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

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Director
Joined: 14 Jan 2007
Posts: 787
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Kudos [?]: 32 [0], given: 0

How many odd three digit integers greater than 800 are there [#permalink]  02 Jun 2007, 15:37
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How many odd three digit integers greater than 800 are there such that all their digits are different?

a. 40
b. 56
c. 72
d. 81
e. 104

Please suggest systematic and quick approach.
CEO
Joined: 17 May 2007
Posts: 2998
Followers: 48

Kudos [?]: 391 [0], given: 210

I solved it systematically but it was anything but quick (especially trying to do it properly under pressure).

Between 800 and 999 there are 200 integers out of which 100 are odd (any answer greater than this can be eliminated straight away). Leaves us with A , B , C , D

Out of this all the ones with 2 digits are either of the following :
811,833,855,...,911,933...999 --> gives us 10 numbers
and
909,919,929,939...989--> gives us 9 numbers (not not counting 999 cos counted it already)

So we can eliminate 19 more numbers

So now its a simple matter of subtracting 19 from 100 leaving us with the answer of 81.

I dont know if I could do it under time pressure though

Ans : D
Director
Joined: 13 Mar 2007
Posts: 555
Schools: MIT Sloan
Followers: 4

Kudos [?]: 3 [0], given: 0

we are considering numbers 801-999 => 100 odd integers

For integers to be different from each other:

801 - 899

(H) - One choice - 8
(T) - eight choices
(U) - five choices (any of the 5 odd numbers)

hence total = 1 x 8 x 5 = 40

900 - 999

(H) - one choice - 9
(T) - eight choices
(U) - 4 choices (9 already gone in 100s place)

hence total = 1 x 8 x 4 = 32

Total = 40 + 32 = 72
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