If a computer program generates three-digit odd numbers using the numbers 9, 7, 0, 4, and 1, and the digits can be repeated, what is the probability that the number formed from the program is a multiple of three?
A. B.
C.
D.
E. Answers
A) 1/2
B) 1/3
C) 1/5
D) 1/6
E) 1/7
1) Lets first find TOTAL ways..   a) first/leftmost digit can be any of 5 except 0, so 4 ways   b) middle digit can be any of the 5. so 5 ways   c) the units digit can be ONLY odd- 1,7, and 9- so 3 ways   TOTAL = 4*5*3= 
<gadget id="calculator">4*5*3</gadget>
<output>60</output>
60 ways   2) lets now find multiple of 3..   a) Digits not div by 3   7,4 and 1 will each leave remainder of 1 so they can be grouped together to form multiples of 3..   first place - 3 ways, second digit - 3 ways, and units digit ONLY 1 and 7 - 2ways   Total ways= 3*3*2= 
<gadget id="calculator">3*3*2</gadget>
<output>18</output>
18   b) digits div by 3   9 and 0- No other digit can be used along with them as no other pair gives us sum of remainder as 3..   first place - 1 way, second digit - 2 ways, and units digit ONLY 9 - 1 way   Total ways= 1*2*1= 
<gadget id="calculator">1*2*1</gadget>
<output>2</output>
2   TOTAL= 2+18= 
<gadget id="calculator">2+18</gadget>
<output>20</output>
20   Prob=2060=132060=13   ANSWER:1/3
<result>B</result>
If a computer program generates three-digit odd numbers using the numbers 9, 7, 0, 4, and 1, and the digits can be repeated, what is the probability that the number formed from the program is a multiple of three?
A. B.
C.
D.
E.