GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Feb 2019, 22:29

### GMAT Club Daily Prep

#### Thank 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
for You

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
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.
• ### Free GMAT RC Webinar

February 23, 2019

February 23, 2019

07:00 AM PST

09:00 AM PST

Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT

# Question of the Week- 16 (Set S contains all the integers from 10...)

Author Message
TAGS:

### Hide Tags

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2594
Question of the Week- 16 (Set S contains all the integers from 10...)  [#permalink]

### Show Tags

Updated on: 05 Oct 2018, 00:27
1
12
00:00

Difficulty:

95% (hard)

Question Stats:

37% (02:28) correct 63% (02:47) wrong based on 91 sessions

### HideShow timer Statistics

Question of the Week #16

A six-digit positive integer, ABCDEF, is divisible by all the integers from 2 to 6, both inclusive. How many such numbers are possible, if all the digits of the number are distinct and less than 7?

A. 24
B. 48
C. 72
D. 96
E. 120

_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Originally posted by EgmatQuantExpert on 28 Sep 2018, 02:41.
Last edited by EgmatQuantExpert on 05 Oct 2018, 00:27, edited 1 time in total.
Intern
Joined: 30 Jun 2017
Posts: 7
Location: India
Schools: Babson '20 (A)
Re: Question of the Week- 16 (Set S contains all the integers from 10...)  [#permalink]

### Show Tags

28 Sep 2018, 04:21
2

It's little tricky question, I solved it but took longer than 2 minute. Pl. suggest quicker methods. Explanation given is lengthy to bring clarity.

Digits possible : 0,1,2,3,4,5, and 6 (Given : digits less than 7)
Divisible by : 2,3,4,5, and 6 (Given)

ABCDEF

F = 0 : For number to be divisible by 5 - unit digit can be 0 or 5, but also divisible by 2 hence only 0 is possible.
E = 2 or 4 or 6 : For number to be divisible by 4, (2E+F) should be divisible by 4, F=0 hence all even numbers are only possible, i.e. 2, 4, and 6.

For digits of ABCDE : Option are : 1,2,3,4,5, & 6

For number to be divisible by 3, The sum of the numbers should be divisible by 3.
i.e. A+B+C+D+E+F = 3k,

Basically there are now 5 digits to be choosen in following manner,

F=0, E=2, : Possible - (1, 3, 4 , & 5) or (1, 4 , 5, & 6); Hence total numbers possible = 4P4 + 4P4 = 24 + 24 =48

F=0, E=4, : Possible - (1, 2, 3, & 5) or (1, 2 , 5, & 6); Hence total numbers possible = 4P4 + 4P4 = 24 + 24 =48

F=0, E=6, : Possible - 1, 2, 4, & 5 Hence total numbers possible = 4P4 = 24

Total = 48+48+24 = 120
Intern
Joined: 15 Jul 2016
Posts: 28
Location: India
Schools: IIMC
GMAT 1: 560 Q46 V21
GMAT 2: 620 Q48 V26
GPA: 2.67
WE: Operations (Manufacturing)
Re: Question of the Week- 16 (Set S contains all the integers from 10...)  [#permalink]

### Show Tags

29 Sep 2018, 04:18
1
ANS - 120 E
As we have 7 digit and the number should be divided by 60 (L.C.M of all digit 0-6), We have to put zero on the right most and from the seven digits we can exclude either 3 or 6. note-Zero cant be excluded.

Excluding 3:

We have 4*3*2*1*3*1= 72 numbers

Excluding 6:

We have 4*3*2*1*2*1= 48 numbers

So Total = 72+48= 120 numbers
Manager
Joined: 12 Mar 2018
Posts: 126
Re: Question of the Week- 16 (Set S contains all the integers from 10...)  [#permalink]

### Show Tags

02 Oct 2018, 04:46
this took 5 mins. I think this is not of 600 pts. scale
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2594
Re: Question of the Week- 16 (Set S contains all the integers from 10...)  [#permalink]

### Show Tags

03 Oct 2018, 03:36

Solution

Given:
• ABCDEF is a six-digit number, and is divisible by all the integers from 2 to 6, both inclusive
• All the six digits are distinct and are less than 7

To find:
• The number of such numbers

Approach and Working:
• ABCDEF is divisible by 2 and 5, which implies that the units digit must be 0
o So, F = 0

• We are given that all the digits are less than 7
• Implies, the possible digits are {0, 1, 2, 3, 4, 5, 6}
o And since, F = 0, no other digit can take that value.
o Sum of the remaining 6 digits of the above set = 1 + 2 + 3 + 4 + 5 + 6 = 21

• Now, A, B, C, D and E can take one value each among these 6 digits, such that their sum is a multiple of 3.
o But, since ABCDEF is divisible by 4, EF must be 20 or 40 or 60
o So, E can take one of the digits among {2, 4, 6}

• The two possible cases for sum to be a multiple of 3 are 21 - 3 and 21 – 6
o That is, (1 + 2 + 4 + 5 + 6) and (1 + 2 + 3 + 4 + 5)

• Case-1: {1, 2, 4, 5, 6}
o Since, E must be even, the number of possibilities for E = 3
o The remaining 4-digits can be arranged in 4! ways = 24
o So, the total number of possibilities = 3 * 24 = 72

• Case 2: {1, 2, 3, 4, 5}
o As E is even, the number of possibilities for E = 2
o The remaining 4-digits can be arranged in 4! ways = 24
o So, the total number of possibilities = 2 * 24 = 48

Therefore, the total number of such numbers = 72 + 48 = 120

Hence, the correct answer is option E.

_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

VP
Status: Learning
Joined: 20 Dec 2015
Posts: 1038
Location: India
Concentration: Operations, Marketing
GMAT 1: 670 Q48 V36
GRE 1: Q157 V157
GPA: 3.4
WE: Engineering (Manufacturing)
Re: Question of the Week- 16 (Set S contains all the integers from 10...)  [#permalink]

### Show Tags

05 Oct 2018, 06:48
EgmatQuantExpert wrote:
Question of the Week #16

A six-digit positive integer, ABCDEF, is divisible by all the integers from 2 to 6, both inclusive. How many such numbers are possible, if all the digits of the number are distinct and less than 7?

A. 24
B. 48
C. 72
D. 96
E. 120

Imo E
The numbers are 1,2,3,4,5 and 6 and this ABCDEF is divisible by each 2,3,4,5 and 6.
So last digit has to 0 to make it even so that it can be divisible by 2.
the number has to be divisible by 5 and 6 so second last digit can be either 3 or 6.
The number will always be divisible by 3 as 1+2+3+4+5+6=21.

So number of such numbers are 2*3*4*5*2*1=120
_________________

Re: Question of the Week- 16 (Set S contains all the integers from 10...)   [#permalink] 05 Oct 2018, 06:48
Display posts from previous: Sort by