How many even 4-digit numbers can be formed, so that the numbers are divisible by 4 and no two digits are repeated?	Choose the correct option:
A) 336
B) 784
C) 1120
D) 1804
E) 1936

Rationale: number is divisible by 4 if the last two digits form a number divisible by 4 . therefore last two digits can be : 00 ; 04 ; 08 ; 12 16 ; . . . 96 . basically multiples of 4 in the range 0 - 96 , inclusive . multiples of 4 in a range 0 - 96 , inclusive are last multiple in the range − first multiple in the range 4 + 1 = 25 last multiple in the range − first multiple in the range 4 + 1 = 25 but 3 numbers out of these 25 are not formed with distinct digits : 00 , 44 , and 88 . hence the numbers we are looking for can have only 22 endings . if there is 0 in the ending ( 04 , 08 , 20 , 40 , 60 , 80 - total 6 such numbers ) , then the first and second digit can take 8 and 7 choices each = 56 choices total . as there are 6 numbers with 0 in ending , hence total 6 * 56 = 336 . if there is no 0 in the ending ( total 22 - 6 with zero = 16 such numbers ) , then the first digit can take 7 choices ( 10 - 2 digits in the ending and zero , so total 3 digits = 7 , as 4 - digit number can not start with zero ) and the second digit can take 7 choices too ( 10 digits - 3 digits we ' ve already used ) = 7 * 7 = 49 choices total . as there are 16 numbers without zero in ending , hence total 16 * 49 = 784 . total : 336 + 784 = 1120 answer : c .
Answer: C