My Unique Twist

Since $\overline{AB}$ is a two digit positive integer, we can write that the value of $\overline{AB}=10A+B$ . Now, putting in the given expression, we get ----

$(10A+B)^2=(20\times A\times B)+425$

$100A^2+B^2+20AB=20AB+425$

$100A^2+B^2=425$

We can observe that if $A=2$ and $B=5$ , the equation is satisfied! So, the required number $=\overline{AB}=\boxed{25}$

Sum of their digits $=A+B=5+2= \boxed{7}$

Easy way (Trail and Error)

As we are adding $425$ to the product, $B$ should be $5$ .

Now, $A$ cannot be $1$ , as $15^2$ is $225$ which is less than $425$ .

Taking $A$ as $2$ , we get $(25^2) = (20 \times A \times B) + 425$

Thus, $AB$ is $25$

(AB)^2 = (20 x A x B) +425 think a perfect square that is near in a 425 and add the product of 20 x A x B... its 525, and the square root of 525 is 25, then A=2 and B=5, so that (20 x 2 x 5) + 425 = 525... A + B = 7 2 + 5 = 7

$let,\quad { (10A+B) }^{ 2 }\quad =\quad 20*A*B\quad +\quad 425\quad =\quad k\\ now\quad we\quad observe\quad that\quad 20*A*B\quad will\\ have\quad unit\quad digit\quad as\quad 0\quad so\quad k\quad will\\ have\quad unit\quad digit\quad 5.\quad since\quad k\quad is\quad a\quad perfect\quad \\ square\quad its\quad square\quad root\quad must\quad end\quad with\quad \\ unit\quad digit\quad 5\quad because\quad k\quad has\quad unit\quad digit\quad 5.\\ so\quad B\quad =\quad 5\quad further\quad solve\quad to\quad get\quad A\quad =\quad 2\\ so,\quad A+B=7$

25 was literally the first number I guessed as AB and it fit as 2 was A and 5 was B and the two sides of the equation matched after solving.

If $AB$ is a 2 digit number then we can write it as $10A+B$ so : $(10A+B)^2=(20AB)+425$ $(10A)^2+20AB+B^2=20AB+425$ $(10A)^2+B^2=425$ Observe that $A=2\;and\;B=5$ satisfies the equation.So sum = $A+B=2+5=\boxed{7}$