In a two-digit number, the units digit is 1 more than the tens digit. The sum of the squares of the two digits is 41. What is the number?

Correct answer: 45

Let the tens digit be \(x\). Then the units digit is \(x+1\). Therefore, \(x^2+(x+1)^2=41\), which gives \(2x^2+2x-40=0\) or \((x-4)(x+5)=0\). Since a digit cannot be negative, \(x=4\); hence the units digit is 5 and the number is 45. Although 56 also has digits differing by 1, its digit-square sum is \(25+36=61\), not 41. Exam tip: reject any negative root when the variable represents a digit.