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 ways
TOTAL = 4*5*3=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=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=2
TOTAL= 2+18=20

Prob =\(\frac{20}{60} =\frac{1}{3}\)
B
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Hey Ammu,

• To find multiples of 3, you need to know the divisibility property of 3.

• So suppose ABC is a three digit number.

• To check if the number of divisible by three, we just need to add the digits and divide it by 3.

• So if (A+B+C) is divisible by 3, the number ABC is divisible by 3.


o Now we have the digits: 9,7,0,4,1
o So, we need to choose 3 digits, the sum of which is divisible by 3,

 If we take say 9,7,0
 This gives us 9 + 7 + 0 = 16 which is not divisible by 3.
o But, if we consider 7,4,1 : 7 +4 +1 = 12 and 12 is divisible by three.

 Hence any number which is formed from the three digits 7,4 and 1 will be divisible by 3.
 The total three digit numbers which can be formed from these digits(7,4,1) = 3*3*2 = 18
 Notice that the last place can be filled only in 2 ways (7 and 1), since we want only ODD numbers.
o As there are no other possible 3 digits using which can form a three digit number divisible by three
 We can conclude that there are only 18 possible multiples of 3.
• Note: If you want to figure out which digits to choose whose sum will be multiple of 3, kindly refer to chetan2u post, he has given a great explanation for that.

Thanks,
Saquib
Quant Expert
e-GMAT

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