info@clay6.com
+91-9566306857
(9am to 6pm)
logo

Ask Questions, Get Answers

X
Want help in doing your homework? We will solve it for you. Click to know more.
 

The number of 4 digit numbers divisible by 3 that can be formed by 4 different even digits (0,2,4,5,6,8) is ?

$\begin{array}{1 1} 18 \\ 24 \\ 36 \\ 48 \end{array}$

1 Answer

Need homework help? Click here.
Since the number is divisible by 3, the sum of the digits should be multiple of 3.
$\therefore$ The digits can be $0,2,4,6\: or \: 0,4,6,8$
Hence the required no. of numbers = $(4!-3!)+(4!-3!)=36$
(0 cannot come in $1^{st}$ position.)
answered Sep 3, 2013 by rvidyagovindarajan_1
 

Related questions

...