A six-digit positive integer, ABCDEF, is divisible by all the integers

Question

e-GMAT          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

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Quant-workshop_Webinar-e1533628475738.jpg

Metwing · Accepted Answer

Answer : E

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

rinkumaa4 · Answer

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

faltan · Answer

this took 5 mins. I think this is not of 600 pts. scale

EgmatQuantExpert · Answer

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.

Answer: E

https://e-gmat.com/blogs/wp-content/uploads/2018/07/FT-GC-FB-event-e1532524398949.jpg

arvind910619 · Answer

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

Kinshook · Answer

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?

Digits = {0,1,2,3,4,5,6} : 7 options and all are distinct

Divisibility rule by 2: even number: last digit = {0,2,4,6} : 4 options
Divisibility rule by 3: Sum of digits is divisible by 3
Divisibility rule by 4: Last 2 digits are divisible by 4
Divisibility rule by 5: Last digit = {0,5}
Divisibility rule by 6: automatic if Divisibility rule by 2 & Divisibility rule by 3 are complied with

Last digit = F = 0; Combining Divisibility rule by 2 & Divisibility rule by 5
Second last digit = E = {2,4,6}: Divisibility rule by 4
Sum of digits = A+B+C+D+E+F(0) is a multiple of 3: Divisibility rule by 3 & Divisibility rule by 6

A+B+C+D+E = 3k; where k is an integer

Case 1: E = 2; A+B+C+D = 3k - 2; 1+3+4+5= 13 <= 3k - 2 <= 6+5+4+3= 18
15 <= 3k <= 20; 5 <= k < 7; k = {5,6}
Case 1a: k = 5; A+B+C+D=13; (A,B,C,D) = (1,3,4,5); Number of ways = 4! = 24
Case 1b: k = 6; A+B+C+D=16; (A,B,C,D) = (1,4,5,6); Number of ways = 4! = 24

Case 2: E=4; A+B+C+D = 3k - 4; 1+2+3+5= 11 <= 3k - 4 <= 6+5+3+2= 16
15 <= 3k <= 20; 5 <= k < 7; k = {5,6}
Case 1a: k = 5; A+B+C+D=11; (A,B,C,D) = (1,2,3,5); Number of ways = 4! = 24
Case 1b: k = 6; A+B+C+D=14; (A,B,C,D) = (1,2,5,6); Number of ways = 4! = 24

Case 3: E=6; A+B+C+D = 3k - 6; 1+2+3+4= 10 <= 3k - 4 <= 5+4+3+2= 14
14 <= 3k <= 18; 5 <= k <= 6; k = {5,6}
Case 1a: k = 5; A+B+C+D=9; No solution since A, B, C & D are distinct
Case 1b: k = 6; A+B+C+D=12; (A,B,C,D) = (1,2,4,5); Number of ways = 4! = 24

Total such numbers = 24*5 = 120

IMO E

A six-digit positive integer, ABCDEF, is divisible by all the integers

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A six-digit positive integer, ABCDEF, is divisible by all the integers

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards