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Imagine P is 5. Then we could have 1, 1, 3 or 1, 2, 2. 

Imagine P is 4. Then we could have 1, 1, 2. That's it

Imagine P is 3. Then we could have 1, 1, 1. That's it

When P is 2, there's no triangle like that.

If it's 5 or above, there will be multiple ways.

Therefore, there are 2 values of P.

Note: Make sure there are no duplications in your counting:

A short computer code shows there are: 24,000 5-digit integers that are NOT divisible by all the elements in the set {1, 2, 3, 4, 5, 6}. They begin like this:

(10001 , 10003 , 10007 , 10009 , 10013 , 10019 , 10021 , 10027 , 10031 , 10033 , 10037 , 10039.......and end like this:

99953 , 99959 , 99961 , 99967 , 99971 , 99973 , 99977 , 99979 , 99983 , 99989 , 99991 , 99997)

Therefore, the probability is: 24,000 / 90,000 ==4 / 15

The total number of 5-digit numbers is 90,000 (from 10,000 to 99,999).
Out of these, numbers divisible by 2, 3, 4, 5, 6 are 30,000, 20,000, 15,000, 12,000, 10,000 respectively.

The number of 5-digit numbers divisible by both 2 and 3 is 6000, and those divisible by 2, 3, and 4 are 2000. The number divisible by 2, 3, 4, 5 is 800, and those divisible by 2, 3, 4, 5, 6 is 400.

Therefore, the number of 5-digit numbers divisible by (1, 2, 3, 4, 5, 6) is 400.
Hence, the probability that the number is NOT divisible by every element of (1,2,3,4,5,6) is (90,000-400)/90,000 = 89,600/90,000 = 224/225.

The LCM of [1,2,3,4,5,6] is going to be 60. The least 5 digit number that is a multiple of 60 is going to be 10020 = 60 x 170, the greatest being 99960 = 60 x 1666. from 170 to 1666, there will be 1666-170+1 multiples of 60, so 1497 multiples. There is a total of 99999-10000+1 = 90000 5 digit numbers so the odds of a 5 digit number not being divisible by all of the numbers in the set is going to be \(\frac{90000-1497}{90000}\) which is going to be 

NOTE:Guest, the problem says that the number is not divisible by EVERY ELEMENT IN THE SET

\(\frac{29501}{30000}\)

The number is divisible by 6 if and only if it is divisible by both 2 and 3. To find the total number of 5-digit numbers, we consider all numbers between 10,000 and 99,999, inclusive, which is 90,000. To find the number of numbers that are not divisible by any of (1,2,3,4,5,6), we can find the number of numbers that are divisible by each of these factors and subtract this number from 90,000.

The number of numbers that are divisible by 2 is 45,000, the number that are divisible by 3 is 30,000, and the number that are divisible by 6 is 15,000. The number of numbers divisible by 4, 5, and 6 can be calculated in a similar fashion.

To find the number of numbers divisible by the least common multiple (LCM) of (2, 3, 4, 5, 6), we find the LCM and then find the number of 5-digit numbers divisible by this LCM. The LCM of (2, 3, 4, 5, 6) is 60, and there are 15,000 numbers divisible by 60.

Therefore, the number of numbers that are divisible by at least one of the numbers in the set (1,2,3,4,5,6) is 45,000 + 30,000 + 15,000 - 15,000 = 60,000.

So, the number of numbers that are not divisible by any of (1,2,3,4,5,6) is 90,000 - 60,000 = 30,000.

Finally, the probability of a random 5-digit number being NOT divisible by any of the numbers in the set (1,2,3,4,5,6) is 30,000/90,000 = 1/3.

Read your question. it makes absolutley no sense and you have no reason to copy paste either alcumus or ur elementary school homework on this. revise your question

The smallest positive integer n so that you can avoid ever having the exact same group of 4 members over all 100 days is n = 16. This is because the total number of possible combinations of 4 members out of a group of n is n!/(n-4)! and for n = 16, this gives 4,096 combinations. Since you have 100 days, this means you can have more than 4 combinations per day and still avoid ever having the exact same group of 4 members.

4-digit permutations==(1, 3, 3, 3),   (2, 2, 3, 3) ==10
5-digit permutations==(1,1,2,3,3), (1, 2, 2, 2, 3), (2, 2, 2, 2, 2)=51

6-digit permutations==(1,1,1,1,3,3), (1,1,1,2,2,3), (1,1,2,2,2,2)=90 

7-digit permutations==(1,1,1,1,1,2,3), (1,1,1,1,2,2,2)==77 

8-digit permutations==(1,1,1,1,1,1,1,3), (1,1,1,1,1,1,2,2)==36 

9-digit permutations==(1,1,1,1,1,1,1,1,2)==9

10-digit permutation ==(1,1,1,1,1,1,1,1,1,1)==1

Grand Total ==10 + 51 + 90 + 77 + 36 + 9 + 1 ==274 permutations.

The answer is 13/56.

ok so the correct answer was -12 and I was overthinking it thank you!

10 - 99 : 11
100 - 999 : 112
1000 - 9999 : 1124
10000 - 99999 : 11248
100000 - 999999 : 112496
1000000 - 9999999 : 1124992
10000000 - 99999999 : 11249985
100000000 - 999999999 : 112499976
1000000000 - 9999999999 : 1124999972

The smallest integer n that satisfies the condition is 23

This can be shown by using the Pigeonhole Principle. If there are 22 or fewer people in the club, then at least two of the groups of 4 must be identical because there are only 100 possible groupings and 100 days, which is not enough to ensure that each possible group of 4 is chosen exactly once. If there are 23 people, then there are exactly 100 possible groupings, ensuring that each group of 4 is chosen exactly once over the 100 days.

n C 4 >=100

The smallest positive integer ==9

Because: 9 C 4 ==126 - unique combinations.

For each day, we have 4 distinct members to pick from. Hence, the number of possible combinations of 4 members is given by n choose 4, or nC4. The smallest positive integer n such that nC4 is greater than or equal to 100 is n = 18. This means that with a club of at least 18 members, it's possible to pick 4 different members every day for 100 days without ever having the exact same group twice.