Question

if we multiply a certain two_digit no. by the sum of its digits , we get 405. if we multiply the no. consisting of the same digits written in the reverse order by the sum of he digits , we get 486 .find the no.

please ans quickly step by step

Answer 1

let two digit number be 10x + y and it's reversed number is 10y + x

we multiply a certain two_digit no. by the sum of its digits , we get 405
so, (10x+y)(x+y) = 405
10x² + 10xy + xy + y² = 405
10x² + 11xy + y² = 405
y² = 405 - 10x² - 11xy...............(1)

we multiply the no. consisting of the same digits written in the reverse order by the sum of it's digits , we get 486
so, (10y + x)(x+y) = 486
10xy + 10y² + x² + xy = 486
11xy + x² + 10y² = 486................(2)

substitute (1) in (2)
11xy + x² + 10(405 - 10x² - 11xy) = 486
11xy + x² + 4050 - 100x² - 110xy = 486
-99x² - 99xy + 4050 = 486
99x² + 99xy = 4050 - 486
99(x² + xy) = 3564
x² + xy = (3564)/99
x² + xy = 36
x(x+y) = 36
x+y = 36/x..............(3)

substitute (3) in (1)
(10x+y)(36/x) = 405
360x+36y = 405x
36y = 405x - 360x
36y = 45x
x/y = 36/45
x/y = 4/5
i.e x:y = 4:5

so the two digit number is 10x + y = 10*4 + 5 = 40 + 5 = 45

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