anirudh alameluvari
Last Activity: 16 Years ago
FOR A NUMBER TO BE DIVISIBLE BY 3 THE SUM OF ITS DIGITS SHOULD BE DIVISIBLE BY 3 . SO, AMONG THE 6 DIGITS , 5 DIGITS
WHOSE SUM IS DIVISIBLE BY 3 MUST BE SELECTED.
0+1+2+3+4+5 = 15 ( a multiple of 3)
SO THE FIVE DIGITS CAN BE OBTAINED BY ELIMINATING A 3 MULTIPLE .
15 - 3q = 3(5 - q) . SO IT IS DIVISIBLE BY 3.
AMONG 0,1,2,3,4,5 ; 0 AND 3 ARE THE MULTIPLES OF 3 .
SO, THE 5 DIGITS MUST BE (1,2,3,4,5) OR (0,1,2,4,5)
FIRST LET US TAKE (1,2,3,4,5) . ACCORDING TO PERMUTATION LAW , IT CAN BE ARRANGED IN 5p5 WAYS
5P5 = 5 x 4 x 3 x 2 x 1 = 120
(0,1,2,4,5) . IF ZERO IS THE FIRST DIGIT IT BECOMES A 4 DIGIT NUMBER .
NUMBER OF NUMBERS HAVING 0 AS FIRST DIGIT ARE
4 x 3 x 2 x 1 = 24
SO, NUMBER OF 5 DIGIT NUMBER FORMED FROM (0,1,2,4,5) ARE 120 - 24 = 96.
TOTAL NUMBER OF 5 DIGIT NUMBER FORMED FROM (0, 1,2,3,4,5) ARE 120 + 96 = 216.