The units digit of the two digit number is one less then the square of the 10 digit. The number with the digits reversed is seven more than twice the original number. Find the original number
MechEng2030
Favorite Answer
Let the original 2 digit number be xy. From the information given in the problem:
y = x^2 - 1
When digits are reversed:
10y + x = 2(10x + y) + 7
10(x^2 - 1) + x = 2(10x + x^2 - 1) + 7
10x^2 + x - 10 = 2x^2 + 20x + 5
8x^2 - 19x - 15 = 0
x = (19 +/- √(361 - 4(8)(-15))/16 = 3 and y = 3^2 - 1 = 8
So original number was 38.
Anonymous
Mmhmm^