Bunuel
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. 1/2
B. 1/3
C. 1/5
D. 1/6
E. 1/7
The Q is not where the answer can be found straightway..
Key words you should look into..
1)If a computer program generates three-digit odd numbers using the numbers 9, 7, 0, 4, and 1
This means that we have to take total ways as possible 3-digit ODD numbers only and NOT all possible 3 - digit numbers
2) Digits can be repeated..Lets solve--
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 waysTOTAL = 4*5*3=60 ways
2) lets now find multiple of 3..a) Digits not div by 37,4 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=18
b) digits div by 39 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=2
TOTAL= 2+18=20
Prob =\(\frac{20}{60} =\frac{1}{3}\)
B