Difficult Math Problem #113 - Permutations - The Beat The GMAT Forum

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Difficult Math Problem #113 - Permutations

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800guy: Master | Next Rank: 500 Posts; Posts: 354; Joined: Tue Jun 27, 2006 9:20 pm; Thanked: 11 times; Followed by:5 members

Difficult Math Problem #113 - Permutations

by 800guy » Mon Apr 02, 2007 12:04 pm

How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
(A) 15
(B) 96
(C) 216
(D) 120
(E) 180

from diff math doc, ans coming when peopple respond with explanations

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Source: Beat The GMAT — Problem Solving | Mon Apr 02, 2007 12:04 pm

800GMAT: Master | Next Rank: 500 Posts; Posts: 174; Joined: Mon Mar 26, 2007 5:51 am; Thanked: 1 times

by 800GMAT » Mon Apr 02, 2007 1:15 pm

A five digit integer divisible by 3 can be formed if:

1. 0 is exluded 1+2+3+4+5= 15, divisible by 3

and

2. 3 is excluded 0+1+2+4+5= 12, divisible by 3

1. 0 is excluded - 1,2,3,4,5
you can make a five digit number with distinct digits in 5! ways = 120

2. 3 is exluded - 0,1,2,4,5
First digit- 4 ways - 0 cannot be the first digit
Second digit- 4 ways
third - 3 ways
fourth - 2 ways
fifth -1 way
Thus, 4*4*3*2*1= 96

Total ways = 120+96 = 216

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800guy: Master | Next Rank: 500 Posts; Posts: 354; Joined: Tue Jun 27, 2006 9:20 pm; Thanked: 11 times; Followed by:5 members

oa

by 800guy » Wed Apr 04, 2007 2:49 pm

here's the oa:

The sum of digits of a multiple of 3 should be div by 3.
for a 5 digit number to be div by 3, the sum of digits (given the digits here) can be only 12 or 15.
For a sum of 12, the digits that can be used : 0,1,2,4,5
for a sum of 15: 1,2,3,4,5
Number of numbers from the first set = 4.4! (0 cannot be the first digit in the numbers)
for the second set : 5!
total = 5! +4.4! = 4!(5+4) = 24*9 = 216

Since 0 cannot be the first digit of a number, for the first position, you have 4 choices (all digits except zero). No such constraints exist for the rest of the positions; hence the next choices are 4,3,2,1 - all multiplying up to give a 4!. Had there been no 0 involved, the choices would've been 5! Instead of 4.4!

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pm; Followed by:6 members

by BTGmoderatorRO » Sun Nov 05, 2017 11:38 am

we are to form 5-digits number which is divisible by 3 from 0,1,2,3,4 and 5.
NOTE: As a rule of divisibility, the sum of digits in the number must be divisible by 3 for the number itself to be divisible by 3.
For a 5-digit numbers, only 1,2,3,4 and 5 can have their sum divisible by 3. This means that we must exclude 0 from our selection, as there is no permutation that will be divisible by 3 if 0 is among.
Therefore , Number of 5-digits number that could be formed, excluding 0 is = 5! = 5*4*3*2! = 120 possible 5-digit numbers

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Tue Oct 29, 2019 6:02 am

800guy wrote:How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
(A) 15
(B) 96
(C) 216
(D) 120
(E) 180

from diff math doc, ans coming when peopple respond with explanations

Recall that a number is divisible by 3 if the sum of the digits of the number is a multiple of 3. Since 1 + 2 + 3 + 4 + 5 = 15 and 0 + 1 + 2 + 4 + 5 = 12, any number formed using the digit set {1, 2, 3, 4, 5} or the digit set {0, 1, 2, 4, 5} is divisible by 3.

In the first digit set, 5! = 120 numbers can be formed. However, in the second digit set, since the first digit can't be 0, the number of numbers can be formed is 4 x 4 x 3 x 2 x 1 = 96. Therefore, the total number of numbers can be formed is 120 + 96 = 216.

Answer: C

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