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

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

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20 Oct 2007, 16:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many odd three-digit integers greater than 800 are there such that all their digits are different?

40
56
72
81
104

please explain your logic. in particular, if we can calculate for it backwards: 200 (total digits between 800 and 999) - # of digits that are the same. i got it right but my understanding of these problems not that strong still. thanks!
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20 Oct 2007, 18:23
beckee529 wrote:
How many odd three-digit integers greater than 800 are there such that all their digits are different?

40
56
72
81
104

please explain your logic. in particular, if we can calculate for it backwards: 200 (total digits between 800 and 999) - # of digits that are the same. i got it right but my understanding of these problems not that strong still. thanks!

1: 800-899:

even integers in tens digits:
hundred = only one integer i.e 8
tens (all even integers except 8: 0, 2, 4, 6 ) = 4 places
unit (all odd integers: 1, 3, 5, 7, 9) = 5 places
so the number of integers = 1x4x5 = 20

odd integers in tens digits:
hundred = only one integer i.e 8
tens (all odd integers: 1, 3, 5, 7, 9) = 5 places
unit (all odd integers: only 4 out of 5 odd integers that is not in tens digit) = 4 places
so the number of integers = 1x5x4 = 20

so total = 40

2: 901-999
similarly,

all even integers in tens digits:
hundred = only one integer i.e 9
tens (all even integers: 0, 2, 4, 6, 8 ) = 5 places
unit (all odd integers except 9: 1, 3, 5, 7) = 4 places
so the number of integers = 1x5x4 = 20

all odd integers in tens digits:
hundreds = only one integer i.e 9
tens (all odd integers except 9: 1, 3, 5, 7) = 4 places
unit (all odd integers except 9 and one more odd integer) = 3 places
so the number of integers = 1x4x3 = 12

so total = 20+20+20+12 = 72

in short: 2(1x4x5) + (1x5x4) + (1x3x4) = 72
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20 Oct 2007, 20:28
i too got 72..

here is how

last term=first term + (n-1)d

where d=2 last term =999 first term =801

999-801=2n-2

198+2=2n

n=100

ok...so there 100 odd number

now we need to elimnate ones where numbers are repeated

801 803 805 807 808 809.. we need to get rid of 808..

there are 10 such numbers from 800-900

then we need to get rid of 881 883...there are 5 such number

for 900-999

901 903 905 907 909..again we need to get rid of 10 such numbers

then 919 929 939 949 959 969 979 989 999 needs to be elimnated 8 such numbers

then we have another 5 number 991 993 995 997 999

we need to get rid of 28 numbers in all

100-28=72
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21 Oct 2007, 00:12
N = 1,000 – 800 = 200

P = 1*9/10*8/10 = 72/100

N*P = 200*72/100 = 144

but we need only the odd numbers (i.e. numbers that end with an odd digit) so (5/10)*144 = 72

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21 Oct 2007, 06:50
brilliant.. thanks so much - the OA is indeed C) 72
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21 Oct 2007, 08:55
[quote="KillerSquirrel"]N = 1,000 – 800 = 200

P = 1*9/10*8/10 = 72/100

N*P = 200*72/100 = 144

but we need only the odd numbers (i.e. numbers that end with an odd digit) so (5/10)*144 = 72

[/quote]

I understand [b]N[/b] ~ numbers of three-digit numbers greater than 800
[b]P [/b]~ probability of these numbers with different digits
I dont understand the part you got [b]N*P[/b] . Could you plz explain??

thanks
alohagirl
[/b]
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21 Oct 2007, 11:03
alohagirl wrote:
KillerSquirrel wrote:
N = 1,000 – 800 = 200

P = 1*9/10*8/10 = 72/100

N*P = 200*72/100 = 144

but we need only the odd numbers (i.e. numbers that end with an odd digit) so (5/10)*144 = 72

I understand N ~ numbers of three-digit numbers greater than 800
P ~ probability of these numbers with different digits
I dont understand the part you got N*P . Could you plz explain??

thanks
alohagirl

see here:

http://www.gmatclub.com/forum/t52697

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04 Nov 2007, 17:44
why cant we solve the question like this? it yields the wrong answer though..

controlling for hundreds digit: 8,9. Therefore 2 possible choices.
controlling for tens digit: 8. Tens digit cannot be whatever hundreds digit is.
controlling for units digit: 5 odd integers.

=2*8*5
=2*40
=80

subtract repeats 808,818 etc...10 of these numbers.
subtract repeats 909, 919, 10 of these

therefore 60.
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05 Nov 2007, 05:44
controlling for hundreds digit: Take 8
controlling for tens digit: 8 possibilities. Tens digit cannot be whatever hundreds or tens digit is.
controlling for units digit: 5 odd integers.
= 8*5 = 40

controlling for hundreds digit: Take 9
controlling for tens digit: 8 possibilities. Tens digit cannot be whatever hundreds digit or tens is.
controlling for units digit: 4 odd integers. (cannot be 9)
= 8*4 = 32

Total = 40 +32 = 72
05 Nov 2007, 05:44
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