The sum of the digits of a two digit number is 9 . Also nine times this number is twice the number obtained by reversing the order of digits . Find the number.
Answers
Answer:
18 is the answer.
Step-by-step explanation:
Let the ten's digit no. be x and one's digit no. be y.
So the no. will be = 10x+y.
Given : x+y=9-----(I)
9(10x+y)=2(10y+x) ⇒88x−11y=0 -----(II)
On solving I and II simultaneously you will get x=1 and y=8.
Therefore your desired no. is 18.
Answer:
Let the unit digit be x and tens digit be y.
So, the number will be = 10y + x
After interchanging the digits the number becomes = 10x + y
It is given that, The sum of the digits of two digit number is 9. Therefore we get :]
x + y = 9.......(Equation i)
It is also given that, Nine times this number is twice the number obtained by reversing the order of the digits :]
9(10y + x) = 2(10x + y)
90y + 9x = 20x + 2y
90y - 2y = 20x - 9x
188y = 11x
x = 8y.......(Equation ii)
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Now,Putting the value of x = 8y in equation (i) we get :]
x + y = 9
8y + y = 9
9y = 9
y = 1
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Now, substitute the value of y in equation (ii) we get :]
x = 8y
x = 8(1)
x = 8
Therefore,
The original number will be 10y + x = 10(1) + 8 = 18