Let m , n be two distinct integers chosen randomly from the set 0,1,2, ., 99 . Then the probability that 4^ m +4^ n +3 is divisible by 5 lies in the interval

Step-by-step solution

possible case Case-1 If unit digit is sum of 4^ m +4^ n is 7 possible case are array l ( m , n )=(0,2)(0,4) (0,98)=49 ( n , m )=(0,2)(0,4) . .(0,98)=49 total =98 Case-2 Unit digit is sum of 4^ m +4^ n is 2 Possible case ( m , n )=(2,4)(2,6) .(2,98)=48 ( ~m , n )=(4,6)(4,8) . .(4.98)=47 ( m , n )=(96,98) 1 C a s e same case repetition ( n , m ) 2 ( 48 49 2 ) & Now, Case 1+ Case 2 2352+98=2450 Total number of ways m , n can be selected =100 99 = & 9900 Probability = 2450 9900 =0.2474 array