Last visit was: 16 Jul 2025, 16:10 It is currently 16 Jul 2025, 16:10
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 16 Jul 2025
Posts: 102,594
Own Kudos:
Given Kudos: 98,202
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,594
Kudos: 741,977
 [40]
3
Kudos
Add Kudos
37
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 15 Jul 2025
Posts: 11,294
Own Kudos:
41,777
 [10]
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,294
Kudos: 41,777
 [10]
6
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
General Discussion
User avatar
chesstitans
Joined: 12 Dec 2016
Last visit: 20 Nov 2019
Posts: 990
Own Kudos:
Given Kudos: 2,562
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
GMAT 1: 700 Q49 V33
Posts: 990
Kudos: 1,899
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 16 Jul 2025
Posts: 102,594
Own Kudos:
Given Kudos: 98,202
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,594
Kudos: 741,977
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chesstitans
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
User avatar
AliciaSierra
Joined: 17 Mar 2014
Last visit: 14 Jun 2024
Posts: 748
Own Kudos:
Given Kudos: 1,350
Products:
Posts: 748
Kudos: 636
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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
User avatar
EgmatQuantExpert
User avatar
e-GMAT Representative
Joined: 04 Jan 2015
Last visit: 02 Apr 2024
Posts: 3,681
Own Kudos:
19,465
 [2]
Given Kudos: 165
Expert
Expert reply
Posts: 3,681
Kudos: 19,465
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
ammuseeru
Bunuel
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
e-GMAT

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts :)

User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 15 Jul 2025
Posts: 11,294
Own Kudos:
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,294
Kudos: 41,777
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Abhi077
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
User avatar
adkikani
User avatar
IIM School Moderator
Joined: 04 Sep 2016
Last visit: 24 Dec 2023
Posts: 1,238
Own Kudos:
Given Kudos: 1,207
Location: India
WE:Engineering (Other)
Posts: 1,238
Kudos: 1,317
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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?
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 16 Jul 2025
Posts: 16,111
Own Kudos:
74,359
 [2]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,111
Kudos: 74,359
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
adkikani
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)
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 16 Jul 2025
Posts: 16,111
Own Kudos:
74,359
 [1]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,111
Kudos: 74,359
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
martinste
Hello. Could someone please verify or correct my approach?

999 is the largest number that can be generated and 101 is the lowest number that can be generated to meet the criteria.

999 - 101 = 898 total possible values in that range.

So if a number is a multiple of 3, every third number in that range will satisfy the criteria.

898*(1/3) = 898/3 = # of values that meet the criteria.

By extension, 898*(2/3) = # of values that do NOT meet the criteria.

(898/3)/3 =

1/3

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.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,422
Own Kudos:
Posts: 37,422
Kudos: 1,013
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
102594 posts
PS Forum Moderator
697 posts