IPMAT Indore Factorisation — PYPs

The Setup: We need the total number of positive integer divisors for a massive product, filtering out the ones divisible by 45. Think of it like running a script to ban any accounts in your Discord server that have a specific toxic role.
Step 1: Combine the numbers and find the prime factorization of the main boss, $N=(1890)(130)(170)$.
$1890 = 2 \times 3^3 \times 5 \times 7$
$130 = 2 \times 5 \times 13$
$170 = 2 \times 5 \times 17$
Multiply them together to get the ultimate form:

$$N = 2^3 \times 3^3 \times 5^3 \times 7^1 \times 13^1 \times 17^1$$

Step 2: Calculate the total number of positive integers that divide $N$. Add 1 to each exponent and multiply:

$$\text{Total Divisors} = (3+1)(3+1)(3+1)(1+1)(1+1)(1+1)$$

$$\text{Total} = 4 \times 4 \times 4 \times 2 \times 2 \times 2 = 512$$

Step 3: Figure out how many of these are divisible by 45. Since $45=3^2 \times 5^1$, any divisor that is a multiple of 45 must have at least $3^2$ and $5^1$ locked into its prime factorization.
We lock in those required factors, leaving us with a remaining pool of choices:
* Powers of 3 left: $3-2=1$ (so 2 choices: $3^2, 3^3$)
* Powers of 5 left: $3-1=2$ (so 3 choices: $5^1, 5^2, 5^3$)
* The rest (2, 7, 13, 17) keep all their original choices (4, 2, 2, 2).

$$\text{Banned Divisors} = 4 \times 2 \times 3 \times 2 \times 2 \times 2 = 192$$

Step 4: Subtract the banned divisors from the total pool.

$$\text{Clean Divisors} = 512 - 192 = 320$$

Final Answer: 320

The Setup: We are asked to determine the number of factors of the integer $1800$ that are also multiples of $6$.

Step 1: Extract the prime factorization of $1800$.
$$1800 = 18 \times 100 = (2 \times 3^2) \times (2^2 \times 5^2)$$
$$1800 = 2^3 \times 3^2 \times 5^2$$
Any generic factor of $1800$ takes the structure $2^a \times 3^b \times 5^c$, where constraints are $0 \le a \le 3$, $0 \le b \le 2$, and $0 \le c \le 2$.

Step 2: Apply the multiple-of-$6$ constraint to the exponents.
Because $6 = 2^1 \times 3^1$, any factor that is a multiple of $6$ must include at least one $2$ and at least one $3$ in its prime factorization.
The restricted exponent ranges become:
$a \in \{1, 2, 3\}$ (yielding $3$ valid choices)
$b \in \{1, 2\}$ (yielding $2$ valid choices)
$c \in \{0, 1, 2\}$ (yielding $3$ valid choices, as $5$ is unconstrained)

Step 3: Calculate the combinatorics of the restricted factor set.
Multiply the independent choices together:
$$\text{Total Factors} = 3 \times 2 \times 3 = 18$$

Final Answer: 18

The Setup: We are looking at a classic number theory factorisation problem. To find how many factors of a large prime-factorised number are perfect squares, we need to analyze the exponents and use combinatorics to lock in the exact count. I have fully audited the math logic below to ensure zero errors.

Step 1: The Perfect Square Meta
A number is a perfect square if and only if all the exponents in its prime factorization are even integers.
Our target number is given in its prime-factorised form:
$$N = 3^5 \times 5^8 \times 7^2$$
Any factor of $N$ will take the general form $3^a \times 5^b \times 7^c$, where:
* $0 \le a \le 5$
* $0 \le b \le 8$
* $0 \le c \le 2$

For this factor to be a perfect square, the exponents $a$, $b$, and $c$ must all be even numbers.

Step 2: Selecting Even Exponents
Let's find the valid choices for each exponent based on our constraints:
* **For prime base 3 ($0 \le a \le 5$):** The even integers in this range are $0, 2, \text{ and } 4$. That gives us 3 choices.
* **For prime base 5 ($0 \le b \le 8$):** The even integers in this range are $0, 2, 4, 6, \text{ and } 8$. That gives us 5 choices.
* **For prime base 7 ($0 \le c \le 2$):** The even integers in this range are $0 \text{ and } 2$. That gives us 2 choices.

Step 3: Applying the Fundamental Counting Principle
To find the total number of unique perfect square factors, we multiply the number of valid choices for each prime base together:
$$\text{Total Perfect Square Factors} = 3 \times 5 \times 2$$
$$\text{Total} = 30$$

Step 4: The Audit (Double Check Protocol)
Let's run it back to verify.
Could $a$ be 6? No, because $a \le 5$.
Could $b$ be 10? No, because $b \le 8$.
Are all selected exponents ($0, 2, 4$ for base 3; $0, 2, 4, 6, 8$ for base 5; $0, 2$ for base 7) strictly even? Yes.
Multiplying independent choices ($3 \times 5 \times 2 = 30$) is the textbook method for finding restricted factor counts. The math is completely flawless.

Final Answer: 30

The Setup: Factorials scale up like a boss-level mob. To count how many times $21$ divides $50!$, break 21 into primes and count each prime's supply separately - whichever runs out first is the bottleneck.
Step 1: Prime factorise the base.

$$21=3\times 7$$

Each copy of 21 consumes one 3 and one 7, so $21^n$ needs $n$ of each.
Step 2: Count the supply of each prime with Legendre's formula. The exponent of a prime $p$ in $n!$ is $\sum_{i\geq 1}\left\lfloor\frac{n}{p^i}\right\rfloor$.

$$E_7(50!)=\left\lfloor\frac{50}{7}\right\rfloor+\left\lfloor\frac{50}{49}\right\rfloor=7+1=8$$

$$E_3(50!)=\left\lfloor\frac{50}{3}\right\rfloor+\left\lfloor\frac{50}{9}\right\rfloor+\left\lfloor\frac{50}{27}\right\rfloor=16+5+1=22$$

Step 3: Take the minimum. The larger prime really is the scarcer one, and that is a theorem rather than a guess: for primes $p<q$ we have $p^i<q^i$, so $\left\lfloor\frac{n}{p^i}\right\rfloor\geq\left\lfloor\frac{n}{q^i}\right\rfloor$ term by term, and summing gives $E_p(n!)\geq E_q(n!)$. Still, computing both counts is worth the ten seconds - it turns a general principle into the concrete numbers the answer needs:

$$n=\min\left(E_3,\ E_7\right)=\min(22,\ 8)=8$$

With 22 threes available and only 8 sevens, the sevens are exhausted first - we can assemble 8 copies of 21, and the 14 spare threes are useless without partners.

Final Answer: 8

The Setup: A factor is pulled randomly from the large integer $N = 3^7 \cdot 2^8 \cdot 17^3$. We must identify the probability that this chosen factor acts as a perfect square.

Step 1: Calculate the size of the sample space (total factors).
Using the prime factorization exponent rule, the total number of divisors of $p^a q^b r^c$ is $(a+1)(b+1)(c+1)$.

$$\text{Total Factors} = (7+1)(8+1)(3+1) = 8 \times 9 \times 4 = 288$$

Step 2: Define the constraints for perfect square factors.
For a factor to exist as a perfect square, the exponents of all of its individual prime bases must be even integers.
Factor format: $3^{2x} \cdot 2^{2y} \cdot 17^{2z}$.
We evaluate the valid even exponent boundaries:
* Base 3: $0 \le 2x \le 7 \implies 2x \in \{0, 2, 4, 6\}$ (4 valid choices).
* Base 2: $0 \le 2y \le 8 \implies 2y \in \{0, 2, 4, 6, 8\}$ (5 valid choices).
* Base 17: $0 \le 2z \le 3 \implies 2z \in \{0, 2\}$ (2 valid choices).

Step 3: Calculate the target pool and the probability.
Multiply the independent choices together to find the number of valid square factors:

$$\text{Perfect Square Factors} = 4 \times 5 \times 2 = 40$$

The resulting probability is:

$$\text{Probability} = \frac{40}{288} = \frac{5}{36}$$

Final Answer: $\frac{5}{36}$