Question

the sum of the digits of a two digit number is 9 if the digits are interchanged and the two numbers are subtracted the difference become 63 find the number I want a picture of the correct answer I will mark the brainliest answer

Answer 1

Number consist of two digit Range[10–99].

→ sum of digits of 9 table is 9…

1. 0 + 9 = 9
2. 1 + 8 = 9
3. 2 + 7 = 9
4. 3 + 6 = 9
5. 4 + 5 = 9
6. 5 + 4 = 9
7. 6 + 3 = 9
8. 7 + 2 = 9
9. 8 + 1 = 9
10. 9 + 0 = 9

Sum of two digit is 9.

Now you can relate it to above logic.

And whats next line no is subtracted from 63.

Here we got the point that number is greater than 63.

Range of solution[72,81,90].

Now pick one by one

72–63=9 {that is not the reverse of 72}

81–63=18 {That is the reverse of 18}

$\boxed{\bold{\therefore{The\:Number\:is\:either\:81\:or\:18}}}$

Answer 2

$\large\bf\underline \blue {To \: \mathscr{f}ind:-}$

• we need to find the number.

$\huge\bf\underline \red{ \mathcal{S}olution..}$

$\bf\underline{\purple{Given:-}}$

• sum of the digits of a two digit number is 9
• Difference of original number and number formed by reversing the digits is 63

${ \green{ \mathscr{ \underline{Let : - }}}}$

• Let units place digit be x
• Let ten's place digit be y

$\underline{ \large \purple{ \mathscr{\dag\:A \bf{ccourding}\: \mathscr{t}o \: \mathscr {Q} \bf{uestion} ....}}}$

≫ Sum of the digits of a two digit number is 9.

⇝ x + y = 9

⇝ x = 9 - y⠀⠀⠀....1)

≫ if the digits are interchanged and the two numbers are subtracted the difference become 63.

• Original number = 10y + x
• Number formed by reversing the digits is 10x + y

⇝ 10x - y -(10y + x) = 63

⇝ 10x - y + 10y - x = 63

⇝ 9x - 9y = 63

• Dividing both sides by 9

⇝ x - y = 7

• From equation ...1)

⇝ 9 - y - y = 7

⇝ - 2y = 7 - 9

⇝ - 2y = -2

⇝ y = 1

• Putting value of y in .....1)

↛ x = 9 - y

↛ x = 9 - 1

↛ x = 8

So,

The original number 10y + x :-

• → 10 × 1 + 8
• → 10 + 8
• → 18

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