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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Thu Oct 06, 2011 1:43 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

Don't have an OA. Detailed explanations would be appreciated.

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Source: Beat The GMAT — Problem Solving | Thu Oct 06, 2011 1:43 pm

GmatMathPro: GMAT Instructor; Posts: 349; Joined: Wed Sep 28, 2011 3:38 pmFollowed by: 54 members; Location: Austin, TX; Thanked: 236 times; GMAT Score:770

by GmatMathPro » Thu Oct 06, 2011 1:52 pm

An integer is divisible by 3 if the sum of its digits is divisible by 3. The sum of 0,1,2,3,4,5 is 15. There are six digits, so we have to choose one digit to omit from the list to form a 5 digit number. To make it a multiple of 3, we can only omit 0 or 3.

Omitting 0:

1,2,3,4,5 can be arranged 5!=120 ways.

Omitting 3:

0,1,2,4,5

Any digit can come first except 0. Using slot method: 4*4*3*2*1=96

[spoiler] 120+96=216 Ans: C[/spoiler]

Pete Ackley
GMAT Math Pro
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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Thu Oct 06, 2011 2:41 pm

Pete,
A slight variation to this problem.

How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 4, without repeating the digits?

Plz help!!

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GmatMathPro: GMAT Instructor; Posts: 349; Joined: Wed Sep 28, 2011 3:38 pmFollowed by: 54 members; Location: Austin, TX; Thanked: 236 times; GMAT Score:770

by GmatMathPro » Thu Oct 06, 2011 3:00 pm

Here is one option along the same lines:

Any integer whose last two digits are divisible by 4 is divisible by 4 itself. For example 53124=53100+24. 4 is a factor of 53100 because 100 is a multiple of 4, and 4 is obviously a factor of 24.

First, find the two digit pairs that are divisible by 4: 52, 40, 32, 24, 12, 04

52, 32, 24, 12:

First three digits can be selected in 3*3*2=18 ways. any of them can be paired with any of those 4 endings so 18*4=72 numbers in this case

40, 04:

4*3*2=24 ways to choose the first three digits. 24*2=48 numbers in this case

72+48=120 numbers

Double check that, but something along those lines should work.

Pete Ackley
GMAT Math Pro
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bhavsk9: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Thu Aug 04, 2011 6:54 am; Location: Seattle, WA; Thanked: 1 times

by bhavsk9 » Thu Oct 06, 2011 3:03 pm

To make the numbers divisible by 4, the last 2 digits of the number must be divisible by 4.

We can have, 12, 24, 32, 52 (set 1) as the last 2 digits. We are left with 4 digits (including a zero) , therefore, 3 * 3 * 3 * 2 are the number of ways in which this can be done. 18 for each combination listed in set 1. Total1 = 18 * 4 = 72.

We can also have 04, 40 (set 2) as the ending. We now have 4 digits which can be arranged in 4 * 3* 2 ways. (0, 4) can be arranged in 2 ways. Total2 = 4 * 3 * 3 * 2 = 48.

We can have 20 as the ending digits (set 3). That leaves us with 4 digits which can be arranged in 4 * 3 * 2 ways = 24.

Total = 72 + 48 + 24 = 144.

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GmatMathPro: GMAT Instructor; Posts: 349; Joined: Wed Sep 28, 2011 3:38 pmFollowed by: 54 members; Location: Austin, TX; Thanked: 236 times; GMAT Score:770

by GmatMathPro » Thu Oct 06, 2011 3:13 pm

Whoops. forgot 20. thanks, bhavsk9.

Pete Ackley
GMAT Math Pro
Free Online Tutoring Trial

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