800 and 900. this way we can really see what is happening here.

Re: How many odd three-digit integers [#permalink]

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24 Sep 2008, 08:12

elmagnifico wrote:

Twoone wrote:

Hey,

it is because 9 is a odd number and it can't come at the unit digit as it is taking place at 100th digit and all digit has to be different.

So I also go with 72.

but 909 is an even integer!, why all digits have to be different??

please read question carefully.. you missed the below part. How many odd three-digit integers greater than 800 are there such that all their digits are different?
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Re: How many odd three-digit integers [#permalink]

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24 Sep 2008, 09:39

Could someone please explain about the 900s part of the problem? I see how it would be 1x8x5 for the 800s, but I keep wanting to do 1x9x4. I see the possible numbers for the tens spot to be 0 through 8.

Re: How many odd three-digit integers [#permalink]

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24 Sep 2008, 12:38

mdavis wrote:

Could someone please explain about the 900s part of the problem? I see how it would be 1x8x5 for the 800s, but I keep wanting to do 1x9x4. I see the possible numbers for the tens spot to be 0 through 8.

What am I doing wrong?

1x9x4. unit digit must be any of these 1,3,5,7 but what if any of these numbers are repeated in tenth digit.)

you can split the solution for 900 part: = when tenth digit is odd + when tenth digit is even = 1*4*3+1*5*4 =32 or other solution = No of ways 100th digit * No of ways for unit digit * No of ways for 10th digit = 1 (only 9 possible) * 4 (1,3,5,7 possible values)* 8 (0 to 9 exlcude 100th and unit digit number to avoid repetition) = 32
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Re: How many odd three-digit integers [#permalink]

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25 Sep 2008, 09:46

x2suresh wrote:

mdavis wrote:

Could someone please explain about the 900s part of the problem? I see how it would be 1x8x5 for the 800s, but I keep wanting to do 1x9x4. I see the possible numbers for the tens spot to be 0 through 8.

What am I doing wrong?

1x9x4. unit digit must be any of these 1,3,5,7 but what if any of these numbers are repeated in tenth digit.)

you can split the solution for 900 part: = when tenth digit is odd + when tenth digit is even = 1*4*3+1*5*4 =32 or other solution = No of ways 100th digit * No of ways for unit digit * No of ways for 10th digit = 1 (only 9 possible) * 4 (1,3,5,7 possible values)* 8 (0 to 9 exlcude 100th and unit digit number to avoid repetition) = 32

x2suresh thanks alot. i am getting it and i think everyone else is also getting this question. However, WAIT A MINUTE here. lets take No of ways for 800. for the unit digit we can have only 1,3,5,7,9 these are the only one that will give us odd. for the tenth digit we can have only 0,2,4,6, (we cannot have 8,1,3,5,7,9, b/c they are already in the hundred of unit digit. ) so we have 1*5*4 = 20!!!! the same logic for the 900......... What is going on here!

Re: How many odd three-digit integers [#permalink]

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25 Sep 2008, 09:53

elmagnifico wrote:

x2suresh wrote:

mdavis wrote:

Could someone please explain about the 900s part of the problem? I see how it would be 1x8x5 for the 800s, but I keep wanting to do 1x9x4. I see the possible numbers for the tens spot to be 0 through 8.

What am I doing wrong?

1x9x4. unit digit must be any of these 1,3,5,7 but what if any of these numbers are repeated in tenth digit.)

you can split the solution for 900 part: = when tenth digit is odd + when tenth digit is even = 1*4*3+1*5*4 =32 or other solution = No of ways 100th digit * No of ways for unit digit * No of ways for 10th digit = 1 (only 9 possible) * 4 (1,3,5,7 possible values)* 8 (0 to 9 exlcude 100th and unit digit number to avoid repetition) = 32

x2suresh thanks alot. i am getting it and i think everyone else is also getting this question. However, WAIT A MINUTE here. lets take No of ways for 800. for the unit digit we can have only 1,3,5,7,9 these are the only one that will give us odd. for the tenth digit we can have only 0,2,4,6, (we cannot have 8,1,3,5,7,9, b/c they are already in the hundred of unit digit. ) so we have 1*5*4 = 20!!!! the same logic for the 900......... What is going on here!

unit digit you can fill only one odd value from the list 1,3,5,7,9..( not all.)

say if unit digit is 7 .. 1,3,5,7 are available to fill in tenth digit. Is it clear?
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Re: How many odd three-digit integers [#permalink]

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25 Sep 2008, 10:02

1

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elmagnifico wrote:

x2suresh thanks alot. i am getting it and i think everyone else is also getting this question. However, WAIT A MINUTE here. lets take No of ways for 800. for the unit digit we can have only 1,3,5,7,9 these are the only one that will give us odd. for the tenth digit we can have only 0,2,4,6, (we cannot have 8,1,3,5,7,9, b/c they are already in the hundred of unit digit. ) so we have 1*5*4 = 20!!!! the same logic for the 900......... What is going on here!

In 800 series, unit digit can be one of (1,3,5,7,9). I will break-up tens digit into two parts - part1 when tens digit is one of (1,3,5,7,9). In this case, units digit will have 5 different ways and tens digit will have only 4 different ways (as digits have to be unique. Hence, total number for this part = 5*4 = 20. (of course hundreds digit will be constant 8.

For part 2, when tens digit is one of (0,2,4,6,8), it cannot have 8 as hundred's place is already occupying 8. Hence, there are only 4 ways for tens place. Hence, total number for this part = 5*4 = 20

Thus, total possible number in 800 series = 20 + 20 = 40.

Similarly, in 900 series, unit digit can be one of (1,3,5,7) as hundred's place is already occupied by 9. Now for part 1 of tens place, ways will be 3 and hence total number will be 4*3 = 12.

For part 2, ways for tens place will be 5 and hence total number will be 4*5 = 20

Thus, total possible number for 900 series = 12 + 20 = 32