The number of 3-digit odd numbers,whose sum o | vedclass

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(A) Let the $3$-$digit$ number be $xyz$,where $x \in \{1, 2, \dots, 9\}$,$y \in \{0, 1, \dots, 9\}$,and $z \in \{1, 3, 5, 7, 9\}$.We require $x + y +

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$63$

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(A) Let the $3$-$digit$ number be $xyz$,where $x \in \{1, 2, \dots, 9\}$,$y \in \{0, 1, \dots, 9\}$,and $z \in \{1, 3, 5, 7, 9\}$.We require $x + y + z = 7k$ for some integer $k$.Since $1 \le x \le 9$,$0 \le y \le 9$,and $1 \le z \le 9$,the sum $S = x + y + z$ ranges from $1+0+1 = 2$ to $9+9+9 = 27$.The possible multiples of $7$ are $7, 14, 21$.Case $1$: $z=1, x+y=6, 13, 20$. For $x+y=6$,pairs are $(1,5), (2,4), (3,3), (4,2), (5,1), (6,0)$ ($6$ values). For $x+y=13$,pairs are $(4,9), (5,8), (6,7), (7,6), (8,5), (9,4)$ ($6$ values). $x+y=20$ is impossible.Case $2$: $z=3, x+y=4, 11, 18$. For $x+y=4$,pairs are $(1,3), (2,2), (3,1), (4,0)$ ($4$ values). For $x+y=11$,pairs are $(2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)$ ($8$ values). For $x+y=18$,pair is $(9,9)$ ($1$ value).Case $3$: $z=5, x+y=2, 9, 16$. For $x+y=2$,pairs are $(1,1), (2,0)$ ($2$ values). For $x+y=9$,pairs are $(1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0)$ ($9$ values). For $x+y=16$,pairs are $(7,9), (8,8), (9,7)$ ($3$ values).Case $4$: $z=7, x+y=0, 7, 14$. For $x+y=0$,impossible. For $x+y=7$,pairs are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (7,0)$ ($7$ values). For $x+y=14$,pairs are $(5,9), (6,8), (7,7), (8,6), (9,5)$ ($5$ values).Case $5$: $z=9, x+y=5, 12$. For $x+y=5$,pairs are $(1,4), (2,3), (3,2), (4,1), (5,0)$ ($5$ values). For $x+y=12$,pairs are $(3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3)$ ($7$ values).Total count $= (6+6) + (4+8+1) + (2+9+3) + (7+5) + (5+7) = 12 + 13 + 14 + 12 + 12 = 63$.

The number of $3$-$digit$ odd numbers,whose sum of digits is a multiple of $7$,is

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