Question

certain number between 10 and 100 is 8 times the sum of its digits if 45 is subtracted from the digit will be reversed find the number​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assumption

Number be p and c

Number = 10p + c

$\fbox{Situation}$

10p + c = 8(p + c) 

10p + c = 8p + 8c

2p - 7c = 0 ......(1) 

$\fbox{Situation}$

10p + c - 45 = 10c + p

9p - 9c = 45 

p - c = 5 [Multiplying by -2]

We get :-

-2p + 2c = -10 ....... (2)

$\fbox{Adding(1)\;and\;(2)}$

-5c = -10

$\tt{\rightarrow c=\dfrac{-10}{-5}}$

c = 2 

$\fbox{Now\:value\:of\:p :-}$

p - c = 5 

p - 2 = 5

p = 5 + 2

p = 7 

$\fbox{Number = 72}$

Answer 2

Answer:

The Number = 72

Step-by-step explanation:

Given :

Certain number between 10 and 100 is 8 times.

The sum of its digits if 45.

To Find :

The number.

Solution :

Let the number be "ab"

Thus, The value of number = 10a + b

Now,

$\sf\ \\\implies 10a+b = 8(a+b)\\ \\\implies 10a+b =8a+8b\\ \\\implies 2a-7b=0\quad...Eq(1) \\ \\\implies 10a+b-45 = 10a+b\\ \\\implies 9a-9b=45\\ \\\implies a-b = 5.\quad.....Eq(2)}$

Now, Just simply multiply it with - 2

⇒ - 2a + 2b = -10

Add this value in Equation (1),

⇒ - 5b = -10

⇒ b = 2  

⇒ a - b = 5  

⇒ a = 7

It forms the value 72.

Hence, The number is 72.