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
It is given thatĀ $$a^m\cdot b^n\ =\ 144^{145}$$, whereĀ $$a>1\ $$ andĀ $$b>1$$.
$$144$$ can be written asĀ $$144\ =\ 2^4\times\ 3^2$$
Hence,Ā $$a^m\cdot b^n\ =\ 144^{145}$$ can be written asĀ $$a^m\cdot b^n\ =\ \left(2^4\times\ 3^2\right)^{^{145}}=2^{580}\times\ 3^{290}$$
We know thatĀ $$3^{290}$$ is a natural number, which implies it can be written asĀ $$a^1$$, whereĀ $$a\ >\ 1$$
Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.
Hence, the largest value of (n-m) is (580-1) = 579
The correct option is D
Create a FREE account and get: