Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.
Find the number of positive integers with three not necessarily distinct digits, \(abc\), with \(a \neq 0\) and \(c \neq 0\) such that both \(abc\) and \(cba\) are multiples of \(4\).
But try the problem first...
Answer: is 40.
AIME, 2012, Question 1.
Elementary Number Theory by David Burton .
Here a number divisible by 4 if a units with tens place digit is divisible by 4
Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b
[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]
and case II 2(mod4) with a pair of a and c for every b
Then combining both cases we get for every b gives a pair of a s and a pair of c s
So for 10 b's with 2 a's and 2 c's for every b gives \(10 \times 2 \times 2\)
Then number of ways \(10 \times 2 \times 2\) = 40 ways.