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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]

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10 Nov 2007, 13:12
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many odd numbers less than 5000 can be formed using the digits 0 1 2 3 4 5 6?

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VP
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10 Nov 2007, 13:25
P = 7/10*7/10*3/10 = 147/1000

N = 5000

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

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VP
Joined: 09 Jul 2007
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Location: London

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10 Nov 2007, 13:31
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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VP
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10 Nov 2007, 13:39
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

What is the OA ?

Last edited by KillerSquirrel on 10 Nov 2007, 22:32, edited 2 times in total.

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VP
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Location: London

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10 Nov 2007, 13:44
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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VP
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10 Nov 2007, 13:45
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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Manager
Joined: 02 Aug 2007
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14 Nov 2007, 19:08
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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Director
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14 Nov 2007, 22:07
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

What is the OA ?

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

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VP
Joined: 08 Jun 2005
Posts: 1144

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14 Nov 2007, 22:09
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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Manager
Affiliations: CFA L3 Candidate, Grad w/ Highest Honors
Joined: 03 Nov 2007
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Location: USA
Schools: Chicago Booth R2 (WL), Wharton R2 w/ int, Kellogg R2 w/ int
WE 1: Global Operations (Futures & Portfolio Financing) - Hedge Fund (\$10bn+ Multi-Strat)
WE 2: Investment Analyst (Credit strategies) - Fund of Hedge Fund (\$10bn+ Multi-Strat)

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14 Nov 2007, 22:46
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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Manager
Joined: 02 Aug 2007
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15 Nov 2007, 06:52
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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VP
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15 Nov 2007, 12:48
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, 12:55, edited 2 times in total.

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VP
Joined: 08 Jun 2005
Posts: 1144

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15 Nov 2007, 12:51
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.
Attachments

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VP
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15 Nov 2007, 13:00
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

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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SVP
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15 Nov 2007, 13:05
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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VP
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15 Nov 2007, 13:20
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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SVP
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15 Nov 2007, 13:39
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!!!

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

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Re: counting basics   [#permalink] 15 Nov 2007, 13:39
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