How many ways two numbers can be chosen from number {1,2, 3, ......... 30}, such that sum of numbers is divisible by 3?
A) 90 B) 100 C) 190 D) 225 E) 435
A) 90 B) 100 C) 190 D) 225 E) 435
To find ways such that a+b is divisible by 3 where a,b:{1,2,3,...30}
Now a number when divided by 3 leaves a remainder of either a 0 or a 1 or a 2 (3n or 3n+1 or 3n+2 where n = 0 to 10)
for remainder 0: 3n
a can be selected in 10 ways and then b can be selected in 9 ways
i.e. if a=6; b can be 3, 9, 12, ...multiples of 3 <=30
- 10*9=90 ways
for remainder 1: 3n+1
a can be selected in 10 ways and then b has to be selected from the remaining to digits of type 3n+2
- 10*10=100 ways
for remainder 2: 3n+2:
a can be selected in 10 ways and then b can be selected in 10 ways from the remaining digits of type 3n+1
but we have already counted all these in case 2.
so total 90+100=190 ways
Thanks for sharing this one... took some time to get hold of it but enjoyed it nonetheless
cheers!!
