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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 the dogits.Find the number.?
Most Upvoted Answer
The sum of the digits of a two digit number is 9.Also nine times,this ...
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.
Community Answer
The sum of the digits of a two digit number is 9.Also nine times,this ...
Problem:
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 the digits. Find the number.

Solution:

Let's assume that the two-digit number is represented by AB, where A represents the tens digit and B represents the units digit.

Condition 1: The sum of the digits of the two-digit number is 9.
A + B = 9

Condition 2: Nine times the number is twice the number obtained by reversing the digits.
9AB = 2(10B + A)
9AB = 20B + 2A
9AB = 18B + 9B + 2A
9AB = 9(2B + A)

We know from condition 1 that A + B = 9, so we can substitute A = 9 - B into the above equation:
9B(9 - B) = 9(2B + 9 - B)
81B - 9B^2 = 18B + 81 - 9B
9B^2 - 27B = 0
9B(B - 3) = 0
B = 0 or B = 3

Since the number is a two-digit number, B cannot be 0. Therefore, B = 3 and A = 6.

So, the number is 63.

Answer: The two-digit number is 63.
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