last two digits

If it is true that $a^2\equiv a$ (mod 100), then it must also be true that $a^2\equiv a$ (mod 10). We know that a is an even number so we can work out the possibilities for the last digit of a.

The only values for which $a^2\equiv a$ (mod 10) holds true (bearing in mind that $a$ (mod 10) is even) are $a\equiv 6$ or $0$ (mod 10).

Let $a$ (mod 10) be $10x+y$ .

If y = 0, $a^2$ (mod 10) $\equiv (10x+0)^2=100x^2$ so is always congruent to 0 (mod 100). This means that x must be 0 as well, and 10x+y, and therefore $a$ must be a multiple of 100.

If y = 6, $a^2$ (mod 10) $\equiv (10x+6)^2=100x^2+120x+36$ (mod 10) $\equiv 20x+36$ (mod 100). The only values for which $10x+y\equiv 20x+36$ (mod 10) is $x = 7$ .

Bearing in mind these constraints and the fact that a is a multiple of 12, the two smallest values that a can be are $276$ and $300$ , the sum of which is $276+300=\mbox{576}$ .

100I(a)(a-1) ; 12Ia
Consider 2 cases one where 25Ia and the other 25I(a-1).U get 300 (obviously the least no.s) for the 1st one & 276 for 2nd case.
So,ANSWER=576

As @Edward Rong did , I first concluded that $a\equiv{0 ;6}\pmod{10}$

When $a\equiv{0}\pmod{10} \Rightarrow a^{2}\equiv{00}\pmod{100} \Rightarrow a\equiv{00}\pmod{100}$ .

When $a\equiv{6}\pmod{10} \Rightarrow a\equiv{06 , 16 , 26 , 36 , 46 , 56 , 66 , 76 , 86 , 96}\pmod{100}$ ,But $12|a \Rightarrow 4|a \Rightarrow a\not\equiv{06 , 26 , 46 , 66 , 86}\pmod{100}$ .

For these cases I just checked them out by squaring them to see whether they satisfied the conditions or not.

And the number which satisfied was $76$ .

So we are left to deal with $00 , 76$ .

$12|a \Rightarrow 3|a$ So , by divisibility by $3$ , we see that the smallest numbers satisfying are $300 , 276$