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What least digit must be assigned to k so that the 7-digit number 86325k6 is divisible by 11?
1
3
4
2
3
Calculate the difference between the sum of digits in odd positions (8, 3, 5, 6) and even positions (6, 2, k). The difference must be a multiple of 11. Here, (8+3+5+6) - (6+2+k) = 22 - (8+k) = 14-k. For this to be 11, k = 3.
A 7-digit number 86325k6 which must be divisible by 11.
∣(Sum of odd-position digits)−(Sum of even-position digits)∣=0 or 11n
Calculate the difference between the sum of digits in odd positions (8, 3, 5, 6) and even positions (6, 2, k). The difference must be a multiple of 11. Here, (8+3+5+6) - (6+2+k) = 22 - (8+k) = 14-k. For this to be 11, k = 3.
Students often misidentify the odd and even positions by counting from left to right instead of using the correct positional indices.
Identify digit positions
The digits are 8, 6, 3, 2, 5, k, 6. Number positions from right to left: 6 is 1st, k is 2nd, 5 is 3rd, 2 is 4th, 3 is 5th, 6 is 6th, 8 is 7th.
87,66,35,24,53,k2,61
Calculate alternating sums
Sum of digits at odd positions (1st, 3rd, 5th, 7th) is So=6+5+3+8=22. Sum of digits at even positions (2nd, 4th, 6th) is Se=k+2+6=8+k.
So=22,Se=8+k
Apply divisibility rule for 11
For the number to be divisible by 11, the difference So−Se must be 0 or a multiple of 11.
22−(8+k)=14−k
Solve for k
If 14−k=0, k=14 (not a digit). If 14−k=11, then k=3. Since 3 is a single digit, it is the correct value.
k=3
B is correct because the difference of alternating sums is 14−k, which equals 11 when k=3.
This divisibility rule by 11 is highly useful in finding missing digits in large numbers, a common pattern in competitive exams like RRB and SSC.
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