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How many odd numbers less than 5000 can be formed using the

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VP
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How many odd numbers less than 5000 can be formed using the [#permalink] New post 10 Nov 2007, 12:12
How many odd numbers less than 5000 can be formed using the digits 0 1 2 3 4 5 6?
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 [#permalink] New post 10 Nov 2007, 12:25
P = 7/10*7/10*3/10 = 147/1000

N = 5000

P*N = 147/1000*5000 = 147*5 = 735

:)
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 [#permalink] New post 10 Nov 2007, 12:31
KillerSquirrel wrote:
P = 7/10*7/10*3/10 = 147/1000

N = 5000

P*N = 147/1000*5000 = 147*5 = 735

:)


KS, where is the thousands digit?
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 [#permalink] New post 10 Nov 2007, 12:39
Ravshonbek wrote:
KillerSquirrel wrote:
P = 7/10*7/10*3/10 = 147/1000

N = 5000

P*N = 147/1000*5000 = 147*5 = 735

:)


KS, where is the thousands digit?


Hello - Ravshonbek

P = 5/10*7/10*7/10*3/10 = 735/10000

N = 10,000

N*P = 10000*735/10000 = 735

The same answer !

What is the OA ?

:)

Last edited by KillerSquirrel on 10 Nov 2007, 21:32, edited 2 times in total.
VP
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 [#permalink] New post 10 Nov 2007, 12:44
thanks KS.

here is the different method that might take longer to go to the end but ur proob method is better i guess.

4 digit odd numbers less than 5000 - 4*7*7*3=588
3 digit odd numbers - 6*7*3=126
2 digit odd numbers - 6*3=18
1 digit odd numbers - 3=3

588+126+18+3=735
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 [#permalink] New post 10 Nov 2007, 12:45
Ravshonbek wrote:
thanks KS.

here is the different method that might take longer to go to the end but ur proob method is better i guess.

4 digit odd numbers less than 5000 - 4*7*7*3=588
3 digit odd numbers - 6*7*3=126
2 digit odd numbers - 6*3=18
1 digit odd numbers - 3=3

588+126+18+3=735


that's a tough problem - I think this is out of GMAT scope. Your approach is valid - use the best method for you.

:)
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 [#permalink] New post 14 Nov 2007, 18:08
KS, why wouldn't the prob of the thousands digit be 5/9 ? Because 0 cannot be chosen here. I've seen this applied to other problems as well. Thoughts?

Thanks
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 [#permalink] New post 14 Nov 2007, 21:07
KillerSquirrel wrote:
Ravshonbek wrote:
KillerSquirrel wrote:
P = 7/10*7/10*3/10 = 147/1000

N = 5000

P*N = 147/1000*5000 = 147*5 = 735

:)


KS, where is the thousands digit?


Hello - Ravshonbek

P = 5/10*7/10*7/10*3/10 = 735/10000

N = 10,000


N*P = 10000*735/10000 = 735

The same answer !

What is the OA ?

:)


KS: Can you please explain how you get N = 10,000? I thought N = 5000? Thanks.
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 [#permalink] New post 14 Nov 2007, 21:09
yuefei wrote:
KS, why wouldn't the prob of the thousands digit be 5/9 ? Because 0 cannot be chosen here. I've seen this applied to other problems as well. Thoughts?

Thanks


If N=10,000 then the P for the thousands digit is 5/10 , since 00001 is a valid number (equal to 1).

If N=5000 then you don't need the P of the thousands digit at all (since its equal to 1).

Can you give me an example for other the problems you saw ?

:)
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 [#permalink] New post 14 Nov 2007, 21:46
KillerSquirrel wrote:
yuefei wrote:
KS, why wouldn't the prob of the thousands digit be 5/9 ? Because 0 cannot be chosen here. I've seen this applied to other problems as well. Thoughts?

Thanks


If N=10,000 then the P for the thousands digit is 5/10 , since 00001 is a valid number (equal to 1).

If N=5000 then you don't need the P of the thousands digit at all (since its equal to 1).

Can you give me an example for other the problems you saw ?

:)


I'm still lost...i still dont understand why n=10,000 with working out the problem with the thousandth digit. Killer Squirrel, you said If N=10,000 then the P for the thousands digit is 5/10 but i thought N=5000 with the 5/10.

Sorry mate! if you dont mind could you (or anyone else for that matter) spell this out nice and easy for a poor fool like myself?

thanks!
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 [#permalink] New post 15 Nov 2007, 05:52
KS, here's an example from Challenge M03

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
---------
My solution, using the probability method:

N = 1000 - 100 = 900
P = (8/9)*(9/10)*(4/10) = 288/900

N*P = 900*(288/900) = 288 D.
----------

For the hundreds digit I've used 8/9. Wouldn't you use the same thinking for this question and use 5/9 ?
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 [#permalink] New post 15 Nov 2007, 11:48
yuefei wrote:
KS, here's an example from Challenge M03

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
---------
My solution, using the probability method:

N = 1000 - 100 = 900
P = (8/9)*(9/10)*(4/10) = 288/900

N*P = 900*(288/900) = 288 D.
----------

For the hundreds digit I've used 8/9. Wouldn't you use the same thinking for this question and use 5/9 ?


This problem can be solved in this way :

N = 1000

P = 8/10*9/10*4/10 = 288/1000

N*P = 288

or it can be solve as you solved it.

The main difference in this problem from the problem we discussed earlier is that in this case you cannot use P = 1 for the hundreds digit since this problem don't allow you to use five in the hundreds digit.

If you could use five then the P for the hundreds digit was 9/9 = 1 and you could just ignore it (the only number you can't choose is 0).

:)

Last edited by KillerSquirrel on 15 Nov 2007, 11:55, edited 2 times in total.
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 [#permalink] New post 15 Nov 2007, 11:51
robertrdzak wrote:
KillerSquirrel wrote:
yuefei wrote:
KS, why wouldn't the prob of the thousands digit be 5/9 ? Because 0 cannot be chosen here. I've seen this applied to other problems as well. Thoughts?

Thanks


If N=10,000 then the P for the thousands digit is 5/10 , since 00001 is a valid number (equal to 1).

If N=5000 then you don't need the P of the thousands digit at all (since its equal to 1).

Can you give me an example for other the problems you saw ?

:)


I'm still lost...i still dont understand why n=10,000 with working out the problem with the thousandth digit. Killer Squirrel, you said If N=10,000 then the P for the thousands digit is 5/10 but i thought N=5000 with the 5/10.

Sorry mate! if you dont mind could you (or anyone else for that matter) spell this out nice and easy for a poor fool like myself?

thanks!


Read the following attachment - If you are still lost PM me.
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A Probability Approach For Solving Counting Problems.pdf [69.11 KiB]
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To download please login or register as a user

VP
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 [#permalink] New post 15 Nov 2007, 12:00
GK_Gmat wrote:
KillerSquirrel wrote:
Ravshonbek wrote:
KillerSquirrel wrote:
P = 7/10*7/10*3/10 = 147/1000

N = 5000

P*N = 147/1000*5000 = 147*5 = 735

:)


KS, where is the thousands digit?


Hello - Ravshonbek

P = 5/10*7/10*7/10*3/10 = 735/10000

N = 10,000


N*P = 10000*735/10000 = 735

The same answer !

What is the OA ?

:)


KS: Can you please explain how you get N = 10,000? I thought N = 5000? Thanks.


N can be either 10,000 or 5,000 but you need to adjust the P accordingly.

:)
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Re: counting basics [#permalink] New post 15 Nov 2007, 12:05
Ravshonbek wrote:
How many odd numbers less than 5,000 can be formed using the digits 0 1 2 3 4 5 6?


i do this way:


thousands place = 5 integers 1, 2, 3, 4, and 5
hundreds place = 7 integers 0, 1, 2, 3, 4, 5, and 6
Tens place = 7 integers 0, 1, 2, 3, 4, 5, and 6
units place = 3 integers 1, 3, and 5

so the possibilities are: 5 x 7 x 7 x 3 = 735
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Re: counting basics [#permalink] New post 15 Nov 2007, 12:20
GMAT TIGER wrote:
Ravshonbek wrote:
How many odd numbers less than 5,000 can be formed using the digits 0 1 2 3 4 5 6?


i do this way:


thousands place = 5 integers 1, 2, 3, 4, and 5 hundreds place = 7 integers 0, 1, 2, 3, 4, 5, and 6
Tens place = 7 integers 0, 1, 2, 3, 4, 5, and 6
units place = 3 integers 1, 3, and 5

so the possibilities are: 5 x 7 x 7 x 3 = 735


Don't you mean 0,1,2,3,4 ? since if thousands place is 5 and the hundreds place is 6 then the number cannot be less then 5,000.

:?
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Re: counting basics [#permalink] New post 15 Nov 2007, 12:39
KillerSquirrel wrote:
GMAT TIGER wrote:
Ravshonbek wrote:
How many odd numbers less than 5,000 can be formed using the digits 0 1 2 3 4 5 6?


i do this way:

thousands place = 5 integers 1, 2, 3, 4, and 5
hundreds place = 7 integers 0, 1, 2, 3, 4, 5, and 6
Tens place = 7 integers 0, 1, 2, 3, 4, 5, and 6
units place = 3 integers 1, 3, and 5

so the possibilities are: 5 x 7 x 7 x 3 = 735


Don't you mean 0,1,2,3,4 ? since if thousands place is 5 and the hundreds place is 6 then the number cannot be less then 5,000.

:?


thats a smart catch as 5 is not possible. thanks man!!! 8-) 8-) 8-)

thousands place = 5 integers 0, 1, 2, 3, and 4
hundreds place = 7 integers 0, 1, 2, 3, 4, 5, and 6
Tens place = 7 integers 0, 1, 2, 3, 4, 5, and 6
units place = 3 integers 1, 3, and 5

so the possibilities are: 5 x 7 x 7 x 3 = 735
Re: counting basics   [#permalink] 15 Nov 2007, 12:39
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