Four-digit numbers are formed using the digit | vedclass

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(C) Let the four-digit number be $N = d_1 d_2 d_3 d_4$. For $N$ to be divisible by $11$,the difference between the sum of digits at odd places and the

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(C) Let the four-digit number be $N = d_1 d_2 d_3 d_4$. For $N$ 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 $11$. That is,$(d_1 + d_3) - (d_2 + d_4) = 11k$,where $k \in \{0, 1, -1\}$. Since $d_i \in \{1, 2, 3, 4\}$,the minimum sum is $1+1=2$ and the maximum sum is $4+4=8$. Thus,the difference $(d_1 + d_3) - (d_2 + d_4)$ can only be $0$. So,$d_1 + d_3 = d_2 + d_4$. Let $S = d_1 + d_3 = d_2 + d_4$. The possible values for $S$ and the number of ways to form them are: If $S=2$: $(1,1) ightarrow 1$ way. Total $= 1 	imes 1 = 1$. If $S=3$: $(1,2), (2,1) ightarrow 2$ ways. Total $= 2 	imes 2 = 4$. If $S=4$: $(1,3), (2,2), (3,1) ightarrow 3$ ways. Total $= 3 	imes 3 = 9$. If $S=5$: $(1,4), (2,3), (3,2), (4,1) ightarrow 4$ ways. Total $= 4 	imes 4 = 16$. If $S=6$: $(2,4), (3,3), (4,2) ightarrow 3$ ways. Total $= 3 	imes 3 = 9$. If $S=7$: $(3,4), (4,3) ightarrow 2$ ways. Total $= 2 	imes 2 = 4$. If $S=8$: $(4,4) ightarrow 1$ way. Total $= 1 	imes 1 = 1$. Summing these up: $1 + 4 + 9 + 16 + 9 + 4 + 1 = 44$.

Four-digit numbers are formed using the digits $1, 2, 3, 4$ (repetition is allowed). The number of such four-digit numbers divisible by $11$ is

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