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# Question from m10 q13

Author Message
Intern
Joined: 18 Jan 2010
Posts: 17
Followers: 0

Kudos [?]: 6 [0], given: 20

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06 Jul 2010, 18:39
How many odd three-digit integers greater than 800 are there such that all their digits are different?

40
56
72
81
104

Manager
Joined: 16 Apr 2010
Posts: 221
Followers: 4

Kudos [?]: 120 [1] , given: 12

Re: Question from m10 q13 [#permalink]

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06 Jul 2010, 22:52
1
KUDOS
Hi,

I think the answer is 72.
For 800:
if the ten's digit is even, we will have 4X5 odd numbers
if the ten's digit is odd, we will have 5X4 odd numbers

For 900:
if the ten's digit is even, we will have 5X4 odd numbers
if the ten's digit is odd, we will have 4X3 odd numbers

So in total we will have 72 odd numbers.

To explain the above numbers, consider:
For 800, if the ten's digit is even, we will have 4X5 odd numbers
we will have
801, 803, 805, 807, 809
821, 823, 825, 827, 829
841, 843, 845, 847, 849
861, 863, 865, 867, 869
881, 883, 885, 887, 889 cannot be considered

For 900:
if the ten's digit is odd, we will have 4X3 odd numbers
913, 915, 917,
931, 935, 937,
951, 953, 957,
971, 973, 975,

regards,
Jack
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 911 [0], given: 334

Re: Question from m10 q13 [#permalink]

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07 Jul 2010, 00:03
See explanations in this thread too:
m10-q13-74071.html
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Re: Question from m10 q13   [#permalink] 07 Jul 2010, 00:03
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# Question from m10 q13

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