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

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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
e-GMAT

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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

Hi chetan2u VeritasKarishma GMATPrepNow EMPOWERgmatRichC

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

Answer (B)

You haven't considered the digits given. Say the given digits were 3, 6, 0 and 9. Then every number you make using these will be a multiple of 3.

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