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 a number to be divisible by $11$,the difference between the sum of digits at odd places and

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(C) Let the four-digit number be $N = d_1 d_2 d_3 d_4$.For a 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 $11$.That is,$(d_1 + d_3) - (d_2 + d_4) = 11k$,where $k \in \{0, 1, -1\}$.Since each digit $d_i \in \{1, 2, 3, 4\}$,the minimum sum of two digits is $1+1=2$ and the maximum sum is $4+4=8$.Thus,the difference $(d_1 + d_3) - (d_2 + d_4)$ can range from $2-8 = -6$ to $8-2 = 6$.The only multiple of $11$ in this range is $0$.Therefore,$(d_1 + d_3) = (d_2 + d_4)$.Let $S = d_1 + d_3 = d_2 + d_4$. Possible values for $S$ are:If $S=2$: $(1,1) \Rightarrow 1$ way.If $S=3$: $(1,2), (2,1) \Rightarrow 2$ ways.If $S=4$: $(1,3), (2,2), (3,1) \Rightarrow 3$ ways.If $S=5$: $(1,4), (2,3), (3,2), (4,1) \Rightarrow 4$ ways.If $S=6$: $(2,4), (3,3), (4,2) \Rightarrow 3$ ways.If $S=7$: $(3,4), (4,3) \Rightarrow 2$ ways.If $S=8$: $(4,4) \Rightarrow 1$ way.The number of ways to choose $(d_1, d_3)$ is the same as the number of ways to choose $(d_2, d_4)$.Total numbers $= 1^2 + 2^2 + 3^2 + 4^2 + 3^2 + 2^2 + 1^2 = 1 + 4 + 9 + 16 + 9 + 4 + 1 = 44$.

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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