Question

In: Math

Determine a two digit number whose value is equal to eight times the sum of its digits and when 45 is subracted from the number, the digits are reversed?

Determine a two digit number whose value is equal to eight times the sum of its digits and when 45 is subracted from the number, the digits are reversed?

Solutions

Expert Solution

Solution:

 

If we consider the digits at the unit's place and at the ten's place be x and y respectively, from the question:

 

the given number, 10 y + x = 8 (x+y)

10y + x = 8x + 8y

⇒ 10y - 8y = 8x - x

⇒ y = (7x)/2 - Eq. I

 

Also, from question:

10 y + x - 45 = 10x + y

⇒ 10y-y-45 = 10x-x

⇒ 9y-45 = 9x

⇒ 9y-9x = 45

y-x = 5 - Eq. II

 

From Eq. I and Eq. II:

 

(7x)/2 - x = 5

⇒ (7x- 2x)/2 = 5

⇒ (7x- 2x) = 5 (2)

⇒ 5x = 5 (2)

⇒ x = 2

 

Substitute x = 2 in Eq. II:

y-2 = 5 or y = 7

 

Thus, the number is 10(7) + 2 = 72.

 


The number is 10(7) + 2 = 72.

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