The value of 2 digit number is four times of the sum of their digits . if the digits are interchanging the place it is 18 more then the original number then find the original number
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Lets say that the original number is represented by 10x + y ; where "x"
is the digit at ten's place and "y" is the digit at one's place.
As per the condition in the question, we can write the following equation:
10y + x = 18 + 10x + y
Solving this, we get y = x + 2 ------------ (i)
Another condition given in the questions shows the following relation;
10x + y = 4(x + y)
From this we get y = 2x ------------------- (ii)
Substituting (ii) into (i), we get x = 2 and hence y =4
Our Original number in that case turns out to be 24
As per the condition in the question, we can write the following equation:
10y + x = 18 + 10x + y
Solving this, we get y = x + 2 ------------ (i)
Another condition given in the questions shows the following relation;
10x + y = 4(x + y)
From this we get y = 2x ------------------- (ii)
Substituting (ii) into (i), we get x = 2 and hence y =4
Our Original number in that case turns out to be 24
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