For any integer ‘n’ greater than 0, n! denotes the product of all the integers from 1 to ‘n’, inclusive. How many multiples of 3 are there between 6!-6 and 6!+6, inclusive ?
A) One
B) Two
C) Three
D) Four
E) Five
GMAT Question of The Day : Defined Function
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ans : D (714,717,720,723,726)
Very less attempt..seems difficult !!!
Answer should be “E” (there are 5 multiples of 3 between 714 and 726 (inclusive)- 714, 717, 720, 723, 726)
6!-6 = 720-6=714
6!+6 = 720+6=726
Quest – I guess you have, by mistake, mentioned D whereas you have mentioned all the 5 multiples correctly.
yeah Correct answer is E.
the answer is 5..
6!-6=714 ; 6!+6=726…
therefore nos divisible by three are 714,717,720,723,726
hence answer is 5…
the option to be chosen is ‘ E’
But is there any shortcut without calculating actually ??
Ya. The answer is E. 6!-6 , 6!-3 , 6!-0, 6!+3 and 6!+6. 6! is always divisible by 3. So second term shud also be a multiple of 3.
Anusha,
….. rocking.fantastic approach !!
E … liked Anushas approach
e
no. 714-726
tot 12/3=4
inclusiv 4+1=5
5
5
Guys, can we not solve the prob like this –
6! + 6 – (6! – 6) = 12
So 12/3 = 4
6! + 6 is divisible by 3. So 1 more to add. So the answer is 4 + 1 = 5.
E
e:5
Ans : E
The numbers are : 714,717,720,723,726
Divisibility by 3 : If sum of digits is divisible by 3 then number is divisible by 3.
e
we donnt need to solve the whole exprssion for deducing the answer.
by logic, we all agree that 6! will be divisible by 3..
now lets assume that 6!= g so 6!-6 = a and 6!+6= m
so we have a b c d e f g h i j k l m, so the factors of three will be at the distance of 3… so from g’s right. we have j and m, from g’s left we will have d,a
so in all we have. a,d,g,j and m.. 5 is the answer. period..
when u do it in ur mind,, it will take less than 3 seconds.. the explaantion is only to limpid the calculations..
imagine doing this stuff with hindi alphabest.. k, kh, g, gh etc…:):):)
love peace
option E
Choice D – Four
option E
Ans E
E it is…
Undoubtedly Ans is D
Previous line was just copy+paste
Undoubtedly Ans is E
E … Anushas approach is the easiest.
E.
E.
I just took a random multiple of 6 (36) and did 36-6=30 and 36+6=42. Between 30 and 42 there are five multiples of 3 which you can extend to any multiple of 6 plus or minus 6.