A two digit number is such that the sum of digitis ten if the digits are reversed the new number formed is less than original number by 18 find the number

Question

A two digit number is such that the sum of digitis ten if the digits are reversed the new number formed is less than original number by 18 find the number

Answer 1

Let the two digits be represented by 10x + y, where x and y are the digits.

Given that the sum of the digits is 10:
x + y = 10 (1)

When the digits are reversed, the new number is 10y + x.

Given that the new number is less than the original number by 18:
10x + y - (10y + x) = 18
9x - 9y = 18
x - y = 2 (2)

Now we have a system of two equations:

x + y = 10
x - y = 2

Adding the two equations together, we get:
2x = 12
x = 6

Substitute x = 6 into equation (1):
6 + y = 10
y = 4

Therefore, the number is 64.

Answer 2

just for future reference, if the reversed digits for a number that is less by 9n, the the digits differ by n.

Answer 3

That is correct! Thank you for pointing that out. If the reversed digits of a two-digit number are less than the original number by a multiple of 9, the digits will differ by that multiple. This is because reversing the digits causes a change in the value of x 10 times and the value of y 1 time, resulting in the difference between the two numbers being a multiple of 9.