Question

QUESTION:-

The sum of the digits of a two-digit number is 10. the number formed by reversing the digits is 18 less than the original number. find the original number...

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Answer 1

$\huge \sf{ \underline{ \underline{Solution :-}}}$

Let the units and tens digit of a number be x and y respectively.

Case I : The sum of the digits of a two digit number is 10.

=> x + y = 10

=> x = 10 - y _____(i)

Case II : The number formed by reversing the digits is 18 less than the original number.

=> 10x + y + 18 = 10y + x

=> 10x - x + y - 10y = - 18

=> 9x - 9y = - 18

=> 9(x - y) = - 18

$=> x - y = \frac{ - 18}{9}$

=> 10 - y - y = - 2 [ from equation (i)]

=> - 2y = - 12

=> y = - 12/-2 = 6

Substituting the value of y in equation (i) we get,

x = 10 - 6 = 4

Hence,

The original number = 10y + x = 10 × 6 + 4 = 64

Answer 2

Answer:

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