Suppose that $$\log_{2} [\log_{3}(\log_{4}a)] = \log_{3} [\log_{4} (\log_{2}b)] = \log_{4} [\log_{2} (\log_{3}c)] = 0$$. Then the value of a + b + c is

We know that $$\log_{2} [\log_{3}(\log_{4}a)] = \log_{3} [\log_{4} (\log_{2}b)] = \log_{4} [\log_{2} (\log_{3}c)] = 0$$

Now Equating, $$\log_2[\log_3(\log_4a)]=0$$

If we take 2 on RHS, it will be $$[\log_3(\log_4a)]=2^0$$

$$\log_3(\log_4a)=1$$

Now, again taking 3 on the RHS,

$$(\log_4a)=3^1$$

Now taking 4 on RHS

$$a=4^3=64$$

Similarly, equating $$\log_3[\log_4(\log_2b)]=0$$

If we take 3 on RHS, it will be $$\log_4(\log_2b)=3^0$$

$$\log_4(\log_2b)=1$$

Now taking 4 on RHS

$$\log_2b=4^1$$

Now taking 2 on the RHS

$$b=2^4=16$$

Similarly, equating $$\log_4[\log_2(\log_3c)]=0$$

Now taking 4 on RHS

$$\log_2(\log_3c)=1$$

Now, taking 2 on RHS

$$\log_3c=2$$

Thus, $$c=9$$

Therefore, a+b+c = 64+16+9 = 89