If an integer $n$ is chosen at random from $1$ to $96$ inclusive ,what is the probability that $n(n+1)(n+2)$ is divisible by 8? In this one If I consider that n is even then probability that the number $n(n+1)(n+2)$ will be divisible by 8 will be 1/2 ,Now if n is odd then for n(n+1)(n+2) to be divisible by 8 ,n+1 should be a multiple of 8 ,Now how to find the probability among 96 numbers such that it is a multiple of 8 ?
 A: Looking at the equation $$n(n+1)(n+2)\equiv 0\pmod{8},$$ we see that this happens when $$n\equiv 0,2,4,6,7,8\pmod{8},$$ and so we get a probability of $\frac{5}{8}.$
A: the total numbers are $96$. for an odd number to be multiple of 8 using expression $n(n+1)(n+2)$ the number $n$ should be a number preceding to a multiple of $8$ starting from $8$ itself so first number is $7$ and such is an AP whose last term is $95$. So total terms which are $1$ less than a multiple of $8$ ie odd are $12$ thus the probability is $\frac{12}{96}=\frac{1}{8}$. Numbers with $4m-2=\frac{1}{4}$ and then $4m=\frac{1}{4}$ so addition is $\frac{5}{8}$
A: *

*$n=8m$

*$n=8m-1$

*$n=8m-2$.

*$n=4m$

*$n=4m-2$


1) is a subset of 4) so we can remove 1).
3) is a subset of 5) so we can remove 3).
So,


*$n=8m-1$ $\rightarrow P=\frac18$

*$n=4m$ $\rightarrow P=\frac14$

*$n=4m-2$ $\rightarrow P=\frac14$


$$\frac18+\frac14+\frac14=\frac58$$
A: HINT:


*

*$n\equiv0\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv0\cdot(0+1)\cdot(0+2)\equiv\color\green0\pmod8$

*$n\equiv1\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv1\cdot(1+1)\cdot(1+2)\equiv\color\red  6\pmod8$

*$n\equiv2\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv2\cdot(2+1)\cdot(2+2)\equiv\color\green0\pmod8$

*$n\equiv3\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv3\cdot(3+1)\cdot(3+2)\equiv\color\red  4\pmod8$

*$n\equiv4\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv4\cdot(4+1)\cdot(4+2)\equiv\color\green0\pmod8$

*$n\equiv5\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv5\cdot(5+1)\cdot(5+2)\equiv\color\red  2\pmod8$

*$n\equiv6\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv6\cdot(6+1)\cdot(6+2)\equiv\color\green0\pmod8$

*$n\equiv7\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv7\cdot(7+1)\cdot(7+2)\equiv\color\green0\pmod8$


Note that the number of elements in the range $[1,96]$ is a multiple of $8$.
A: All 48 even values of $n$ in the range qualify, as values of $(n+1)$ from $8\times 1$ thru $8\times 12$,
thus there are $60$ favorable cases,
and $ Pr = \dfrac{60}{96} = \dfrac{5}{8}$  
