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Re: By using the numbers 1, 2, 3, 5 and 7 only once, how many [#permalink]
06 Sep 2013, 17:41

2

This post received 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 12321. 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.

Re: By using the numbers 1, 2, 3, 5 and 7 only once, how many [#permalink]
14 May 2014, 01:06

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

gmatclubot

Re: By using the numbers 1, 2, 3, 5 and 7 only once, how many
[#permalink]
14 May 2014, 01:06

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