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

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Math Expert
Joined: 02 Sep 2009
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If a computer program generates three-digit odd numbers using the numb  [#permalink]

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07 Apr 2016, 07:41
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39% (03:03) correct 61% (03:19) wrong based on 84 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?

A. 1/2
B. 1/3
C. 1/5
D. 1/6
E. 1/7

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Posts: 8249
Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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

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29 Apr 2017, 16:01
do you know any kind of similar problem
Math Expert
Joined: 02 Sep 2009
Posts: 61396
Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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30 Apr 2017, 02:33
chesstitans wrote:
do you know any kind of similar problem

You can look for similar problems in Probability questions: https://gmatclub.com/forum/search.php?s ... &tag_id=54
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If a computer program generates three-digit odd numbers using the numb  [#permalink]

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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 ?

Regards,
Ammu
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Joined: 04 Jan 2015
Posts: 3250
Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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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.

Thanks,
Saquib
Quant Expert
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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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?

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

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

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09 Jan 2020, 15:57

I did not follow either of two approaches mentioned above by chetan2u.
I could easily understand how the expert got the finding probability of denominator,
but am unable to understand the link between grouping / remainders of numbers
to divisibility rule: the sum of all digits must be divisible by 3. Can you elaborate the same?
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Re: If a computer program generates three-digit odd numbers using the numb  [#permalink]

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09 Jan 2020, 19:05
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I did not follow either of two approaches mentioned above by chetan2u.
I could easily understand how the expert got the finding probability of denominator,
but am unable to understand the link between grouping / remainders of numbers
to divisibility rule: the sum of all digits must be divisible by 3. Can you elaborate the same?

The sum of all digits logic can be broken down to make it quicker n easy.

Say 435
4 + 3 + 5 = (3 + 1) + 3 + (3 + 2)
So overall remainder will be 1 + 2 = 3 which means a remainder 0.

Say 154426
1 + 5 + 4 + 4 + 2 + 6 = 1 + (3 + 2) + (3 + 1) + (3 + 1) + 2 + 6
1 + 2 = 3 Ignore
1 + 2 = 3 Ignore
Remainder is 1

Now coming back to the original question:
9, 7, 0 4, 1

Total such numbers = 4 * 5 * 3 = 60 (no 0 in hundreds place, any digit in tens place and only 9, 7 or 1 in units place)

Now 7, 4 and 1 leave remainder 1.
So to make the number divisible by 3, we need three of these digits (either repetition or distinct) if even one of them is used.
Number of such 3 digit numbers = 3 * 3 * 2 = 18

Or we can use 9 and 0 to make the 3 digit numbers = 1 * 2 * 1 = 2

Total 3 digit odd numbers = 18 + 2 = 20

Probability = 20/60 = 1/3

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Re: If a computer program generates three-digit odd numbers using the numb   [#permalink] 09 Jan 2020, 19:05
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