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Solution We have 0 + 2 + 3 + 4 + 6 = 15. Since number is to be divisible by 6 meaning it need to be divisible by 2 and 3. So, sum of digits need to be multiple of 3 and unit digit should be an even number. Possible combination: 1. 2, 3, 4, 6 2. 0, 2, 3, 4 3. 0, 2, 4, 6 For 1st combination: Units place can be taken by 3 numbers, tens can be taken by 3, hundreds by 2 and thousand s place by 1 way. Numbers possible in this case = 1 x 2 x 3 x 3 = 18 For 2nd combination: Total number of numbers possible without any condition = 4! = 24. Now out of 24, 6 will have 3 at units place and hence to be eliminated. 6 numbers will have 0 at thousands place, hence need to be eliminated. Out of 12 eliminated numbers, 2 numbers which are 0243 and 0423 are deleted twice and hence need to be added. Therefore, numbers possible in this cases = 24 – 6 – 6 + 2 = 14. For 3rd combination: Total number of numbers possible without any condition = 4! = 24. Out of 24, 6 will have 0 at thousands place and hence need to be eliminated. Therefore, numbers possible in this cases = 24 – 6 = 18 Total numbers possible = 18+ 14 + 18 = 50.Math Expert Joined: 02 Sep 2009 Posts: 86787 How many four digit numbers that are divisible by 4 can be formed usin [#permalink]
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Hide Show timer StatisticsHow many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number? A. 520 Are You Up For the Challenge: 700 Level Questions _________________ DS Forum Moderator Joined: 19 Oct 2018 Posts: 1938 Location: India Re: How many four digit numbers that are divisible by 4 can be formed usin [#permalink]
Unit digit must be even CASE 1- unit digit is 2 or 6 Total possible numbers= 5*5*4*2=200 Case 2- unit digit is 0 Total possible numbers= 6*5*3*1= 90 Case 3- unit digit is 4 1. when tens digit is 0 2. when tens digit is 2 or 6 Total possible numbers in all cases= 200+90+30+50=370 Bunuel wrote: How many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number? A. 520 Are You Up For the Challenge: 700 Level Questions GMAT Club Legend Joined: 18 Aug 2017 Status:You learn more from failure than from success. Posts: 7205 Location: India Concentration: Sustainability, Marketing GPA: 4 WE:Marketing (Energy and Utilities) Re: How many four digit numbers that are divisible by 4 can be formed
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total possible combinations Bunuel wrote: How many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number? A. 520 Are You Up For the Challenge: 700 Level Questions Target Test Prep Representative Joined: 14 Oct 2015 Status:Founder & CEO Affiliations: Target Test Prep Posts: 16176 Location: United States (CA) Re: How many four digit numbers that are divisible by 4 can be formed usin
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Bunuel wrote: How many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number? A. 520 Are You Up For the Challenge: 700 Level Questions In order for a number to be divisible by 4, the last two digits of the number must be divisible by 4. Since we can choose only digits 0 to 7 and no digits can repeat, the last two digits of the number must be 04, 12, 16, 20, 24, 32, 36, 40, 52, 56, 60, 64, 72 or 76. Let’s separate these numbers into two groups - those with the digit 0 and those without the digit 0. With the digit 0: 04, 20, 40, 60 Without the digit 0: 12, 16, 24, 32, 36, 52, 56, 64, 72, 76 Since these are the last two digits of the number and the first digit of the number can’t be 0, there are 5 choices for the first digit and 5 choices for the second digit. Therefore, there are 5 x 5 x 10 = 250 such numbers if the last two digits do not have the digit 0. Therefore, there are a total of 120 + 250 = 370 such numbers. Answer: C See why Target Test Prep is the top rated GMAT course on GMAT Club. Read Our Reviews Intern Joined: 08 Nov 2020 Posts: 12 Re: How many four digit numbers that are divisible by 4 can be formed usin [#permalink]
Hi all, I reasoned as following, please let me know if it makes sense: total possible combinations of 4-digit numbers out of non-repeating values from 0->7: 7*7*6*5=1470 (i.e., not counting zero as a first digit) Now, 1470 takes into account odd and even values. I'm interested in values divisible by 4, so I will split in halves 1470 twice, i.e.: once to get only even values, and again to get only values divisible by 4. This leads to something close to 370, so my approach seems to work! Thanks all Non-Human User Joined: 09 Sep 2013 Posts: 24376 Re: How many four digit numbers that are divisible by 4 can be formed usin
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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 four digit numbers that are divisible by 4 can be formed usin [#permalink] 24 Feb 2022, 04:22 Moderators: Senior Moderator - Masters Forum 3085 posts How many 4 digit numbers that can be formed using digits 2 3 4 6 7 8 which are divisible by 4 if no digit occurs more than once in each number?Last 2 digits can be chosen from 4 possibilities. First 2 digits can be chosen from remaining 6 numbers. =120.
How many 4 digit numbers can be formed from the digits 2 3 5 6 and 8 which are divisible by 2 if none of the digits are repeated?Answer. Here ur answer is 72.
How many four digit number that can be formed using the digits 2 3 4 5 6 7 9 which are divisible by 9 if no digit occur more than once in each number?∴ 60 four-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9.
How many 4 digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number?So there are 4∗30+10∗25=370 total choices.
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