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Question: A five digit number divisible by 3 is to be formed using the digits 0,1,2,3,4 and 5, without repetition. The total number of ways this can be done, isA. 216B. 240C. 600D. 3125Answer 
Hint: A number to be divisible by 3, the sum of all the digits should be divisible by 3. Complete step-by-step answer: Let us observe the digits given to us, we have 0,1,2,3,4 and 5, to make a
five digit number we only need 5 digits out of the given 6 digits, Case 2: Using the digits 1,2,3,4 and 5 Therefore, Option A is the correct answer. Note: Make sure that you do not take 0 in the first place because that will make the number a 4-digit number which will be considered wrong as we are supposed to form a 5-digit number. A five digit number divisible by 30 is to be formed using the digits 0,1,2,3,4,5 without repetition of the digits. The number of ways it can be done is ..... No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Solution The correct option is B 24Last digit is zero and the remaining from digits are 1,2,4,5. Number of arrangements = 4! =24How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated? Five-digits numbers divisible by 3 are to be formed using the digits 0, 1, 2, 3, 4, and 5 without repetition. Number of 5-digit numbers = 5 × 4 × 3 × 2 × 1 = 120 Now, consider the digits: 0, 1, 2, 4 and 5 Sum of the digits = 0 + 1 + 2 + 4 + 5 = 12 which is divisible by 3. ∴ Any 5-digit number formed using the digits 0, 1, 2, 4, and 5 will be divisible by 3. Starting with the most significant digit, 4 digits are available for this place (since 0 cannot be used). Since, repetition is not allowed, for the next significant place, 4 digits are available (since 0 can now be used). Similarly, all the places can be filled as: Number of 5-digit numbers = 4 × 4 × 3 × 2 × 1 = 96 Next, consider the digits: 0, 1, 2, 3, 4 Sum of the digits = 0 + 1 + 2 + 3 + 4 = 10 which is not divisible by 3. ∴ None of the 5-digit numbers formed using the digits 0, 1, 2, 3, and 4 will not be divisible by 3. Further, no other selection of 5 digits (out of the given 6) will give a 5-digit number, which is divisible by 3. ∴ Total number of 5adigit numbers divisible by 3 = 120 + 96 = 216 GMAT Club Daily PrepThank 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 we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice 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.Hello Guest!It appears that you are browsing the GMAT Club forum unregistered! Signing up is free, quick, and confidential. Join 700,000+ members and get the full benefits of GMAT ClubRegistration gives you:
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Manager Joined: 29 May 2008 Posts: 91 How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
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00:00 Question Stats:
Hide Show timer StatisticsHow many five digit numbers can be formed using digits 0, 1, 2, 3, 4, 5, which are divisible by 3, without any of the digits repeating? A. 15 Originally posted by TheRob on 22 Oct 2009, 13:20. Renamed the topic and edited the question. Math Expert Joined: 02 Sep 2009 Posts: 86798 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
TheRob wrote: How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating? A. 15 First step: We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3. We have six digits: 0,1,2,3,4,5. Their sum=15. For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. Second step: We have two set of numbers: {1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120. {0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96 120+96=216 Answer: E. Manager Joined: 15 Sep 2009 Posts: 73 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
Only 2 sets are possible case (1) 1,2,3,4,5 case (1) : there will 5! ways to form the number = 120 case (2) ; there will 4*4*3*2*1 = 96 ways So total no.of ways = 120+96 = 216 ways Senior Manager Joined: 03 Sep 2006 Posts: 495
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
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By the property of divisibility by 3 i.e "a no: is divisible by 3, if the sum of the digits is divisible by 3"(e.g= 12-->1+2=3) so from 0,1,2,3,4,5 the set of 5 digit no:s that can be formed which is divisible by 3 are 0,1,2,4,5(sum=12) & 1,2,3,4,5(sum=15) from first set(0,1,2,4,5) no:s formed are 96 i.e first digit can be formed from any 4 no: except 0, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed. from second set(1,2,3,4,5) no:s formed are 120 i.e first digit can be formed from any 5 digits, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed. so total 120+96=216 Manager Joined: 28 Aug 2010 Posts: 128 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. i understood the first part but did not get the second part 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. ..Could you please explain it in a little bit more detail. Thanks Math Expert Joined: 02 Sep 2009 Posts: 86798 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
ajit257 wrote: For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. i understood the first part but did not get the second part 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. ..Could you please explain it in a little bit more detail. Thanks The sum of the given digits is already a multiple of 3
(15), in order the sum of 5 digits to be a multiple of 3 you must withdraw a digit which is itself a multiple of 3, otherwise (multiple of 3) - (non-multiple of 3) = (non-multiple of 3). Manager Joined: 28 Aug 2010 Posts: 128 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
so lets say we were asked a multiple of 5 so in that case we would have to withdraw the digit 5 ..is that correct ? Math Expert Joined: 02 Sep 2009 Posts: 86798 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
ajit257 wrote: so lets say we were asked a multiple of 5 so in that case we would have to withdraw the digit 5 ..is that correct ? 5 or 0, as 0 is also a multiple of 5. AGAIN: we have (sum of 6 digits)=(multiple of 3). Question what digit should we withdraw so that the sum of the remaining 5 digits remain a multiple of 3? Answer: the digit which is itself a multiple of 3. Below might help to understand this concept better. If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by
\(k\)): If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): If integers \(a\) and
\(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Hope it's clear. Retired Moderator Joined: 20 Dec 2010 Posts: 1252 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
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0,1,2,3,4,5 One digit will have to remain out for all 5 digit numbers; if 0 is out; Leftover digits will be 1,2,3,4,5 =
Sum(1,2,3,4,5)=15. if 1 is out; Leftover digits will be 0,2,3,4,5 = Sum(0,2,3,4,5)=14. Ignore(Not divisible by 3) if 3 is out; Leftover digits will be 0,1,2,4,5 = Sum(0,1,2,4,5)=12. if 4 is out; Leftover digits will be 0,1,2,3,5 = Sum(0,1,2,3,5)=11. Ignore Total count of numbers divisible by 3 = 120+96 = 216 Ans: "E" Manager Joined: 20 Aug 2011 Posts: 72 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
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A number is divisible by 3 if sum of its digits is a multiple of 3. With the given set of digits, there are two possible combinations of 5 digits each- A. [1,2,3,4,5] No. of possible 5 digit numbers: 5!= 120 A+B= 120+96= 216 E Originally posted by blink005 on 06 Jan 2012, 06:48. Senior Manager Joined: 13 Aug 2012 Posts: 356 Concentration: Marketing, Finance GPA: 3.23
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
TheRob wrote: How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating? A. 15 0 + 1 + 2 + 3 + 4 + 5 = 15 To form 5-digit number, we can remove a digit and the sum should still be divisible by 3. 15 - 1 = 14 Possible = {5,4,3,2,1} and {5,4,0,2,1} There are 5! = 120 ways to arrange {5,4,3,2,1} 120 + 96 = 216 Answer: E Manager Joined: 12 Jan 2013 Posts: 55 Location: United States (NY) GPA: 3.89
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
I did in 1 min 18 sec. Then I thought that the last digits could always be chosen in only two ways so as to ensure divisibility by three - however, I quickly realized that I would not get all different digits. Then I realized that once I get a number I can keep permuting the digits while still getting valid numbers. In an attempt to avoid the leading zero I tried 12345 and noticed that it was divisible by 3. Thus, I've got 5!=120 answers and immediately eliminated two answers, A and B. Then I addressed the case of a leading zero. Since I wanted to preserve divisibility by 3, I quickly saw that I could only use 0 instead of 3. Thus, the only other possible set was {0, 1, 2, 4, 5}. I tried adding another 5! and got 240, so the answer was slightly less than that. After that I knew I had to subtract 4!=24 to account for all the possibilities with a leading zero, which left me with 240-24=216. This is how I do such problems... Sergey Orshanskiy,
Ph.D. Intern Joined: 09 Jul 2013 Posts: 20 Location: United States (WA) GPA: 3.65 WE:Military Officer (Military & Defense)
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
E. I'm offering up a way to apply the "slot method" to this problem below. First, as everyone else has identified above, you need to find the cases where the 6 numbers (0, 1, 2, 3, 4, 5) create a 5-digit number divisible by 3. Shortcut review: a number is divisible by 3 if the sum of the digits in the number is divisible by 3. So, 12345 would be divisible by 3 (1+2+3+4+5 = 15, which is divisible by 3). Analyzing the given numbers, we can conclude that only the following two groups of numbers work: 1, 2, 3, 4, 5 (in any order, they would create a five digit number divisible by 3 - confirmed by the shortcut above), and 0, 1,2, 4, 5. Now we need to count the possible arrangements in both cases, and then add them together. To use the slot method with case 1 (1,2,3,4,5): Now consider case 2 (0,1,2,4,5): Now the final step is to add all the possible arrangements together from case 1 and case 2: Hope this alternate "slot" method helps! This is how I try to work these combinatoric problems instead of using formulas... in this case it worked out nicely. Here, order didn't matter (we are only looking for total possible arrangements) in the digits, so we didn't need to divide by the factorial number of slots. Senior Manager Joined: 07 Apr 2012 Posts: 277 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
Bunuel wrote: TheRob wrote: How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating? A. 15 First step: We have six digits: 0,1,2,3,4,5. Their sum=15. For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. Second step: We have two set of numbers: {1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120. {0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96 120+96=216 Answer: E. I tried to do as follows: What is wrong with this logic? Math Expert Joined: 02 Sep 2009 Posts: 86798 Re: How
many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
ronr34 wrote: Bunuel wrote: TheRob wrote: How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating? A. 15 First step: We have six digits: 0,1,2,3,4,5. Their sum=15. For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. Second step: We have two set of numbers: {1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120. {0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96 120+96=216 Answer: E. I tried to do as follows: What is wrong with this logic? Because the numbers divisible by 3 are not 1/3rd of all possible numbers. {0, 1, 2, 3, 4} --> 96 5-digit
numbers possible with this set. Total = 5*5*4*3*2 = 600 but the numbers which are divisible by 3 come
from third and sixth sets: 96 + 120 = 216. GMAT Expert Joined: 16 Oct 2010 Posts: 13164 Location: Pune, India
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
ronr34 wrote: I tried to do as follows: What is wrong with this logic? We cannot do this because we have the asymmetric 0 as one of the digits. The number of 5 digit numbers that can be formed with 0, 1, 2, 3 and 4 is different from the number of 5 digit numbers that can be formed with 1, 2, 3, 4 and 5 (because 0 cannot be the first digit). Had the digits been 1, 2, 3, 4, 5 and 6, then your method would have been correct. If 0 is included: If 0 is not included: Karishma For Individual GMAT Study Modules, check Study Modules > CrackVerbal Representative Joined: 03 Oct 2013 Affiliations: CrackVerbal Posts: 4989 Location: India Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
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We need to form five digit numbers with distinct digits which are divisible by 3. A number is divisible by 3 when the sum of its digits is divisible by 3. When we observe the digits, we see that the sum of the given digits is 15. Typical of a GMAT kind of question – data given is very precise and rarely vague. So, there are only two cases that can be considered to fit the constraints given Case 1: A 5 digit number with the digits {1,2,3,4,5}. Since 0 is not a part of this set and there are 5 different digits, we can form a total of 5P5 = 5! = 120 numbers. All of these will be divisible by 3. At this stage, we can eliminate answer options A, B and C. Case 2: A 5 digit number with the digits {0,1, 2, 4, 5}. Since 0 is a part of this set, we need to use Counting methods to find out the number of 5-digit numbers. The ten thousands place can be filled in 4 ways, since 0 cannot come here; the thousands place can be filled in 4 ways, the hundreds in 3 ways, the tens in 2 ways and the units place in 1 way. Total number of 5-digit numbers with distinct digits, divisible by 3 = Case 1 + Case 2 = 120 + 96. Answer option D can be eliminated. The correct answer option is E. Hope that helps! Non-Human User Joined: 09 Sep 2013 Posts: 24386 Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
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. Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink] 01 Jul 2022, 03:04 Moderators: Senior Moderator - Masters Forum 3084 posts How many 5Answer : `=2xx5xx6xx6xx6=2160` ways. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
How many 5Therefore, there are a total of 192 numbers which can be formed using the digits 1,2,3,4,5,6 without repetition such that the number is divisible by 4.
How many numbers of 5 digits can be formed with the digits 0 1,2 3 4 if the digits can repeat themselves?Since, repetition is not allowed, for the next significant place, 4 digits are available (since 0 can now be used). Sum of the digits = 0 + 1 + 2 + 3 + 4 = 10 which is not divisible by 3. ∴ None of the 5-digit numbers formed using the digits 0, 1, 2, 3, and 4 will not be divisible by 3.
How many 6 digit numbers are possible by using the digits 0 1,2 3 4 5 which are divisible by 5 if the repetition is allowed?So the number of 6 digit numbers divisible by the number in the units place is: 384+24+18=426. Was this answer helpful?
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