If an integer p is to be chosen at random from 1 to 96 (both inclusive), what is the probability that p(p+1)(p+2) will be divisible by 8?

62.5%

Step-by-step explanation:

p is either even or row.

IF P IS EVEN

If even then p(p+1)(p+2) will be always be divisible by 8. p and p+2 are even numbers an at least one of those two numbers is divisible by 4.

let say that p/2=a and (p+2)/2=b, then a is even or b is b is even (b=a+1, therefore one of those two numbers must be even), so if b is divisible by 2 then (p+2) is divisible by 4, or if a is divisible by 2 then p is divisible by 4.

p(p+1)(p+2) = 2*a*(p+1)*2*b

since a or b are divisible by 2 then p(p+1)(p+2) is divisible by 8

IF P IS ROW

If p is row, then p+2 is also row, and p(p+2) is also row (and row numbers are never divisible by 8), so p(p+1)(p+2) is divisible if and only if (p+1) is divisible by 8. The list of p between 1 and 96 where (p+1) is divisible by 8 is: 7,15,23,31,39,47,55,63,71,79,87,95. A total of 12 numbers.

IN CONCLUSION

for all even numbers, between 1 and 96, p(p+1)(p+2) is divisible by 8. there are 43 even numbers.

For all the row numbers, between  1 and 96, just (p+1) divisible by 8 will be divisible by 8 . There are 12 numbers like that.

d

step-by-step explanation:

d

answer: 2530

hope this