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If a computer program generates three-digit odd numbers using the numb
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07 Apr 2016, 07:41
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Question Stats:
39% (02:49) correct 61% (03:33) wrong based on 82 sessions
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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?
Re: If a computer program generates three-digit odd numbers using the numb
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07 Apr 2016, 19:23
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Bunuel wrote:
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 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
_________________
If a computer program generates three-digit odd numbers using the numb
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30 Apr 2017, 07:52
Bunuel wrote:
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
I couldn't understand how to find 3 digit numbers, which is multiple of 3. Bunuel/Karishma could you help me ?
Re: If a computer program generates three-digit odd numbers using the numb
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30 Apr 2017, 09:07
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ammuseeru wrote:
Bunuel wrote:
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
I couldn't understand how to find 3 digit numbers, which is multiple of 3. Bunuel/Karishma could you help me ?
Regards, Ammu
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.
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Re: If a computer program generates three-digit odd numbers using the numb
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13 Oct 2018, 05:50
Abhi077 wrote:
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?
The fastest way is to know the remainder all leave.. 1) 0 as remainder - 0 and 9 So first and last place by 9 and middle value by any of 0 or 9, so 1*2*1=2 2) remainder as 1- 7,4,1 First and second place can be any of the three and the units digit any of two odd numbers that is 7 or 1, so 3*3*2=18 Total 2+18=20
Total ways - all except 0 at first place , any of 5 in second place and any of three in units digit So 4*5*3=60
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39% (02:49) correct 
