2.10 Divisibility by 11

Rule: Take the alternating sum of digits (sum of digits at odd places from the right) minus (sum of digits at even places from the right). If the result is 0 or a multiple of 11 (including negative multiples), the number is divisible by 11. Reasoning: 10 ≡ -1 (mod 11), so 10^k ≡ (-1)^k (mod 11). Thus N= d_i 10^i ≡ d_i(-1)^i (mod 11), which is exactly the alternating digit sum.

Example 1: Check 9,28,389. From the right: digits are 9(1st),8(2nd),3(3rd),8(4th),2(5th),9(6th). Sum at odd positions (1st,3rd,5th) = 9+3+2 = 14. Sum at even positions (2nd,4th,6th) = 8+8+9 = 25. Difference = 14 - 25 = -11, a multiple of 11 → divisible by 11. (Check: 928389 = 11 × 84399 ✓)

Example 2: Check 1,002,001. Digits from right: 1,0,0,2,0,0,1. Odd positions (1,3,5,7): 1+0+0+1=2. Even positions (2,4,6): 0+2+0=2. Difference = 0 → divisible by 11.

Example 3: Find the value of x so that 3x5 is divisible by 11 for the number 46x27. Positions from right: 7(1),2(2),x(3),6(4),4(5). Odd sum = 7+x+4 = 11+x. Even sum = 2+6=8. Difference = (11+x) - 8 = 3+x. For divisibility by 11, 3+x = 0 or 11. Since x is a digit (0–9), 3+x=11 ⇒ x=8.

Example 4: Verify that 9,18,082 is divisible by 11. Digits from right: 2,8,0,8,1,9. Odd positions (1,3,5): 2+0+1=3. Even positions (2,4,6): 8+8+9=25. Difference =3-25=-22, a multiple of 11 → divisible by 11 (check: 918082÷11=83462 exact ✓).

Example 5 (unique missing digit): Find x so that 5x2879 is divisible by 11. Digits from right: 9(1),7(2),8(3),2(4),x(5),5(6). Odd sum (1,3,5): 9+8+x=17+x. Even sum (2,4,6): 7+2+5=14. Difference =(17+x)-14=3+x. For divisibility by 11: 3+x=0 (invalid, gives x=-3) or 3+x=11 ⇒ x=8. x=8, giving the number 5,82,879 (check: 582879÷11=52989 exact ✓).