CAT 2023 Slot 2 QA Question 4

First off, when dealing with equations involving exponents, prime factorization is usually a good approach. So, let me factorize 144. I know that 144 is 12 squared, which is $12^2$. But breaking it down further, 12 is 2 times 6, and 6 is 2 times 3. So, putting that together, 144 is $2^4 \times 3^2$. Let me write that down:

Therefore, $144^{145}$ would be:

Using the exponent rule $(x × y)^n = x^n × y^n , this becomes:

$2^{4 \times 145} \times 3^{2 \times 145}$ = 2^{580} × 3^{290}$

So, the equation $a^m b^n = 144^{145}$ simplifies to:

Now, since $a$ and $b$ are natural numbers greater than 1, they must be composed of the prime factors 2 and 3. Let me express $a$ and $b$ in terms of their prime factors. Let's say:

$a = 2^{x} \times 3^{y}$ $b = 2^{p} \times 3^{q}$

Where $x, y, p, q$ are non-negative integers. Since $a > 1$ and $b > 1 , at least one of$x$or$y$must be positive for$a , and similarly for $b$.

Substituting these into the equation $a^m b^n$:

Simplifying the exponents:

Since the prime bases must be equal on both sides, we can set up the following equations:

Our goal is to maximize $n - m$. To do this, we need to find suitable values of $x, y, p, q$ such that both equations are satisfied, and $n - m$ is as large as possible.

Let me think about how to approach this. Since we want to maximize $n - m , we should try to make$n$as large as possible and$m$ as small as possible, given the constraints.

Looking at equation 1: $x m + p n = 580$. To minimize $m , we can set$x$and$p$to be as large as possible, but they are constrained by the prime factors of$a$and$b$. However, since$a$and$b$can only have 2 and 3 as prime factors,$x, y, p, q$ can be any non-negative integers.

Wait, but actually, $a$ and $b$ can have other prime factors as well, but since the right-hand side only has 2 and 3, $a$ and $b$ can't have any other prime factors. So, $a$ and $b$ must be of the form $2^k \times 3^l , where$k$and$l$ are non-negative integers.

Therefore, $x, y, p, q$ can be any non-negative integers, but $a$ and $b$ must be greater than 1, so at least one of $x$ or $y$ must be positive for $a , and similarly for$b$.

So, going back to the equations:

We need to find $m$ and $n$ such that these equations hold, with $a, b > 1$.

To maximize $n - m , we can try to express$m$and$n$ in terms of the coefficients.

Let me consider equation 2 first: $y m + q n = 290$. Since 290 is half of 580, maybe there's a relationship here.

If I can set $y = 2$ and $q = 1 , then equation 2 becomes:

Similarly, equation 1 would be:

If I can set $x = 1$ and $p = 2 , then equation 1 becomes:

So now we have a system of equations:

Let me solve this system.

From equation 2: $2m + n = 290$. Let's solve for $n$:

Substitute this into equation 1:

$m + 2(290 - 2m) = 580$ $m + 580 - 4m = 580$ - 3m + 580 = 580$- 3m = 0$ $m = 0$

But $m$ must be a natural number greater than 1, so $m = 0$ is invalid. Hmm, that didn't work. Maybe my choice of \( x