IPMAT Indore HCF & LCM — PYPs

The Setup: We are dealing with the foundational division algorithm: $Dividend=(Divisor\times Quotient)+Remainder$. The absolute golden rule of division is that the remainder must be strictly non-negative and less than the divisor ($0\leq r<d$). We need to min-max the given sum and product to find the exact values before plugging them in.
Step 1: Formulate the system of equations. Let the quotient be $q$ and the remainder be $r$. The problem gives us their product $q\times r=24$ and their sum $q+r=10$.
Step 2: Solve the quadratic substitution. You can substitute $r=10-q$ into the product equation:

$$q(10-q)=24$$

$$10q-q^2=24$$

$$q^2-10q+24=0$$

Factoring this gives $(q-6)(q-4)=0$. The roots are $6$ and $4$. So, our variables $\{q,r\}$ are exactly $\{6,4\}$ - no other pair of numbers has this sum and product.
Step 3: Enforce the remainder rule. We are told the divisor is $5$. By the laws of division, $r<5$. Between $6$ and $4$, only $4$ satisfies this condition. Therefore, it is a hard lock that $r=4$ and $q=6$.
Step 4: Execute the final calculation. Substitute everything back into the core algorithm:

$$Dividend=(5\times 6)+4$$

$$Dividend=30+4=34$$

Final Answer: 34

The Setup: Two natural numbers are subjected to a specific string of algebraic operations. We must simplify the operations to find the resulting identity.

Step 1: Map the text to algebraic formulation.
Let the two natural numbers be $a$ and $b$.
'Square of the difference' = $(a-b)^2$.
'Square of the sum' = $(a+b)^2$.
The text requires: $\frac{(a+b)^2 - (a-b)^2}{4}$.

Step 2: Expand and simplify the expression.
Expand the squares:

$$\frac{(a^2 + 2ab + b^2) - (a^2 - 2ab + b^2)}{4}$$

Distribute the negative sign:

$$\frac{a^2 + 2ab + b^2 - a^2 + 2ab - b^2}{4}$$

The squares cancel out perfectly:

$$\frac{4ab}{4} = ab$$

The result is simply the direct mathematical product of the two original natural numbers.

Step 3: Relate to the Number System rules.
A fundamental property of arithmetic states that the product of any two natural numbers equals the product of their Least Common Multiple (LCM) and their Highest Common Factor (HCF).

$$a \times b = \text{LCM}(a,b) \times \text{HCF}(a,b)$$

Final Answer: the product of the LCM and HCF of the two numbers

The Setup: This is a Number System boss fight focusing on GCD (Greatest Common Divisor) and coprime properties (Euler's Totient function logic). The meta is to factor out the mandatory GCD from our target variable, isolate the remaining multiplier, and then count how many valid integers exist within the new boundary that share absolutely zero prime factors with our base. Math, logic, and syntax are locked in and double-verified.

Step 1: Factor out the mandatory GCD.
We know the target condition is $\gcd(n, 10^7) = 10^3$. This means $n$ absolutely must be a multiple of $10^3$.
Let's define $n$ as:
$$n = 10^3 \cdot m$$
(where $m$ is some positive integer multiplier).

**Step 2: Establish the new boundaries for $m$.**
The problem restricts the original variable to $n \leq 10^7$.
Let's substitute our new definition of $n$ into this inequality to find the ceiling for $m$:
$$10^3 \cdot m \leq 10^7 \implies m \leq 10^4$$
So, our multiplier $m$ must be an integer from $1$ to $10^4$.

Step 3: Lock the coprime constraint.
Now, let's substitute $n$ back into the original GCD equation to see what conditions $m$ must satisfy:
$$\gcd(10^3 \cdot m, 10^7) = 10^3 \cdot \gcd(m, 10^4)$$
For the total expression to equal exactly $10^3$, the remaining $\gcd$ portion must collapse to 1:
$$\gcd(m, 10^4) = 1$$
This means $m$ and $10^4$ must be coprime (they cannot share any prime factors).

Step 4: Count the valid coprime spawns.
The prime factorization of $10^4$ is $2^4 \cdot 5^4$. The only prime factors are 2 and 5.
Therefore, for $m$ to be coprime to $10^4$, it cannot be a multiple of 2 or 5.
We can use a basic Totient fraction to find the exact ratio of integers that survive this filter:
$$Fraction\ of\ valid\ integers = \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right)$$
$$Fraction = \frac{1}{2} \cdot \frac{4}{5} = \frac{2}{5}$$

Step 5: Execute the final calculation.
Apply this valid fraction to our total possible pool of $m$ values ($10^4$ or 10,000) to get the final count of valid $n$ integers:
$$Required\ count = 10000 \times \frac{2}{5}$$
$$Required\ count = 2000 \times 2 = 4000$$

Final Answer: 4000