Latest Elitmus Aptitude Question SOLUTION: Q. g(n)= 2 *sum of first n digit no+41 then wîch is the smallest non prime no. A)n=6 b)n=7 c)n=40 d) g(n) is always prime no.

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Q. g(n)= 2 *sum of first n digit no+41 then wîch is the smallest non prime no. A)n=6 b)n=7 c)n=40 d) g(n) is always prime no.

Asked In Elitmus ankit chauhan (10 years ago)

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pawann raj chauhan  (10 years ago)

i think answer should be c)40 put n=40 in eqnt n(n+1)+41= 40*41+41= 1640+41=1681 which is not a prime no...because it square of 41...41*41=1681..and prime no are those no which is divisible by 1 or itself...g(n) is non prime at n=40

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RAKESH  (10 years ago)

d) g(n) is always prime no

for any value n=1,2,3,... ,value of g(n) is always greater than 41
so option a,b,c are incorrect

g(n)= 2 *sum of first n digit no + 41
g(1)=2*1+41=43
g(2)=2*(10+11)+41 =83
g(3)=3*(100+101+102)+41=647
so g(n) is always prime no.



however g(n) is not clear?

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deep chandra bhatt  (10 years ago)

ans is d
g(n)=2*n(n+1)/2+41
put the value n=1,2,3,4,4,10...
2*1*2/2+41=43 its prime no
2*10*11/2+41=151 its prime no

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Bhaskar Kumar Singh  (10 years ago)

a)n=6

(n^2+2n+86)/2==>if n=6,then 134/2=67 which is prime no.

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Naveen Agarwal  (10 years ago)

n=40.... g(n)=1681 it is not prime

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arijit roy  (10 years ago)

40 will be the ans
g(n)=40*41+41
then ans should be divided by 41
so 40 will be the ans guys :)

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T.VIGNESWARA REDDY  (9 years ago)

given g(n)=n*(n+1)+41 is prime number and n should small from given options n=6 6*7+42=83 so ans is n=6

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