The number of 7-digit numbers which are multi | vedclass

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(C) The given digits are $S = \{1, 2, 3, 4, 5, 7, 9\}$. The sum of these digits is $1 + 2 + 3 + 4 + 5 + 7 + 9 = 31$.Let the $7$-$digit$ number be $abc

Vedclass · Accepted Answer

$576$

Vedclass · Answer

(C) The given digits are $S = \{1, 2, 3, 4, 5, 7, 9\}$. The sum of these digits is $1 + 2 + 3 + 4 + 5 + 7 + 9 = 31$.Let the $7$-$digit$ number be $abcdefg$. For the number to be a multiple of $11$,the difference between the sum of digits at odd places and the sum of digits at even places must be a multiple of $11$.Let $O = a + c + e + g$ and $E = b + d + f$. We have $O + E = 31$ and $O - E = 11k$.Since $O + E = 31$,$O - E$ must be odd,so $11k$ must be odd. Possible values for $O - E$ are $11$ or $-11$.Case $1$: $O - E = 11$. Adding $O + E = 31$ gives $2O = 42 \Rightarrow O = 21$ and $E = 10$.Sets for $E = \{b, d, f\}$ summing to $10$: $\{1, 2, 7\}, \{1, 4, 5\}, \{2, 3, 5\}$. There are $3$ sets. For each set,there are $3! = 6$ arrangements. For each,the remaining $4$ digits form $4! = 24$ arrangements. Total $= 3 	imes 6 	imes 24 = 432$.Case $2$: $O - E = -11$. Adding $O + E = 31$ gives $2O = 20 \Rightarrow O = 10$ and $E = 21$.Sets for $E = \{b, d, f\}$ summing to $21$: $\{5, 7, 9\}$. There is $1$ set. Arrangements $= 1 	imes 3! = 6$. The remaining $4$ digits form $4! = 24$ arrangements. Total $= 6 	imes 24 = 144$.Total numbers $= 432 + 144 = 576$.

The number of $7$-$digit$ numbers which are multiples of $11$ and are formed using all the digits $1, 2, 3, 4, 5, 7,$ and $9$ is

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