MATHCOUNTS Exercises – 14

Solution Lets assume the bigger 2-digit number is \(10x+y\), and the smaller 2-digit number is \(10z+y\), we have: $$(10x + y)^2-(10z+y)^2=2340$$ $$100x^2+20xy+y^2-(100z^2+20zy+y^2)=2340$$ Simplify the above equation, we have: $$(x-z)(5x+y+5z)=117=1\cdot 3\cdot 3\cdot 13$$ $$ \because 10 > x > z > 0 $$ $$\therefore x-z=1 \ \ or \ \ x-z=3 $$ If \(x-z=1\), we have $$ 5x + y + 5z=117 $$ which is impossible, as the maximum possible value of \(5x+y+5z\) is 99. Therefore, $$x-z=3\tag{1}$$ $$5x+y+5z=39\tag{2}$$ From (1), we have $$x=z+3\tag{3}$$ Based on (2) and (3), we have $$5(z+3)+y+5z=39$$ Simplify the above equation, we have $$y+10z=24$$ Since \(y\) and \(z\) are single digit numbers, the only solution is $$ y=4 \ \ \ and \ \ \ z=2$$ Therefore \(x=5\). The sum of \(10x+y\) and \(10z+y\) is $$ 54+24=78$$