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How many 4-digit numbers begin with an even digit and end

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Director
Joined: 28 Oct 2003
Posts: 503
Location: 55405
Followers: 1

Kudos [?]: 10 [0], given: 0

How many 4-digit numbers begin with an even digit and end  [#permalink]  12 Dec 2003, 21:41
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:55) correct 0% (00:00) wrong based on 2 sessions
How many 4-digit numbers begin with an even digit and end
with an odd digit?

A 2000
B 2001
C 2499
D 2500
E 2501
Senior Manager
Joined: 12 Oct 2003
Posts: 251
Location: USA
Followers: 1

Kudos [?]: 5 [0], given: 0

A

Director
Joined: 13 Nov 2003
Posts: 971
Location: Florida
Followers: 1

Kudos [?]: 31 [0], given: 0

9000-1000 = 8000, in total.

numbers starting with odd and ending even = 2500
numbers starting with odd and ending odd = 2500
left us with 4000
numbers starting with even and ending even = 2000
numbers starting with even and ending odd = 2000
Senior Manager
Joined: 12 Oct 2003
Posts: 251
Location: USA
Followers: 1

Kudos [?]: 5 [0], given: 0

is this the optimum approach dj? Aren't you solving potentially 3 questions to come up with the ans for a 4th one?

Also I did not understand "left us with 4000 " .. you are saying total = 8000; first two cases sum up to 5000 .. so ..?

my approach ...
first digit can be any of 4 = 2,4,6,8
2nd and 3rd = 10 each
4th can be any of 5 = 1,3,5,7,9

total = 4 * 10 * 10 * 5 = 2000

Stoolfi's approach (explained in another problem of the same type) is probably much faster for similar problems for GMAT

Any thoughts?
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

Re: PS [#permalink]  12 Dec 2003, 22:50
stoolfi wrote:
How many 4-digit numbers begin with an even digit and end
with an odd digit?

A 2000
B 2001
C 2499
D 2500
E 2501

10 digits... 5 even and 5 odd

first digit can be selected in 4 ways ( 0 cant be considered)

second and third in 10 ways each ...

the fourth in 5 ways

total = 4*10*10*5 = 2000
Re: PS   [#permalink] 12 Dec 2003, 22:50
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