Sam's Gym Dilemma

The number of divisors in a number $n$ , which can be factored into $p_1^a \times p_2^b \times p_3^c \dots$ , where $p_1 , p_2, p_3 \dots$ are primes, and $a, b, c ..$ are non-negative integers, is given by $(a + 1) \times (b + 1) \times (c + 1) \dots$ .

Finding the number of ways in which $300$ students can be divided is equivalent to finding the number of divisors of $300$ , since each divisor $d$ has a complement $\big(\frac{300}{d}\big)$ , and is thus capable of forming a pair of sides of the rectangle formed. Now $300 = 3^1 \times 2^2 \times 5^2$ , so number of divisors of $300$ is $(1 + 1) \times (2 + 1) \times (2 + 1) = 18$ However, we cannot consider factors $1$ and $300$ , due to the restrictions imposed in the question. So the number of divisors, or the number of ways $300$ students can be divided = $18 - 2$ which is $16$

The problem is equivalent to finding the number of divisors of 300 (excluding 1 and 300), as each divisor will correspond to one specific rectangular arrangement - for example, the divisor "3" corresponds to a 3 x 100 rectangle, "5" corresponds to a 5 x 60 rectangle, and so on.

The first step is to find the prime factorisation of 300. This can be done by repeatedly dividing by 2, then by 3, then by increasingly larger primes until the result is 1, giving $2^2 3^1 5^2$ .

The number of factors of an integer is then given by $(q_{1}+1)(q_{2}+1)...(q_{n}+1)$ , where $q_{n}$ represents the exponents of 2, 3, 5 and so on in the prime factorisation above.

Therefore, 300 has (2+1)(1+1)(2+1)=18 divisors. Remembering to exclude 1 and 300, the solution to the original problem is 16.

300=2^2 3 5^2 Thus, it has 3 2 3=18 different factors Each of the factor 300 has can be the number of rows, and there will be a corresponding number of columns. Apart from 1 300 and 300 1, all other formations work. Thus, there are 18-2=16 formations

we have to count how many positive divisors $300$ has. $300$ can be expresses as $2^2 \cdot 5^2 \cdot 3$ , which means that $300$ has $(2+1)(2+1)(1+1)=18$ positive factors. We must not include 1 and 300 from the list, hence there are $16$ factors all in all.

the divisors of 300 are 2,3,4,5,6,10,15,20,25,30,50,60,75,100,150(total 16)

                              rectangles may of  

                              2x150 and 150x2  specify(row x column)
                              3x100 and 100x3
                              4x75   and 75x4
                              5x60   and  60x5
                              6x50   and 50x6
                              10x30 and 30x10
                               12x25 and 25x12
                               15x20 and 20x15

so there are 16 possible way.

function isInt(n) { return typeof n === 'number' && parseFloat(n) == parseInt(n, 10) && !isNaN(n); } // 6 characters

isInt(300);

rowColumnCombo = function(numStudents){ var totalStudents = numStudents; var cases = 0; for(var i=1; i<=numStudents; i++){ var temp = totalStudents/i; if(isInt(temp)){ console.log(i+" rows and "+temp+" columns"); cases++; } } console.log("There are "+cases+" possible cases."); }

rowColumnCombo(300);

Running this script will result in 18 cases. Removing the cases that satisfy the case where there is one student in either row or column, we then have 16 cases.

Any formation with $A$ rows and $B$ columns corresponds to a factorization of $300 = A \times B$ . From divisors of an integer , we know that $300 = 2 ^2 \times 3 \times 5^2$ has $(2+1) \times (1+1) \times (2+1) = 18$ positive factors. After accounting for the formations of 1 row and 1 column, there will be $18 - 2 = 16$ allowable formations.