Number y is defined as the sum of the digit of the number x, and z as the sum of the digits of the number y. How many natural numbers x satisfy the equation x+y+z = 60?

Let x be the original number, y be the sum of its digits, and z be the sum of digits of y. We need x + y + z = 60. The maximum possible sum of digits for any number grows slowly. Testing valid natural numbers: x=48 gives y=12, z=3, so 48+12+3=63. x=47 gives y=11, z=2, so 47+11+2=60 (valid). x=38 gives y=11, z=2, so 38+11+2=51. x=29 gives y=11, z=2, so 29+11+2=42. x=19 gives y=10, z=1, so 19+10+1=30. Continuing this systematic search, we find exactly 3 solutions exist.