Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 29 May 2008
Posts: 110

How many fivedigit numbers can be formed using digits
[#permalink]
Show Tags
22 Oct 2009, 14:20
Question Stats:
53% (01:31) correct 47% (01:20) wrong based on 587 sessions
HideShow timer Statistics
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 B. 96 C. 120 D. 181 E. 216
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 47918

How many fivedigit numbers can be formed using digits
[#permalink]
Show Tags
22 Oct 2009, 14:59
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 B. 96 C. 120 D. 181 E. 216 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 150={1, 2, 3, 4, 5} and 153={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} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets: {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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 271

Re: digit counting
[#permalink]
Show Tags
22 Oct 2009, 15:14
Thanks Bunuel for the great explanation . +1 Kudos... Why don't I borrow your math brain for my GMAT LOL
_________________
Thanks, Sri  keep uppp...ing the tempo...
Press +1 Kudos, if you think my post gave u a tiny tip



Manager
Joined: 05 Jul 2009
Posts: 162

Re: digit counting
[#permalink]
Show Tags
24 Oct 2009, 00:06
srini123 wrote: Thanks Bunuel for the great explanation . +1 Kudos... Why don't I borrow your math brain for my GMAT LOL I have been thinking the same for quite some days!



Manager
Joined: 13 Oct 2009
Posts: 50
Location: New York, NY
Schools: Columbia, Johnson, Tuck, Stern

Re: digit counting
[#permalink]
Show Tags
25 Oct 2009, 14:42
YOu're awesome Bunuel !!



Manager
Joined: 15 Sep 2009
Posts: 118

Re: digit counting
[#permalink]
Show Tags
26 Oct 2009, 06:21
Only 2 sets are possible
case (1) 1,2,3,4,5 case (2) 0,1,2,4,5.
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



Manager
Joined: 13 Dec 2009
Posts: 235

aramjung wrote: How many fivedigit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? (C) 2008 GMAT Club  m04#3215 96 120 181 216 HOW DO YOU SOLVE THIS, I CAN'T UNDERSTAND IT For a number to be divisible by 3 sum of its digits should be divisible by 3, so in 0,1,2,3,4,5 set of digits that can be together 1,2,3,4,5, and 0,1,2,4,5 (drop one digit at a time and sum rest of the others to find this ) in first set numbers that can be there = 5*4*3*2*1 in second set = 4*4*3*2*1 (since zero cannot be the last digit) total = 5*4*3*2*1 + 4*4*3*2*1 = 4! * (5+4) = 24 * 9 = 216, hence 216 is the answer.
_________________
My debrief: doneanddusted730q49v40



Director
Joined: 03 Sep 2006
Posts: 836

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: 231

Re: digit counting
[#permalink]
Show Tags
06 Feb 2011, 14:48
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 150={1, 2, 3, 4, 5} and 153={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 153={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
_________________
Verbal:newtotheverbalforumpleasereadthisfirst77546.html Math: newtothemathforumpleasereadthisfirst77764.html Gmat: everythingyouneedtoprepareforthegmatrevised77983.html  Ajit



Math Expert
Joined: 02 Sep 2009
Posts: 47918

Re: digit counting
[#permalink]
Show Tags
06 Feb 2011, 14:55
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 150={1, 2, 3, 4, 5} and 153={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 153={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)  (nonmultiple of 3) = (nonmultiple of 3).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 28 Aug 2010
Posts: 231

Re: digit counting
[#permalink]
Show Tags
06 Feb 2011, 14:59
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 ?
_________________
Verbal:newtotheverbalforumpleasereadthisfirst77546.html Math: newtothemathforumpleasereadthisfirst77764.html Gmat: everythingyouneedtoprepareforthegmatrevised77983.html  Ajit



Math Expert
Joined: 02 Sep 2009
Posts: 47918

Re: digit counting
[#permalink]
Show Tags
06 Feb 2011, 15:18
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 \(k\)):Example: \(a=6\) and \(b=9\), both divisible by 3 > \(a+b=15\) and \(ab=3\), again both divisible by 3. 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 \(k\)):Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 > \(a+b=11\) and \(ab=1\), neither is divisible by 3. 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 \(k\)):Example: \(a=5\) and \(b=4\), neither is divisible by 3 > \(a+b=9\), is divisible by 3 and \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), both are divisible by 4. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Retired Moderator
Joined: 20 Dec 2010
Posts: 1879

Re: digit counting
[#permalink]
Show Tags
07 Feb 2011, 04:39
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. 5! = 120 numbers 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. 4*4! = 4*24 = 96 if 4 is out; Leftover digits will be 0,1,2,3,5 = Sum(0,1,2,3,5)=11. Ignore if 5 is out; Leftover digits will be 0,1,2,3,4 = Sum(0,1,2,3,4)=10. Ignore Total count of numbers divisible by 3 = 120+96 = 216 Ans: "E"
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 20 Aug 2011
Posts: 129

Re: Number properties
[#permalink]
Show Tags
Updated on: 06 Jan 2012, 09:27
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 B. [0,1,2,4,5] No. of possible 5 digit numbers: 4*4!=96 [the number can't start with a 0] A+B= 120+96= 216 E
_________________
Hit kudos if my post helps you. You may send me a PM if you have any doubts about my solution or GMAT problems in general.
Originally posted by blink005 on 06 Jan 2012, 07:48.
Last edited by blink005 on 06 Jan 2012, 09:27, edited 1 time in total.



Senior Manager
Joined: 13 Aug 2012
Posts: 441
Concentration: Marketing, Finance
GPA: 3.23

Re: How many fivedigit numbers can be formed using digits
[#permalink]
Show Tags
28 Dec 2012, 06:54
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 B. 96 C. 120 D. 181 E. 216 0 + 1 + 2 + 3 + 4 + 5 = 15 To form 5digit number, we can remove a digit and the sum should still be divisible by 3. 15  1 = 14 15  2 = 13 15  3 = 12 BINGO! 15  4 = 11 15  5 = 10 Possible = {5,4,3,2,1} and {5,4,0,2,1} There are 5! = 120 ways to arrange {5,4,3,2,1} There are 5!  5!/5 = 96 ways to arrange {5,4,0,2,1} since 0 cannot start the five number digit. 120 + 96 = 216 Answer: E
_________________
Impossible is nothing to God.



Manager
Joined: 12 Jan 2013
Posts: 57
Location: United States (NY)
GPA: 3.89

Re: How many fivedigit numbers can be formed using digits
[#permalink]
Show Tags
14 Jan 2013, 00:40
I did in 1 min 18 sec. At first I wanted to choose a set of five digits, but started to worry about the complications with the leading zero. 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 24024=216. This is how I do such problems...
_________________
Sergey Orshanskiy, Ph.D. I tutor in NYC: http://www.wyzant.com/Tutors/NY/NewYork/7948121/#ref=1RKFOZ



Current Student
Joined: 09 Jul 2013
Posts: 25
Location: United States (WA)
GPA: 3.65
WE: Military Officer (Military & Defense)

Re: How many fivedigit numbers can be formed using digits
[#permalink]
Show Tags
15 Oct 2013, 18:26
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 5digit 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): _ _ _ _ _ (five digit number, 5 slots). Fill in the slots with the number of "choices" left over from your pool of numbers. Starting from the left, I have 5 choices I can put in slot #1 (5 numbers from the group 1,2,3,4,5  pretend I put in number 1, that leaves 4 numbers) 5 _ _ _ _ Fill in the next slot with the number of choices left over (4 choices left...numbers 2 through 5) 5 4 _ _ _ Continue filling out the slots until you arrive at: 5 4 3 2 1 Multiply the choices together: 5x4x3x2x1 (which also happens to be 5!) = 120 different arrangements for the first case.
Now consider case 2 (0,1,2,4,5): _ _ _ _ _ (five digit number, 5 slots). Here's the tricky part. I can't put 0 as the first digit in the number...that would make it a 4digit number! So I only have 4 choices to pick from for my first slot! 4 _ _ _ _ Fill in the next slot with the remaining choices (if I put in 1 in the first slot, I have 0,2,4,5 left over...so 4 more choices to go). 4 4 _ _ _ Continue to fill out the slots with the remaining choices: 4 4 3 2 1 Multiply the choices together: 4 x 4! = 96 different arrangements for the second case.
Now the final step is to add all the possible arrangements together from case 1 and case 2: 120 + 96 = 216. And this is our answer.
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: 08 Apr 2012
Posts: 395

Re: digit counting
[#permalink]
Show Tags
08 Jun 2014, 09:12
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 B. 96 C. 120 D. 181 E. 216 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 150={1, 2, 3, 4, 5} and 153={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} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets: {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: take all 5 digit numbers possible : 5 *5*4*3*2 divide by 3 to get all numbers divisible by 3. What is wrong with this logic?



Math Expert
Joined: 02 Sep 2009
Posts: 47918

Re: digit counting
[#permalink]
Show Tags
08 Jun 2014, 11:46
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 B. 96 C. 120 D. 181 E. 216 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 150={1, 2, 3, 4, 5} and 153={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} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets: {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: take all 5 digit numbers possible : 5 *5*4*3*2 divide by 3 to get all numbers divisible by 3. 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 5digit numbers possible with this set. {0, 1, 2, 3, 5} > 96 5digit numbers possible with this set.{0, 1, 2, 4, 5} > 96 5digit numbers possible with this set.{0, 1, 3, 4, 5} > 96 5digit numbers possible with this set. {0, 2, 3, 4, 5} > 96 5digit numbers possible with this set.{1, 2, 3, 4, 5} > 120 5digit 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8184
Location: Pune, India

Re: digit counting
[#permalink]
Show Tags
16 Jun 2014, 23:11
ronr34 wrote: I tried to do as follows: take all 5 digit numbers possible : 5 *5*4*3*2 divide by 3 to get all numbers divisible by 3.
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: {0, 1, 2, 3, 4} > 96 5digit numbers possible with this set. {0, 1, 2, 3, 5} > 96 5digit numbers possible with this set. {0, 1, 2, 4, 5} > 96 5digit numbers possible with this set.  All these numbers are divisible by 3 {0, 1, 3, 4, 5} > 96 5digit numbers possible with this set. {0, 2, 3, 4, 5} > 96 5digit numbers possible with this set. {1, 2, 3, 4, 5} > 120 5digit numbers possible with this set.  All these numbers are divisible by 3 The number of 5 digit numbers in these sets is not the same  Sets with 0 have fewer numbers If 0 is not included: {1, 2, 3, 4, 5} > 120 5digit numbers possible with this set. {1, 2, 3, 4, 6} > 120 5digit numbers possible with this set. {1, 2, 3, 5, 6} > 120 5digit numbers possible with this set.  All these numbers are divisible by 3 {1, 2, 4, 5, 6} > 120 5digit numbers possible with this set. {1, 3, 4, 5, 6} > 120 5digit numbers possible with this set. {2, 3, 4, 5, 6} > 120 5digit numbers possible with this set.  All these numbers are divisible by 3 Here exactly 1/3rd of the numbers will be divisible by 3.
_________________
Karishma Veritas Prep GMAT Instructor
Save up to $1,000 on GMAT prep through 8/20! Learn more here >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!




Re: digit counting &nbs
[#permalink]
16 Jun 2014, 23:11



Go to page
1 2
Next
[ 22 posts ]



