Question

since ( (3 a+2007)^{2} ) is divisible by ( 9, ) we see that ( 4 b 85924 ) has to be divisible by ( 9 . )
So ( 4+b+8+5+9+2+4=32+b )
has to be divisible by ( 9 . )
It follows from this that ( b=4, a=37 )
After knowing ( b=4 ), one way to find ( a ) is as follows :
since the right-most digit of ( 3 a ) is either 1 or 5 , the right-most digit of ( a ) is either 5 or
7
The right-most two digits of ( (3 a+2007)^{2} ) are
84,84,04,64,24,44,44,24 for
( a=5,7,15,17,25,27,35,37 )
respectively.
The right-most three digits of ( (3 times 25+2007)^{2} ) is ( 724 . ) So, ( a=37 )

# 7. If 'a' is a natural number and 'b' a digit such that (3a + 2007)2 = 4b85924, then a - b equals to (A) 41 (B) 35 (C) 37 (D) 33

Solution