My, what large numbers you are

Let the given numbers are: $p=2^{12} \times 3^{23} \times 5^{45}$ $q=2^{23} \times 3^{45} \times 5^{12}$ $r=2^{45} \times 3^{12} \times 5^{23}$

So, the L.C.M(Least Common Multiple) of them must contain the 2 , 3 and 5 of the highest order present{i.e, the highest power of 2 here is 45, so, the L.C.M must contain at least 45 2's(as factors) or else the L.C.M would be divisible by the number containing 12 and 23 2's but not by the number containing 45 2's. I hope you get my point}. Similarly, the highest powers of 3 and 5 here are also 45 . So, the required number must contain at least 45 3's and 5's as factors as well.

So, The required number, n , is given as follows: $n=2^{a} \times 3^{b} \times 5^{c}=2^{45} \times 3^{45} \times 5^{45}$ Comparing powers of same bases, we get: $a=45, b=45, c=45$ So, $a+b+c=45+45+45=\boxed{135}$

Add the highest powers of a,b,c You get 45+45+45 = 135

the least common multiple is 2^45 x 3^45 X 5^45. so, a = b = c = 45. a + b + c = 135.

The least common multiple is $2^{45}\times 3^{45}\times 5^{45}$ ; $45 \times 3=135$