I would like it if someone could give me pointers on solving problems like these. And why was 4 the answer here ?
If $a=4b+26$ and $b$ is positive , then a could be divisible by all of following except
a)2 b)4 c)5 d)6 e)7
Edit:. I know by taking b=4 its divisible by 7 . However is there any other way by which we could tell its divisible by 7 other than randomly plugging in a no and testing ? Also I am using the current method to check if the expression is divisible by other options , is this method correct
Check by 2: 4b mode 2 and 26 mod 2 = 0 hence divisible by 2
For numbers greater than 4 the expression is doubled to $a=8b + 52$
Check by 5: 8b mode 5 = 3 while 52+3 mod 5 = 0 hence divisible by 5
Check by 6: 8b mode 6 = 2 while 52+2 mod 6 = 0 hence divisible by 6
Check by 4: 4b mode 2 =0 while 26 mod 2 $\not=$ 0 hence not divisible by 4
I need to know if this checking mechanism is correct ? is so why doesn't it work for 7 ? How could I check for 7 without plugging in and testing values ?