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# By using the numbers 1, 2, 3, 5 and 7 only once, how many

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Intern
Joined: 28 Jan 2011
Posts: 22

Kudos [?]: 17 [0], given: 15

By using the numbers 1, 2, 3, 5 and 7 only once, how many [#permalink]

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06 Sep 2013, 13:54
1
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Difficulty:

25% (medium)

Question Stats:

78% (01:25) correct 22% (01:42) wrong based on 79 sessions

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By using the numbers 1, 2, 3, 5 and 7 only once, how many five digit numbers can be made that are divisible by 25?

(a) 10
(b) 11
(c) 12
(d) 13
(e) 14
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Sep 2013, 14:31, edited 1 time in total.
RENAMED THE TOPIC.

Kudos [?]: 17 [0], given: 15

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 129067 [1], given: 12194

Re: By using the numbers 1, 2, 3, 5 and 7 only once, how many [#permalink]

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06 Sep 2013, 14:36
1
KUDOS
Expert's post
salsal wrote:
By using the numbers 1, 2, 3, 5 and 7 only once, how many five digit numbers can be made that are divisible by 25?

(a) 10
(b) 11
(c) 12
(d) 13
(e) 14

A number to be divisible by 25 must end with 00, 25, 50, or 75.

So, there are the following cases possible:

137-25 --> the first three digits can be arranged in 3=6 ways.
123-75 --> the first three digits can be arranged in 3=6 ways.

Total = 6 + 6 = 12.

_________________

Kudos [?]: 129067 [1], given: 12194

Manager
Joined: 31 Jul 2013
Posts: 51

Kudos [?]: 29 [2], given: 4

Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.6
WE: Law (Entertainment and Sports)
Re: By using the numbers 1, 2, 3, 5 and 7 only once, how many [#permalink]

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06 Sep 2013, 18:41
2
KUDOS
salsal wrote:
By using the numbers 1, 2, 3, 5 and 7 only once, how many five digit numbers can be made that are divisible by 25?

(a) 10
(b) 11
(c) 12
(d) 13
(e) 14

Given the restrictions, and divisibility by 25, we know that the last digit must be a 5 and the penultimate digit must be either 2 or 7. Setting out the digits in a row to think of possibilities, and going from right to left (most to least restrictive), we have 1 2 3 2 1. 1x2x3x2x1 = 12.

1 option for the last digit (5), 2 options for the penultimate digit (2 or 7), 3 then 2 then 1 option for the remaining 3 digits.

Kudos [?]: 29 [2], given: 4

Manager
Joined: 17 Jul 2013
Posts: 100

Kudos [?]: 7 [0], given: 67

Re: By using the numbers 1, 2, 3, 5 and 7 only once, how many [#permalink]

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14 May 2014, 02:06
To solve this :
A number is divisible by 25 if the last two digits are multiple of 25 i.e 00, 25 , 50 or 75.

Hence from the given digits 7 or 2 must followed by 5.
number of ways we can select numbers of the first digit : 3 (1,3,7/2)
number of ways to select the second digit : 2 (since we can't repeat the digits)
third digit selected : 1
Fourth selection : 2 (2/7)
last digit must be 5 : 1

hence all selection is dependent we need to perform the product of all possible cases : 3*2*1*2*1 = 12

Kudos [?]: 7 [0], given: 67

Re: By using the numbers 1, 2, 3, 5 and 7 only once, how many   [#permalink] 14 May 2014, 02:06
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