Prove that:
ab√x^a/x^b . bc√x^b/x^c . ca√x^c/x^a = 1
Answers
\huge\mathfrak{Answer}Answer
65
\huge\textsf{Explanation}Explanation
Let the digits be x and y.
SO, the original number would be 10x + y (two digits numbers are of this form, when expanded)
Given, sum of digits = 11
⇒ x + y = 11
And, when the digits are interchanged, then the original number is 9 more than the new number.
So, ⇒ 10y + x + 9 = 10x + y
⇒ 9x - 9y = 9
⇒ 9(x - y) = 9
⇒ x - y = 9/9
⇒ x - y = 1
So, now we get that, x + y = 11 and x - y = 1
So, add these two equations to get the value of x
x + y + x - y = 11 + 1
⇒ 2x = 12
⇒ x = 6
Now, put the value of x in any equation to get the value of y.
⇒ x - y = 1
⇒ 6 - y = 1
⇒ - y = 1 - 6
⇒ - y = - 5
⇒ y = 5
So, the digits are 6 and 5, hence, the number would be :-
⇒ 10x + y
⇒ 10(6) + 5
⇒ 60 + 5
⇒ 65
Hence, the original number is 65.