The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is \_\_\_\_\_\_\_.
Answer: 64
Solutions · 1
Official solution
xyz
Let x=2⇒y+z=13(4,9),(5,8),(6,7),(7,6),(8,5),(9,4),→6
Let x=3→y+z=12(3,9),(4,8),……..,(9,3)→7
Let x=4→y+z=11(2,9),(3,8),………,(9,1)→9
Let x=5→y+z=10(1,9),(2,8),……..,(9,1)→10
Let x=6→y+z=9(0,9),(1,8),……..,(9,0)→9
Let x=7→y+z=8(0,9),(1,7),……..,(8,0)→9
Let x=8→y+z=7(0,7),(1,6),……..,(7,0)→8
Let x=9→y+z=6(0,6),(1,5),………,(6,0)→7
Total =6=7+8+9+10+9+8+7=64