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-

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

a n i n t e g e r n b e t w e e n 1 a n d 98,

i n c l u s i v e i s t o b e c h o s e n a t

r a n d o m . w h a t i s t h e p r o b a b i l i t y

t h a t n (n + 1) w i l l b e d i v i s i b l e b y 3

Commented by bemath last updated on 16/Jul/20

t h a n k y o u a l l

Answered by OlafThorendsen last updated on 16/Jul/20

1 st case :

n = 3 p, p = 1 . . 32

32 possibilities

n (n + 1) can be divisible by 3

2 nd case :

n = 3 p + 1 then n + 1 = 3 p + 2

n (n + 1) = (3 p + 1) (3 p + 2)

n (n + 1) {can}^{'} t be divisible by 3

3 rd case :

n = 3 p + 2, p = 0 . . 32

33 possibilities

then n + 1 = 3 p + 3 = 3 (p + 1)

n (n + 1) = 3 (3 p + 2) (p + 1)

n (n + 1) can be divisible by 3

Total result :

32 + 0 + 33 = 65

probability = \frac{65}{98} \approx 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] \to 98 terms

1) AP : (3, 6, 9, \dots, 96) \to {a_{n} = 3 n}

3 n = 96 \Rightarrow n = 32 (the AP has 32 terms)

so the probability that n is a multiple of

3 is : P_{1} = \frac{32}{98} = \frac{16}{49}

2) AP : (2, 5, 8, \dots, 95) \to 32 terms

probability of n + 1 multiple of 3 :

P_{2} = \frac{32}{98} = \frac{16}{49}

\Rightarrow probability that n (n + 1) is divisible

by 3 :

P_{1} + P_{2} = \frac{32}{49} \approx 65 %

Commented by OlafThorendsen last updated on 16/Jul/20

Why n = 98 cannot be chosen in

2) ?

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

yeah {you}^{'} re right i miss that : /

Answered by bobhans last updated on 16/Jul/20

n o t e t h a t n (n + 1) i s n o t d i v i s i b l e b y 3

p r e c i s e l y w h e n n \equiv 1 (m o d 3); t h a t i s n

h a s r e m a i n d e r 1 u p o n d i v i s i o n b y 3 . f r o m

1 t o 98 i n c l u s i v e t h e r e a r e 33 s u c h v a l u e s

o f n . h e n c e t h e d e s i r e d p r o b a b i l i t y e q u a l s

t o 1 - \frac{33}{98} = \frac{65}{98} .

[p (A) = 1 - p (A^{c})]

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