If 738 A 6 A is divisible by 11, then the val | vedclass

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(C) number is divisible by $11$ if the difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a mult

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) number is divisible by $11$ if the difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a multiple of $11$.For the number $738A6A$:Sum of digits at odd places (from right to left) = $A + 8 + 7 = A + 15$.Sum of digits at even places (from right to left) = $6 + A + 3 = A + 9$.Difference = $(A + 15) - (A + 9) = 6$.Since the difference is $6$,which is not $0$ or a multiple of $11$,let's re-evaluate the alternating sum based on positions:Sum of digits at odd positions (1st,3rd,5th) = $A + 8 + 7 = A + 15$.Sum of digits at even positions (2nd,4th,6th) = $6 + A + 3 = A + 9$.Wait,the standard rule is (Sum of digits at odd places) - (Sum of digits at even places) = $0$ or $11k$.$(A + 8 + 7) - (6 + A + 3) = 6$. This is constant. Let's check the positions again: $7(1), 3(2), 8(3), A(4), 6(5), A(6)$.Sum of odd positions = $A + 8 + 7 = A + 15$.Sum of even positions = $A + 6 + 3 = A + 9$.Difference = $(A + 15) - (A + 9) = 6$. This implies the number is never divisible by $11$ for any $A$. Re-reading the sequence: $7, 3, 8, A, 6, A$. Odd positions: $A, A, 3$. Even positions: $6, 8, 7$.Sum of odd positions = $2A + 3$.Sum of even positions = $6 + 8 + 7 = 21$.Difference = $(2A + 3) - 21 = 2A - 18$.For divisibility by $11$,$2A - 18 = 0 \Rightarrow 2A = 18 \Rightarrow A = 9$.

If $738 A 6 A$ is divisible by $11,$ then the value of $A$ is

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