Very useful explanations in this post. My rewording and working through the problems:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?We have 100 consec ints, half are even and half are odd.
Out of the 50 even numbers, half are divisible by 4 (can check quickly by writing out: {1,2,3,
4} {2,3,
4,5} {3,
4,5,6} {
4,5,6,7} or use the formula
Last multiple of 4 (100) - First multiple of 4 (4) / 4 + 1 = 24 + 1 = 25.
BUT, this is the pattern for n alone. If we have n(n+1) that means either n or n+1 can be one out of 4 numbers divisible by 4. So 25*2 --> 50/100 = 1/2.
In other words, we have a repeating pattern of 4 numbers where 1n out of 4n is divisible by 4. {1,2,3,4} choose 1, 2, 3 or 4.
This pattern is repeated 25 times out of 100 from 1-100 {1,2,3,4} ... {97,98,99,100} when we have 1 choice.
So we have 25/100 = 1/4 choices for just n.
For n(n+1) we have 2 choices from {1,2,3,4} choose 1,2 or 2,3 or 3,4, 4,5
The product of 2 out of those 4 is divisible by 8. That means 50 out of 100 possible numbers or 1/2.
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?Given: 96 consec ints from 1-96
1) Out of every 8 consecutive ints, for just n have 1/8 that is acceptable.
2) But, we have 3 choices to make a number divisible by 8, n(n+1)(n+2)
Unlike the previous example, there isn't symmetry since there's 2 YES cases : Where n(n+1)(n+2) is E#O#E#, and when n(n+1)(n+2) is O#E#O# AND n+1 = a multiple of 8. The single NO case is when O#E#O# and the E# does not have enough factors of 2.
From #1-8 we can pick and see which 3 numbers have enough factors of 2 to make the product divisible by 8.
n,n+1,n+2
{1,2,3} no
{2,3,4} yes
{3,4,5} no
{4,5,6} yes
{5,6,7} no
{6,7,8} yes
{7,8,9}yes
{8,9,10} yes
It is symmetrical (Case 1 E#O#E#) until we get to {7,8,9} which is our yes Case 2, where even though n = O#, we have enough factors of 2 in the E# n+1.
So every 8 numbers, we have 5/8 choices where it is divisible by 8.
If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ? Given: 99-20+1 = 80 consec ints from 20 to 99.
1) Out of every 12 ints, 1 is divisible by 12
2) We have (n-1)n(n+2) so it is slightly different from the problem above.
Question: do the 3 numbers have 2 factors of 2? (3 consec ints are always divisible by 3, but may not be divisible by 4)
We have 2 Yes cases (all E#O#E# are at least 2 factors of 2, and O#E#O# where n has at least 2 factors of 2). The No case is when O#E#O# and n doesn't have enough factors of 2.
Sequence of 12 numbers where n = 20 through 31
n-1, n, n+1
{19,20,21} yes, case 2
{20,21,22} yes, case 1
{21,22,23} no
{22,23,24} yes, case 1
{23,24,25} yes, case 2
{24,25,26} yes, case 1
{25,26,27} no
{26,27,28} yes, case 1
{27,28,29} yes, case 2
{28,29,30} yes, case 1
{29,30,31} no
{30,31,32} yes, case 1
So 9/12 or cases are YES, 3/12 are NO.
VeritasKarishma I hope that is all correct, my brain is fried from typing all that out.
I don't understand the logic behind how 1 - number of choices/sequence gives us the right answer all 3 times here. I can understand finding the probability from 1-Not A = A but how does Not A relate to number of choices/sequence? Can you give an example of when this shortcut doesn't work?