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How many five-digit numbers can be formed using digits

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Intern
Intern
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Joined: 10 Apr 2013
Posts: 13
Location: United States (TX)
Concentration: Finance, Entrepreneurship
GMAT 1: 740 Q50 V41
GPA: 3.3
WE: Analyst (Investment Banking)
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Re: How many five-digit numbers can be formed using digits [#permalink]

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New post 17 Sep 2015, 09:42
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15
B. 96
C. 120
D. 181
E. 216


Key Knowledge:
(Divisibility Rules, Number Theory) math-number-theory-88376.html
For a number to be divisible by 3, the sum of it's digits must be a multiple of 3


Case Method:

Case#1:
\(\sum 0,1,2,4,5 = 12\)

\(5!=120\) = Number of arrangements of 5 digits
Less:
\(4!=24\) = Number of arrangements of 4 digits
(Removes arrangements where 0 is the leading digit (i.e. fix 0 as leading, arrange the remaining 4 digits 4! ways))

\(5!-4!=96\)

Case #2:
\(\sum 1,2,3,4,5 = 15\)

\(5!=120\) = Number of arrangements of 5 digits

\(120+96=216\)
Senior Manager
Senior Manager
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G
Joined: 31 Jul 2017
Posts: 341
Location: Malaysia
WE: Consulting (Energy and Utilities)
Re: How many five-digit numbers can be formed using digits [#permalink]

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New post 26 Jan 2018, 22:41
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15
B. 96
C. 120
D. 181
E. 216


As the property for divisibility by 3 states that sum of all digits a number should be Divisible by 3. So, from the above sets only two sets is possible {1,2,3,4,5} & {0,1,2,4,5}.
\(Set A = 5! = 120\), \(Set B = 96\). \(Hence, 216.\)
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Re: How many five-digit numbers can be formed using digits   [#permalink] 26 Jan 2018, 22:41

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