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If a computer program generates threedigit odd numbers using the numb
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27 Oct 2018, 09:10
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13% (01:41) correct 87% (02:38) wrong based on 85 sessions
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If a computer program generates threedigit 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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If a computer program generates threedigit odd numbers using the numb
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27 Oct 2018, 09:10



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Re: If a computer program generates threedigit odd numbers using the numb
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Updated on: 04 Nov 2018, 13:46
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



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Re: If a computer program generates threedigit odd numbers using the numb
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27 Oct 2018, 13:00
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 threedigit odd numbers using the numb
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04 Nov 2018, 01:39
gmatbusters wrote: If a computer program generates threedigit 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 addup 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 threedigit odd numbers using the numb
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04 Nov 2018, 07:07
quocbt93 wrote: gmatbusters wrote: If a computer program generates threedigit 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 addup 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 threedigit odd numbers using the numb
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11 Nov 2018, 01:24
gmatbusters wrote: If a computer program generates threedigit 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 threedigit odd numbers using the numb
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11 Nov 2018, 01:32
OA = B See the explanation posted. Aamirso wrote: gmatbusters wrote: If a computer program generates threedigit 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 threedigit odd numbers using the numb
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11 Nov 2018, 01:33
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 threedigit odd numbers using the numb
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11 Nov 2018, 01:37



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If a computer program generates threedigit odd numbers using the numb
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11 Nov 2018, 03:24
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 = \(\frac{5*4*3*2*1}{3!(53)!} = \frac{60}{6} =10\) why am i getting 10 ? i would like to know the precise formula with repetitions allowed OR if I rewrite like this \(\frac{5*4*3*2*1}{(53)!} = 60\) 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 threedigit odd numbers using the numb
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11 Nov 2018, 03:48
Hi dave13The 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 = \(\frac{5*4*3*2*1}{3!(53)!} = \frac{60}{6} =10\) why am i getting 10 ? i would like to know the precise formula with repetitions allowed OR if I rewrite like this \(\frac{5*4*3*2*1}{(53)!} = 60\) 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 Posted from my mobile device
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Re: If a computer program generates threedigit odd numbers using the numb
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11 Nov 2018, 04:00
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 did you solve it under 2 minutes ?



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Re: If a computer program generates threedigit odd numbers using the numb
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11 Nov 2018, 04:11
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 0.501 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 did you solve it under 2 minutes ? Posted from my mobile device
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