Power of 2
Level
pending
Three numbers x,y and z are selected from the set {1,2,3....10^100-1} with repetition allowed . The probability that 2^x+2^y+2^z is divisible by 7 is a/b, where a and b are co-prime, then a+b is equal to ...?
The answer is 11.
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1 solution
2^n leaves either 1,2 or 4 as remainder when divided by 7, if n leaves 1,2 or 0 as remainder when divided by 3,respectively.
divide the set into 3 sets of sizes (10^100-1)/3,depending on the remainder the numbers leave when divided by 3. 2^x+2^y+2^z will be divisible by 7 if and only if x,y and z belong to different sets. This is because only 1+2+4 is divisible by 7(and 2+1+4,and 4+2+1... ,and not 1+1+4, or any other combination that has repeated value). Hence the number of such triplets (x,y,z) are 3!*((10^100-1)/3)^3 and the required probability is 6/27,or 2/9.