Question

A two digit number is 7 times the sum if its digits. The number fomed by reversing its digits the 18 less than the original number. Find the number.​

Answer 1

Answer:

Step-by-step explanation:

let the two-digit number be 10x+y

given,the number is 7 times the sum of its digits

⇒10x+y=7(x+y)

10x+y=7x+7y

3x=6y

x=2y

the number formed by reversing the digits is 18 less than original number

⇒10x+y-(10y+x)=18

  10x+y-10y-x=18

  9x-9y=18

 9(x-y)=18

 (x-y)=18/9=2

⇒x-y=2

x=2y

⇒2y-y=2

y=2

x=4

∴the number is 42

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Answer 2

Answer:

Answer is 42

Step-by-step explanation:

Let the ten's digit be x and the unit's digit be y

Then, number = 10x + y

∴10x+y=7(x+y)⇔3x=6y⇔x=2y∴10x+y=7(x+y)⇔3x=6y⇔x=2y

Number formed by reversing the digits = 10y + x

∴(10x+y)−(10y+x)=18⇔9x−9y=18⇔x−y=2⇔2y−y=2⇔y=2So, x=2y=4∴(10x+y)−(10y+x)=18⇔9x−9y=18⇔x−y=2⇔2y−y=2⇔y=2So, x=2y=4

Hence,

∴ Required number

= 10x + y

= 40 + 2

= 42

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