A three - digit number has digits in strictly descending order and divisible by 10. By changing the places of the digits a new three - digit number is constructed in such a way that the new number is divisible by 10. The difference between the original number and the new number is divisible by 40. How many numbers will satisfy all these conditions?

Since the three digit number is divisible by 10, then the unit's digit is 0

Let the three digit number = $$a b 0$$

After the digits are interchanged, the new number is also divisible by 10, thus only a and b are interchanged.

=> New number = $$b a 0$$

Difference between number is divisible by 40

=> $$(100a + 10b) - (100b + 10a) = 40 k$$    (k is constant)

=> $$90a - 90b = 90 (a - b) = 40 k$$

=> k = $$\frac{9\left(a-b\right)}{4}$$

Since k is a natural number (a-b) should be a multiple of 4

If a = 9 , the values of b that satisfies the given equation are 1,5

If a = 8 , the value of b that satisfies the given equation is 4

If a = 7 , the values of b that satisfies the given equation is 3

If a = 6 , the values of b that satisfies the given equation is 2

If a = 5 , the values of b that satisfies the given equation is  1

The number could be = 510,620,730,840,950, 910

Thus, there are 6 numbers that satisfy these conditions.