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.

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Solution

Let the ten's digit no. be x and one's digit no. be y.

So the no. will be = $10 x + y$ .

Given : $x + y = 9$ -----(I)

$9 (10 x + y) = 2 (10 y + x)$ $\Rightarrow 88 x - 11 y = 0$ -----(II)

On solving I and II simultaneously you will get $x = 1$ and $y = 8$ .
Therefore your desired no. is 18.

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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 digits. Find the number.

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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 digits. Find the number.

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.

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.

Let the ten's digit no. be x and one's digit no. be y.

So the no. will be = $10 x + y$ .

Given : $x + y = 9$ -----(I)

$9 (10 x + y) = 2 (10 y + x)$ $\Rightarrow 88 x - 11 y = 0$ -----(II)

On solving I and II simultaneously you will get $x = 1$ and $y = 8$ .
Therefore your desired no. is 18.