If you have the sume of squares of the digits of a positive two-digit number is 20, the tens digit is 2 more than the units digit.
answer is 2^2 and 4^2 (?)
2 answers
Yes, that looks right.
a b
a^2+b^2 = 20
a = b+2
(b+2)^2 + b^2 = 20
b^2 + 4 b + 4 + b^2 = 20
2 b^2 + 4 b -16 = 0
b^2 + 2 b - 8 = 0
(b-2)(b+4 = 0
b = 2
a = 4
so number is indeed
42
a^2+b^2 = 20
a = b+2
(b+2)^2 + b^2 = 20
b^2 + 4 b + 4 + b^2 = 20
2 b^2 + 4 b -16 = 0
b^2 + 2 b - 8 = 0
(b-2)(b+4 = 0
b = 2
a = 4
so number is indeed
42
9 answers