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How many odd numbers less than 5000 can be formed using the [#permalink]
 10 Nov 2007, 13:12
How many odd numbers less than 5000 can be formed using the digits 0 1 2 3 4 5 6?
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VP
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P = 7/10*7/10*3/10 = 147/1000
N = 5000
P*N = 147/1000*5000 = 147*5 = 735
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VP
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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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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, 22:32, edited 2 times in total.
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VP
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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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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
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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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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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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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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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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
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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.
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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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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.
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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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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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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]
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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Re: counting basics [#permalink]
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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Re: counting basics [#permalink]
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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