Question

The sum of the digits of a two digit number is 17 . If the number formed by reversing the digits is less than the original number by 9 find the original number

Answer 1

Correct QuestioN :

The sum of the digits of a two digit number is 17. If the number formed by reversing the digits is more than the original number by 9. find the original number ?

To FinD :

Find the original number.

SolutioN :

Let's

• Ten Place digit be x.
• And Unit place digit be y.
• Original number be 10x + y.
• Reversing the number 10y + x.

☛ Case 1.

• The sum of the digits of a two digit number is 17.

→ x + y = 17.

$\rule{200}2$

☛ Case 2.

• If the number formed by reversing the digits is more than the original number by 9.

→ 10x + y + 9 = 10y + x.

→ 9x - 9y = 9.

→ x - y = - 1 ____ ( 2 )

✎ Now, Taking Equation ( 2 )

→ x - y = - 1.

→ x = y - 1 ____ ( 3 )

✎ Putting the value of x in Equation ( 1 )

→ x + y = 17.

→ y - 1 + y = 17.

→ 2y = 18.

→ y = 9.

✎ Putting the value of y = 9 in Equation ( 3 )

→ x = y - 1.

→ x = 9 - 1.

→ x = 8.

✡ Our Number become → 10x + y → 10(8) + 9

→ 80 + 9.

→ 89.

Answer 2

Given that:

• Sum of the digits of a two digit number is 17.

• Number formed by reversing the digits is less than the original number by 9.

To Find:

• The original number.

Solution:

Let the original number be 10x+y.

So, Sum of digits = x+y = 17....(1)

Also, 10x+y-10y-x = 9

x-y = 1.....(2)

Solving eq(1) and eq(2), we get

2x = 18

x = 9

Putting x=18 in the above equation, we get

y = 9-1 = 8

So, the original number is 10(9)+8 = 98.

Therefore, the original number is 98.

Hope this helps