Let $a$ , $b$ , $m$ and $n$ be natural numbers such that $a > 1$ and $b > 1$ . If $a^m b^n = 144^{145}$ , then the largest possible value of $n - m$ is

289

580

579

290

✅ Correct Option: 3

We know: $a^m b^n = 144^{145}$ (where $a,b,c,d \in \text{N}$ ... and $a,b>1$ )

$144 = 12^2 = (4 \times 3)^2 = (2^2 \times 3)^2 = 2^4 \times 3^2$

$144^{145} = (2^4 \times 3^2)^{145} = 2^{4 \times 145} \times 3^{2 \times 145} = \boxed{2^{580} \times 3^{290}}$

Now our equation becomes:

$a^m \times b^n = 2^{580} \times 3^{290}$

We need to maximise: $n-m$

a. Make $n$ the largest value

b. Make $m$ the smallest value (preferably 1)

Let the equation become: $\boxed{a^m \times b^n = ({3^{290}})^1 \times {2^{580}}}$

Here,

$a = 3^{290}$

$m = 1$

$b = 2$

$n = 580$

Now, $n-m = 580-1 = 579$ .

Let the options be a hint. Think about how you can access the available options somehow or another. Don't just mark the answer and move forward without thinking properly.

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Let aaa, bbb, mmm and nnn be natural numbers such that a>1a > 1a>1 and b>1b > 1b>1. If ambn=144145a^m b^n = 144^{145}ambn=144145, then the largest possible value of n−mn - mn−m is

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CAT Past Year Questions (Topic-Wise)

Let $a$ , $b$ , $m$ and $n$ be natural numbers such that $a > 1$ and $b > 1$ . If $a^m b^n = 144^{145}$ , then the largest possible value of $n - m$ is