Question Number 103606 by bemath last updated on 16/Jul/20

an integer n between 1 and 98 ,  inclusive is to be chosen at  random. what is the probability  that n(n+1) will be divisible by 3

Commented bybemath last updated on 16/Jul/20

thank you all

Answered by OlafThorendsen last updated on 16/Jul/20

1st case :  n = 3p, p = 1..32  32 possibilities  n(n+1) can be divisible by 3  2nd case :  n = 3p+1 then n+1 = 3p+2  n(n+1) = (3p+1)(3p+2)  n(n+1) can′t be divisible by 3  3rd case :  n = 3p+2, p = 0..32  33 possibilities  then n+1 = 3p+3 = 3(p+1)  n(n+1) = 3(3p+2)(p+1)  n(n+1) can be divisible by 3  Total result :  32+0+33 = 65  probability = ((65)/(98)) ≈ 66,3%

Answered by floor(10²Eta[1]) last updated on 16/Jul/20

for n(n+1) be divisible by 3, we have 2  cases:  1)n is a multiple of 3  2)n+1 is a multiple of 3  [1, 98]→98 terms  1)AP: (3, 6, 9, ..., 96)→{a_n =3n}  3n=96⇒n=32(the AP has 32 terms)  so the probability that n is a multiple of  3 is: P_1 =((32)/(98))=((16)/(49))  2)AP: (2, 5, 8, ..., 95)→32 terms  probability of n+1 multiple of 3:  P_2 =((32)/(98))=((16)/(49))  ⇒probability that n(n+1) is divisible  by 3:  P_1 +P_2 =((32)/(49))≈65%

Commented byOlafThorendsen last updated on 16/Jul/20

Why n = 98 cannot be chosen in  2) ?

Commented byfloor(10²Eta[1]) last updated on 16/Jul/20

yeah you′re right i miss that :/

Answered by bobhans last updated on 16/Jul/20

note that n(n+1) is not divisible by 3  precisely when n ≡ 1 (mod 3) ; that is n  has remainder 1 upon division by 3. from  1 to 98 inclusive there are 33 such values  of n. hence the desired probability equals  to 1−((33)/(98)) = ((65)/(98)) .   [ p(A) = 1−p(A^c ) ] □