A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is

Question

Accepted Answer

#### Step 1: Key Conditions  
Let the total number of students in the school be $$ N $$.  
- $$ N < 5000 $$  
- $$ N $$ leaves a remainder of $$ 4 $$ when divided by $$ 9, 10, 12, $$ and $$ 25 $$.  
- $$ N $$ is divisible by $$ 11 $$.

#### Step 2: LCM Condition for Remainder  
Since $$ N $$ leaves a remainder of $$ 4 $$ for all these divisors, it must satisfy:  
$$
N \equiv 4 \pmod{	ext{LCM}(9, 10, 12, 25)}.
$$

Calculate the LCM:  
$$
	ext{LCM}(9, 10, 12, 25) = 	ext{LCM}(3^2, 2 	imes 5, 2^2 	imes 3, 5^2) = 900.
$$

Thus:  
$$
N \equiv 4 \pmod{900}.
$$

#### Step 3: Divisibility by $$11$$ and Range of $$N$$  
Express $$ N $$ as:  
$$
N = 900x + 4,
$$  
where $$ x $$ is an integer such that $$ N < 5000 $$.

Solve for possible values of $$ x $$:  
$$
900x + 4 < 5000 \implies x < \frac{4996}{900} \approx 5.55.
$$  
Thus, $$ x $$ can take values from $$0$$ to $$5$$.

Additionally, since $$ N $$ is divisible by $$11$$:  
$$
900x + 4 \equiv 0 \pmod{11}.
$$

Simplify modulo $$11$$:  
- Since $$900 \equiv -1 \pmod{11}$$:  
$$
(-x) + 4 \equiv 0 \pmod{11} \implies x \equiv -4 \pmod{11} \implies x \equiv 7 \pmod{11}.
$$

From the valid range of values for $$ x $$(0 to 5), only $$ x = 2 $$ satisfies this condition.

#### Step 4: Calculate Total Students ($$N$$)  
Substitute $$ x = 2 $$ into the equation for $$ N $$:  
$$
N = (900)(2) + 4 = 1804.
$$

#### Step 5: Maximum Teams of Size $$12$$  
Divide the total number of students into teams of size $$12$$:  
$$
	ext{Teams} = \frac{1804}{12}.
$$

Perform division:  
$$
1804 = (12)(150) + 4.
$$

Thus, the maximum number of teams is **150**.

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### Final Answer:  
The maximum number of teams of size **12** that can be formed is **150**.