Question

One of the digits of a two-digit number is 3 less than the other digit. When the digits are interchanges, we get a new number. The sum of the squares of the new number and the initial number is 1877. Find the two numbers.

I really don't know how to systematically go about doing this. I think a quadratic is involved though.

Answer 1

let this numbers be 3x, x3
3x=30+x and x3=10x+3
then lets take square of this numbers
(30+x)^2+(10x+3)^2=1877
900+60x+x^2+100x^2+60x+9=1877
101x^2+120x=968

Answer 2

but i didnt find x , please check whether there is a mistake or not

Answer 3

Thank you!

Answer 4

yes. he has made a mistake..

let one digit=a and the other=a+3

Answer 5

@simplethoughts

so one number is 10(a+3)^2 + a
the other is 10a+a+3

Answer 6

ignore the ^2 up there.

so now

[10(a+3)+a]^2 + [10a+a+3]^2= 1877
distribute
(10a+30+a)^2 + (10a+a+3)^2 = 1877
combine like terms
(11a+30)^2 +(11a+3)^2=1877
FOIL
121a^2+660a+900+121a^2+66a+9=1877
combine like terms and move everything to the left hand side

242a^2 +726a - 968 = 0

use quadratic formula and a=1 1+3=4, so other digit is 4

which means ur numbers are 14 and 41

Answer 7

let number be xy
either x = y-3 or x=y+3
(xy)^2 + (yx)^2 = 1877

Now it does not matter which one is bigger, because we are interchanging them tens place is gotten by mult by 10

[10(y-3)+y)]^2 + [10y+(y-3)]^2 = 1877

Perhaps this will work, but @nikato has solved it already.

Answer 8

@douglaswinslowcooper yup. got carried away. just got to continue and finish it with all my hard work

Answer 9

Impressive solution.

Answer 10

yeah i read the question wrongly but it is the logic anyway