Question

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

Answer 1

$x$ y $z \leftarrow$ odd number
$z =1,3,5,7,9$
$x + y + z =7,14,21$ [sum of digit multiple of $7$]
$\underset{1 \text { to 9 }}{x}+\underset{0 \text { to 9 }}{y}=6,4,2,13,11,9,7,5,20,18,16,14,12$
$x+y=6 \Rightarrow(1,5),(2,4),(3,3),(4,2),(5,1),(6,0)$
$\rightarrow$ T.N. $=6$
$x+y=4 \Rightarrow(1,3),(2,2),(3,1),(4,0)$
$\rightarrow T$. $=4$
$x + y =2 \Rightarrow(1,1),(2,0)$
$\rightarrow$ T.N. $=2$
$x+y=13 \Rightarrow(4,9),(5,8),(6,7),(7,6),(8,5),(9,4)$
$\rightarrow$ T.N. $=6$
$x+y=11 \Rightarrow(2,9),(3,8),(4,7),(5,6),(6,5)$,
$(6,5),(7,4),(8,3),(9,2)$
$\rightarrow$ T.N. $=8$
$x+y=9 \Rightarrow(1,8),(2,7),(3,8),(4,5),(5,4), \ldots .(8,1),(9,0)$
$\rightarrow$ T.N. $=9$
$x+y=7 \Rightarrow(1,8),(2,5),(3,4), \ldots .(8,1),(7,0)$
$\rightarrow$ T.N. $=7$
$x+y=5 \Rightarrow(1,4),(2,3),(3,2),(4,1),(5,0)$
$\rightarrow$ T.N. $=5$
$x + y =20 \Rightarrow$ Not possible
$x + y =18 \Rightarrow(9,9) \rightarrow$ T.N. $=1$
$x+y=16 \Rightarrow(7,9),(8,8),(9,7)$
$\rightarrow$ T.N. $=3$
$x+y=14 \Rightarrow(5,9),(6,8),(7,7),(8,6),(9,5)$
$\rightarrow$ T.N. $=5$
$x+y=12 \Rightarrow(3,9),(4,8),(5,7),(6,6) \ldots .(9,3)$
$\rightarrow$ T.N. $=7$