IIT JEE , AIEEE Forum - Mathematics , Algebra

man111

(1) find the probability that 2^a+3^b+4^c is divisable by 4 if a,b,c

{0,1,2,3,4,5,6,7,8,9}

(a) divisable by 5

(c) divisable by 7

chimanshu_007

is ans ( divisible by 4) is 89/1000?

akhil_o

Hey Himanshu!
How do u figure that?
plz post ur soln!am quite stuck

hsbhatt

2^a+3^b+4^cis to be divisible by 4.

If a>=2 and c>0 it is not div by 4.

Now, if c = 0, we have to examine 2^a+3^b+1.

Now, if a=0, since 3^b

1 no cases occur

If a = 1, 3^b+3 is divisible by 4 if b is even gives 4 cases........1

If a

2, 2^a+3^b+1 is div by 4 when b is odd giving 8*5 = 40 cases .......2

If b = 0, 2^a+2 is divisible by 4 when a is odd giving 4 cases ....3

If c>0,

If a=0 and b =0, gives no cases

If a =0, b is odd gives 5*9 = 45 favourable cases ....4

If a = 1 no cases.

Thus, there are 93 favourable cases out of 1000 giving a prob of 93/1000.