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

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New post 27 Oct 2018, 10:10
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If a computer program generates three-digit odd numbers using the numbers 9, 7, O, 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) 1/4
D) 1/5
E) 1/6

Weekly Quant Quiz #6 Question No 3


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New post 27 Oct 2018, 10:10
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post Updated on: 04 Nov 2018, 14:46
Total possible Odd nos 4*5*3 = 60.(You can't have 0 as the first number possiblities - 4. Second number - 5 possibilities. Third number 3 possibilties removing the even numbers)
Total possible Multiples of 3 - 15(999,777,444,111,990,414,441,144,174,741,147,417,177,771,717)
Probability = 15/60 = 1/4

Originally posted by sriramsundaram91 on 27 Oct 2018, 11:31.
Last edited by sriramsundaram91 on 04 Nov 2018, 14:46, edited 1 time in total.
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 27 Oct 2018, 14:00
Solved it using sum of digits. Min possible sum = 3 ; Max = 27
Total possible 3 digit odd nos (digits can be repeated) = 4 * 5 * 3 = 60
No of 3 odd 3 digit no which are multiple of 3 = 15

Probability = 15/60 = 1/4
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 04 Nov 2018, 02:39
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gmatbusters wrote:
If a computer program generates three-digit odd numbers using the numbers 9, 7, O, 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) 1/4
D) 1/5
E) 1/6

Weekly Quant Quiz #6 Question No 3



Hi Gmatbusters, in your explanation, 2 of the possibilities are missing: 477 and 747, whose digits add-up to 18, therefore there's 3 outcomes for numbers whose digits add up to 18, including 909.

Should the answer be 17/60?
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 04 Nov 2018, 08:07
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quocbt93 wrote:
gmatbusters wrote:
If a computer program generates three-digit odd numbers using the numbers 9, 7, O, 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) 1/4
D) 1/5
E) 1/6

Weekly Quant Quiz #6 Question No 3



Hi Gmatbusters, in your explanation, 2 of the possibilities are missing: 477 and 747, whose digits add-up to 18, therefore there's 3 outcomes for numbers whose digits add up to 18, including 909.

Should the answer be 17/60?


The same problem with the numbers 471, 441, 417. I think the answer should be 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 11 Nov 2018, 02:24
gmatbusters wrote:
If a computer program generates three-digit odd numbers using the numbers 9, 7, O, 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) 1/4
D) 1/5
E) 1/6

Weekly Quant Quiz #6 Question No 3




9,7,0,4, 1

Total three digit numbers possible-
4x5x5 = 100 (note that 0 cannot take the hundredth place)

A number is multiple of 3 if sum of its digits are multiple of 3.
Favorable cases:
7,7,7- 1 case
4,4,4- 1
1,1,1- 1
9,9,9- 1
9,0,0- 1

7,4,1- 3! = 6 cases
4,4,1: 3!/2! = 3
7,7,1: 3!/2! = 3
1,1,4: 3!/2! = 3
7,1,1: 3!/2! = 3
9,0,9: 2

Total: 25

Probability: 25/100 = 1/4

C is my answer.
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 11 Nov 2018, 02:32
1
OA = B

See the explanation posted.

Aamirso wrote:
gmatbusters wrote:
If a computer program generates three-digit odd numbers using the numbers 9, 7, O, 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) 1/4
D) 1/5
E) 1/6

Weekly Quant Quiz #6 Question No 3




9,7,0,4, 1

Total three digit numbers possible-
4x5x5 = 100 (note that 0 cannot take the hundredth place)

A number is multiple of 3 if sum of its digits are multiple of 3.
Favorable cases:
7,7,7- 1 case
4,4,4- 1
1,1,1- 1
9,9,9- 1
9,0,0- 1

7,4,1- 3! = 6 cases
4,4,1: 3!/2! = 3
7,7,1: 3!/2! = 3
1,1,4: 3!/2! = 3
7,1,1: 3!/2! = 3
9,0,9: 2

Total: 25

Probability: 25/100 = 1/4

C is my answer.

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

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New post 11 Nov 2018, 02:33
The argument says that the computer generates the ODD numbers. If I dnot make mistake there are 60 possible combinations, cuz we should exclude the even numbers. Thus, 4*5*3=60
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 11 Nov 2018, 02:37
You are right, total no of combinations = 60

Elekber wrote:
The argument says that the computer generates the ODD numbers. If I dnot make mistake there are 60 possible combinations, cuz we should exclude the even numbers. Thus, 4*5*3=60

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

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New post 11 Nov 2018, 04:24
gmatbusters wrote:
You are right, total no of combinations = 60

Elekber wrote:
The argument says that the computer generates the ODD numbers. If I dnot make mistake there are 60 possible combinations, cuz we should exclude the even numbers. Thus, 4*5*3=60



hi gmatbusters

choosing 3 of 5 = \(\frac{5*4*3*2*1}{3!(5-3)!} = \frac{60}{6} =10\)

why am i getting 10 ?

i would like to know the precise formula with repetitions allowed :)

OR if I rewrite like this \(\frac{5*4*3*2*1}{(5-3)!} = 60\) now I get 60 , so with repetitions allowed I need to omit 3! in denominator, am I Right ? is it correct approach to find total number of ways with repetitions allowed ?


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

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New post 11 Nov 2018, 04:48
Hi dave13

The first formula you used is for selection of 3 items from 5 items.

Second formula you used is for selection and arrangement of 3 items out of 5 items.

Both the formulas are irrelevant here.

In general there is no fixed formula to make cases of numbers/words and other complex problems, we have to do it appropriately as per question.

For this particular question, please refer the sketch attached in my second post.

Thanks
dave13 wrote:
gmatbusters wrote:
You are right, total no of combinations = 60

Elekber wrote:
The argument says that the computer generates the ODD numbers. If I dnot make mistake there are 60 possible combinations, cuz we should exclude the even numbers. Thus, 4*5*3=60



hi gmatbusters

choosing 3 of 5 = \(\frac{5*4*3*2*1}{3!(5-3)!} = \frac{60}{6} =10\)

why am i getting 10 ?

i would like to know the precise formula with repetitions allowed :)

OR if I rewrite like this \(\frac{5*4*3*2*1}{(5-3)!} = 60\) now I get 60 , so with repetitions allowed I need to omit 3! in denominator, am I Right ? is it correct approach to find total number of ways with repetitions allowed ?


thank you :)


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New post 11 Nov 2018, 05:00
gmatbusters thanks! i actually got your solution, just was wondering if there are alterative ways to solve the problem, also writing on paper all possibilities is quite time consuming , and there is probability of missing some number :grin: did you solve it under 2 minutes ?
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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New post 11 Nov 2018, 05:11
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It is definitely 700+ level question.
One doesn't need to do all questions under 2 minutes,
Some questions are easy enough to be done under 0.5-01 minutes. Hence, one can devote more than 2 minutes to some questions.

Hence, TIME MANAGEMENT is important.
dave13 wrote:
gmatbusters thanks! i actually got your solution, just was wondering if there are alterative ways to solve the problem, also writing on paper all possibilities is quite time consuming , and there is probability of missing some number :grin: did you solve it under 2 minutes ?


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Re: If a computer program generates three-digit odd numbers using the numb   [#permalink] 11 Nov 2018, 05:11
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