A five-digit number divisible by 3 is to be formed using numerical 0, 1, 2, 3, 4 and 5 without repetition. The total number R of ways this can be done is: Answers
A) 122  B) 210  C) 216  D) 217  E) 225
We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.   We have six digits: 0, 1, 2, 3, 4, 5. Their sum=15.   For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.   Second step:   We have two set of numbers:   1, 2, 3, 4, 5 and 0, 1, 2, 4, 5. How many 5 digit numbers can be formed using these two sets:   1, 2, 3, 4, 5 --&gt; 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.   0, 1, 2, 4, 5 --&gt; here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --&gt; 5!-4!=4!(5-1)=4!*4=96   120+96= 
<gadget id="calculator">120+96</gadget>
<output>216</output>
216=R   Answer: 216.
<result>C</result>
A five-digit number divisible by 3 is to be formed using numerical 0, 1, 2, 3, 4 and 5 without repetition. The total number R of ways this can be done is: