Whose camp are you in?

The number 40 can be written as 2^3*5 so it is divisible by the prime numbers 2 and 5. Let's suppose to divide the students in 2 groups: it's evident that if K is even then we could put K/2 student in each group so K must be odd. Now let's divide the students in 5 groups: K can't be divisible by 5 otherwise we could put K/5 students in each group. The other divisors of 40 (greater than 1) are all even so we could only formate an even number of groups dividing 40 by them: K should be even to have the same number of math students in each group but we said that K is odd. Then the possible values ​​of k are the odd numbers between 0 and 40 (which are 20) to which we must remove the odd multiples of 5 (5,15,25,35). In conclusion there are 16 possible values of K.

The total number of students are 40. Since they have to be partitioned in groups of equal sizes, it is needed that the the number of groups that are made must be a factor of 40. So the number of possible groups is 2, 4, 5, 8, 10, 20 and 40. (1 is not taken because it is mentioned clearly in the question that the students were partitioned into 2 or more groups ).

It is given that k number of students are working on math problems and it was found by the camp mentor that when they were partitioned into groups of equal sizes, the k students could not be partitioned equally among those groups. So, k must not be divisible by the number of groups , i.e the numbers listed above, namely 2, 4, 5, 8, 10, 20 and 40.

Now, of these numbers, we see that a number divided by 40 is also divisible by the others (namely 2, 4, 5, 8, 10 and 20), a number divisible by 20 is also divisible by 2, 4 , 5 and 10. Similarly, it can be observed that a number divisible by 10 is also divisible by 2 and 5. So, it is needed to just find the prime factors of these numbers (2, 4, 5, 8, 10, 20, 40). It can be found that their prime factors include 2 and/or 5. So, any number k which is divisible by either 2 or 5 can be partitioned into equal groups. From the numbers 1-40, if we remove the numbers which are divisible by either 2 or 5, we get the required possible values of k .

Since 40 = 2 X 20, so there are 20 numbers in 1-40 which are divisible by 2. Also, since 40 = 5 X 8, there are 8 numbers which are divisible by 5. But, also 40 = 10 X 4 = 2 X 5 X 4, so there are 4 numbers which are divisible by both, so these are getting counted twice. Hence these are to be removed. So the number of numbers in 1-40, which are divisible by either 2 or 5 are - 20 + 8 - 4 = 24

So, remaining numbers which are not divisible by 2 or 5 and hence which are possible values of k are - 40 - 24 = 16. Hence, 16 is the answer.

Note that the groups to be divided are divisors of 40, so if is not possible to form groups of mathematicians, this means that the dividers of 40 do not divide K. Therefore, GCD (40, K) = 1. Thus, all possible values for K are φ (40) = 16.

There are 40 students on the camp, so they can be divided into only few equal groups. These are 2 groups of 20 people, 4 groups of 10, 5 groups o 8, 8 groups of 5, 10 groups of 4 and 20 groups of 2. So k needs to be relatively prime with all the group numbers. Group numbers are divisible by only 2 and 5, so k can be every odd number not divisible by 5. There are 16 such numbers.

$40 = 2^3*5$

Since if all the students were partitioned into 2 or more groups of equal size, these groups could never have the same number of math students in each, k must not be a multiple of 2 or 5.

Using the Principle of Inclusion and Exclusion, the number of integer from 1 to 40 which is a multiple of 2 or 5 is: $20 + 8 - 4 = 24$

Hence, the answer is $40 - 24 = 16$

The factors of 40 are : $1,2,4,5,8,10,20,40$

Hence in order to ensure that the mathematics students cannot be partitioned evenly, the number k must not be divisible by all of these factors of 40, except 1.

Hence, the possible values of k are $1,3,7,9,11,13,17,19,21,23,27,29,31,33,37,39$ 1 should be included because the question asks for more than 1 group to be formed.

Hence, the answer is 16

We consider case in which we have same number of math students in every group(opposite of what is required). For this we only need to consider partitioning of 40 into groups of 2 and 5. Because if number of math students are even then they can be partitioned equally into group of two equally. And if number of math students are multiple of 5 then they can be partitioned into group of 5 equally. So number of math students which can be partitioned equally are: 2,4,6,8,10,12 ,..., 40 ==> total 20 5,15,25,35 ==> total 4 Total number of values of k for which we can divide math students equally are 20 + 4 = 24. So number of values of k which are left = 40 -24 = 16.

If we want to divide 40 students into several groups with equal size, then we have to divide it with one of the factors of 40. - 40 = 2^3 \times 5 We have k , which could never divide into the same number in those groups, so that we have to ignore the integers that have the same factor of the 40. Then we could count the rest of the integers as the different values that k could have, which is 16.

Solution 1: If $\gcd(k,40) = 1$ , then the students can not be split into groups so that the same number of math students are in the group, since the number of groups must be a divisor of $40$ . If $\gcd(k,40) = m > 1$ , then there is a way dividing the students into $m$ groups so that the same number of math students are in each group. The number of numbers less than or equal to 40 which are relatively prime to 40 is $\phi(40) = 40 \times \frac{4}{5} \times \frac{1}{2} = 16$ , so there are $16$ possible values for $k$ .

Solution 2: The number of groups of equal size the students can be divided into is one of $2,4,5,8,10,20,40$ . The number of math students in each group can be the same if the number of students is a multiple of one of these numbers. Notice that every number is a multiple of 2 or 5, so the number of students must not be a multiple of 2 or 5. We can count the number of numbers less than 40 that are not multiples of 2 or 5 by the principle of inclusion and exclusion. There are $20$ multiples of $2$ , $8$ multiples of $5$ , and $4$ multiples of $10$ . So in total, there are $20 + 8 - 4 = 24$ numbers which are multiples of $2$ or $5$ and $40 - 24 = 16$ which are not.

There are 40 people in the group and if k isn't relatively prime to 40, split the people into n groups where n divides both k and 40. So k is relatively prime to 40. The prime factors of 40 are 2 and 5 so count the amount of numbers less than 40 which are not multiples of 2 or 5. There are 20 of 2 and 8 of 5 for 28. For both 2 and 5 there are 4 so by PIE there are 24 total. 40-24=16.