Animated Solution for Mathematics - Permutations and Combinations: Let $n$ be a non-negative integer. Then the number of divisors of the form "$4n+1$" of the number $(10{)}^{10}\cdot (11{)}^{11}\cdot (13{)}^{13}$ is equal to

Welcome, fellow traveler in the realm of number theory! Today, we are going to unravel a beautiful problem that might look like a daunting counting exercise, but is actually a masterclass in the elegance of modular arithmetic.

We are tasked with finding the number of divisors of $N=10^{10}\cdot 11^{11}\cdot 13^{13}$ that take the specific form $4n+1$. Let us embark on this journey step by step.

The Foundation of Prime Factorization

Before we can count anything, we must understand the building blocks of our number $N$. The number is given as $N=10^{10}\cdot 11^{11}\cdot 13^{13}$.

While $11$ and $13$ are proud, indivisible primes, $10$ is a composite number waiting to be broken down. We know that $10=2\cdot 5$.

Substituting this back into our expression, we get:

Now, we have the complete prime factorization. This is our map; every divisor of $N$ must be constructed from these specific prime factors.

The Parity Trap

A general divisor $d$ of $N$ will take the form $d=2^a\cdot 5^b\cdot 11^c\cdot 13^d$, where the exponents are bounded by the powers available in $N$: $0\le a\le 10$, $0\le b\le 10$, $0\le c\le 11$, and $0\le d\le 13$.

The problem imposes a strict condition: $d$ must be of the form $4n+1$. Pause for a moment and think about the nature of this form.

Any number $4n$ is a multiple of $4$, which is inherently even. Adding $1$ to an even number always yields an odd number.

This is the crucial insight: our divisor $d$ must be odd. If $d$ is odd, it cannot contain the prime factor $2$. Thus, the exponent $a$ must be $0$. We have effectively eliminated one variable from our search!

The Elegance of Modulo 4

Now, we turn our attention to the remaining factors: $5^b\cdot 11^c\cdot 13^d$. We need this product to be congruent to $1(\mathrm{mod}4)$.

Let us analyze the bases modulo $4$:

- $5=4(1)+1$, so $5\equiv 1(\mathrm{mod}4)$. - $11=4(2)+3$, which is equivalent to $11\equiv -1(\mathrm{mod}4)$. - $13=4(3)+1$, so $13\equiv 1(\mathrm{mod}4)$.

Substituting these into our expression for $d$, we get:

We require $d\equiv 1(\mathrm{mod}4)$, which means we need $(-1{)}^c=1$. This only happens when $c$ is an even number.

The Final Assembly

We have reduced the problem to counting the valid combinations of exponents.

For $c$, which ranges from $0$ to $11$, we need even values: $c\in \{0,2,4,6,8,10\}$. That gives us $6$ choices.

For $b$, which ranges from $0$ to $10$, there are no restrictions, giving us $11$ choices. For $d$, which ranges from $0$ to $13$, there are also no restrictions, giving us $14$ choices.

The total number of divisors is the product of these independent choices:

We have arrived at our destination. The beauty of this problem lies in how the constraints, which initially seemed complex, collapsed into a simple parity check for the exponent $c$.

The final answer is 924. Keep practicing this kind of reduction, and you will find that even the most intimidating problems have a simple, elegant heart.