How many even 4-digit numbers can be formed, so that the numbers are divisible by 4 and no two digits are repeated?	Pick one.
A) 336	B) 784	C) 1120	D) 1804	E) 1936
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 range4+1=25last multiple in the range − first multiple in the range4+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= 
<gadget id="calculator"> 6*56</gadget>
<output>336</output>
336.   If there is no 0 in the ending (total 22 - 6= 
<gadget id="calculator">22 - 6</gadget>
<output>16</output>
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= 
<gadget id="calculator">7*7</gadget>
<output>49</output>
49 choices total. As there are 16 numbers without zero in ending, hence total  16*49= 
<gadget id="calculator"> 16*49</gadget>
<output>784</output>
784.   TOTAL:  336+784= 
<gadget id="calculator"> 336+784</gadget>
<output>1_120</output>
1120   Answer: 1120.
<result>C</result>
How many even 4-digit numbers can be formed, so that the numbers are divisible by 4 and no two digits are repeated?