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Question of the Week 16 (Set S contains all the integers from 10...)
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Updated on: 05 Oct 2018, 01:27
Question Stats:
51% (02:27) correct 49% (03:05) wrong based on 73 sessions
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eGMAT Question of the Week #16A sixdigit 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
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Re: Question of the Week 16 (Set S contains all the integers from 10...)
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28 Sep 2018, 05:21
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



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Re: Question of the Week 16 (Set S contains all the integers from 10...)
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29 Sep 2018, 05:18
ANS  120 E As we have 7 digit and the number should be divided by 60 (L.C.M of all digit 06), We have to put zero on the right most and from the seven digits we can exclude either 3 or 6. noteZero 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



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Re: Question of the Week 16 (Set S contains all the integers from 10...)
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02 Oct 2018, 05:46
this took 5 mins. I think this is not of 600 pts. scale



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Re: Question of the Week 16 (Set S contains all the integers from 10...)
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03 Oct 2018, 04:36
Solution Given:• ABCDEF is a sixdigit 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
• 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) • Case1: {1, 2, 4, 5, 6}
o Since, E must be even, the number of possibilities for E = 3 o The remaining 4digits 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 4digits 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
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Re: Question of the Week 16 (Set S contains all the integers from 10...)
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05 Oct 2018, 07:48
EgmatQuantExpert wrote: eGMAT Question of the Week #16A sixdigit 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
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