The number of natural numbers,between 212 and | vedclass

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(A) Let the three-digit number be represented as $xyz$,where $x, y, z \in \{0, 1, \dots, 9\}$ and $x \in \{2, 3, \dots, 9\}$. We require $x+y+z=15$. S

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$64$

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(A) Let the three-digit number be represented as $xyz$,where $x, y, z \in \{0, 1, \dots, 9\}$ and $x \in \{2, 3, \dots, 9\}$. We require $x+y+z=15$. Since the number is between $212$ and $999$,we check cases for $x$:$1$. If $x=2$,then $y+z=13$. Possible pairs $(y, z)$ are $(4,9), (5,8), (6,7), (7,6), (8,5), (9,4)$. Total = $6$.$2$. If $x=3$,then $y+z=12$. Possible pairs $(y, z)$ are $(3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)$. Total = $7$.$3$. If $x=4$,then $y+z=11$. Possible pairs $(y, z)$ are $(2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)$. Total = $8$.$4$. If $x=5$,then $y+z=10$. Possible pairs $(y, z)$ are $(1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1)$. Total = $9$.$5$. If $x=6$,then $y+z=9$. Possible pairs $(y, z)$ are $(0,9), (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0)$. Total = $10$.$6$. If $x=7$,then $y+z=8$. Possible pairs $(y, z)$ are $(0,8), (1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1), (8,0)$. Total = $9$.$7$. If $x=8$,then $y+z=7$. Possible pairs $(y, z)$ are $(0,7), (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (7,0)$. Total = $8$.$8$. If $x=9$,then $y+z=6$. Possible pairs $(y, z)$ are $(0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0)$. Total = $7$.Summing these: $6+7+8+9+10+9+8+7 = 64$. Note: The number $212$ itself has a digit sum of $5$,so it is not included in the count.

The number of natural numbers,between $212$ and $999$,such that the sum of their digits is $15$,is . . . . . .

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