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If a computer program generates three-digit odd numbers using the numb

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If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 07 Apr 2016, 08:41
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 07 Apr 2016, 20:23
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 =\(\frac{20}{60} =\frac{1}{3}\)
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 29 Apr 2017, 17:01
do you know any kind of similar problem
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 30 Apr 2017, 03:33
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If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 30 Apr 2017, 08:52
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 ?

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Ammu
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 30 Apr 2017, 10:07
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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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Saquib
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 13 Oct 2018, 06:50
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?

A)\(\frac{1}{2}\)
B)\(\frac{1}{3}\)
C)\(\frac{1}{5}\)
D)\(\frac{1}{6}\)
E)\(\frac{1}{7}\)



Hi rahulkashyap

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

Prob = 20/60=1/3

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Re: If a computer program generates three-digit odd numbers using the numb   [#permalink] 13 Oct 2018, 06:50
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