The number of 6-digit numbers that can be for | vedclass

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(B) Let the $6$-digit number be $abcdef$. For the number to be divisible by $11$,the difference between the sum of digits at odd places and the sum of

Vedclass · Accepted Answer

$60$

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(B) Let the $6$-digit number be $abcdef$. For the number to be divisible by $11$,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 $S_1 = a+c+e$ and $S_2 = b+d+f$. The sum of all digits is $0+1+2+5+7+9 = 24$.We have $S_1 + S_2 = 24$ and $S_1 - S_2 = 11k$. Since $S_1+S_2$ is even,$S_1-S_2$ must be even,so $k=0$. Thus $S_1 = S_2 = 12$.Possible sets for ${a, c, e}$ and ${b, d, f}$ such that their sum is $12$:Case $I$: ${a, c, e} = {7, 5, 0}$ and ${b, d, f} = {9, 2, 1}$.For ${a, c, e}$,$a$ cannot be $0$,so there are $2 	imes 2! = 4$ ways. For ${b, d, f}$,there are $3! = 6$ ways. Total $= 4 	imes 6 = 24$.Case $II$: ${a, c, e} = {9, 2, 1}$ and ${b, d, f} = {7, 5, 0}$.For ${a, c, e}$,there are $3! = 6$ ways. For ${b, d, f}$,there are $3! = 6$ ways. Total $= 6 	imes 6 = 36$.Total numbers $= 24 + 36 = 60$.

The number of $6$-digit numbers that can be formed using the digits $0, 1, 2, 5, 7,$ and $9$ which are divisible by $11$ and no digit is repeated,is

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