Question

the sum of the digits of a 2-digit number is 9. the number is 6 times the unit digit. find the number .

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Answer 1

$\large\underline{\sf{Given- }}$

• The sum of the digits of a 2-digit number is 9.

• The number is 6 times the unit digit.

$\large\underline{\sf{To\:Find - }}$

• The number

$\red{\large\underline{\sf{Solution-}}}$

Given that,

The sum of the digits of a 2-digit number is 9.

Let assume that,

Digit at unit place = x

So,

Digit at ten's place = 9 - x

So,

Original number = 1 × x + (9 - x) × 10 = x + 90 - 10x = 90 - 9x.

Reverse number = 10 × x + (9 - x)× 1 = 10x + 9 - x = 9x + 9

According to statement again,

Number is 6 times the unit digit.

$\rm :\longmapsto\:90 - 9x = 6x$

$\rm :\longmapsto\:90 = 6x + 9x$

$\rm :\longmapsto\:90 = 15x$

$\bf\implies \:x = 6$

So, number = 90 - 9x = 90 - 9 × 6 = 90 - 54 = 36

Answer 2

Answer:-

Given,

The sum of the digits of a 2-digit number is 9.

Let's assume that,

Digit at unit place = x

So,

Digit at ten's place = 9-x

So,

Original number ⤵

$= 1 \times x + (9 - x) \times 10 = x + 90 - 10x = 90 - 9x$

Reverse number ⤵

$= 10 \times x + (9 - x) \times 1 = 10x + 9 - x = 9x + 9$

According to the statement,

Number is 6 times the unit digit.

$90 - 9x = 6x$

$90 = 6x + 9x$

$90 = 15x$

$x = 6$

So, number =

↳ 90 - 9x = 90 - 9 × 6 = 90 - 54 = 36