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How many different three-digit multiples of 5 can be

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Director
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How many different three-digit multiples of 5 can be [#permalink] New post 28 Oct 2007, 12:19
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

75% (01:52) correct 25% (00:50) wrong based on 57 sessions
How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3
B. 6
C. 10
D. 12
E. 18
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Dec 2013, 00:58, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Director
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 [#permalink] New post 28 Oct 2007, 12:31
never mind i think i got it.

since we can only use 2,3,4, and 5. units digit must be 5

so for possible combinations of xy5

it would be 235, 325, 245, 425, 345, 435

answer 6
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Re: counting multiple [#permalink] New post 19 Nov 2007, 08:37
GMAT TIGER wrote:
beckee529 wrote:
How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

3
6
10
12
18


= 3c2 x 2 = 6


can you explain your method. i solved it differently

XYZ, where Z must be 5. therefore 1 variation of digit in Z.
Y can be any of the 3 possible choices.
X can be any of the 2 possible choices.

2+3+1= 6
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Re: How many different three-digit multiples of 5 can be [#permalink] New post 16 Dec 2013, 19:52
I don't get why the combination is multiplied by two? i solved by writing out the different possibilities
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Re: How many different three-digit multiples of 5 can be [#permalink] New post 16 Dec 2013, 20:46
Expert's post
legitpro wrote:
I don't get why the combination is multiplied by two? i solved by writing out the different possibilities


Available: 2, 3, 4, 5

You need to make 3 digit numbers which are multiples of 5.
___ ___ ___

For the number to be a multiple of 5, it must end with either 0 or 5. We only have 5 available so the number must end with 5.

___ ___ 5

Now you select 2 numbers out of the leftover 3 numbers (2, 3 and 4) in 3C2 ways and arrange them in the 2 places in 2! ways which gives you 3C2*2! = 6

or you can say that you can select a number for the hundreds place in 3 ways and select a number for the tens place in 2 ways (since numbers cannot be repeated). So you can make the number in 3*2 = 6 ways
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Re: How many different three-digit multiples of 5 can be   [#permalink] 16 Dec 2013, 20:46
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