Question

the sum of the digits of a two digit number is 4. The number got by interchanging the digits is 18 Less than the original number. what is the original number?
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Answer 1

SOLUTION :-

Let,

Digit in ten's place = x

Digit in one's place = y

Two digit number = 10x + y $\longrightarrow\sf 2x=6 \\\\\longrightarrow\bf x = 3$

According to the first condition,

$\longrightarrow \sf x+y = 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(1)$

According to the second condition,

$\longrightarrow\sf 10x+y-18 = 10y+x\\\\\longrightarrow 10x-x+y-10y= 18 \\\\\longrightarrow 9x-9y = 18\\\\\longrightarrow x-y = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(2)$

Eq (2) + Eq (1),

$\longrightarrow\sf 2x = 6 \\\\\longrightarrow\bf x = 3$

Substitute the value of x in eq (1),

$\longrightarrow\sf 3+y= 4 \\\\\longrightarrow\bf y = 1$

Two digit number = 31