NCERT Exemplar Solutions
Class 6 - Mathematics - Unit 1: Number System - Problems and Solutions

Question 198

Rule for 11: Find the sum of digits in odd places (from the right) and the sum of digits in even places. Take the difference. If the difference is 0 or a multiple of 11, the number is divisible by 11.

(a) 5335

1. Write places from the right: 5 (1st), 3 (2nd), 3 (3rd), 5 (4th).
2. Odd places: 1st and 3rd → 5 and 3.
3. Even places: 2nd and 4th → 3 and 5.
4. Odd-place sum:
   \(5 + 3 = 8\)
5. Even-place sum:
   \(3 + 5 = 8\)
6. Difference:
   \(8 - 8 = 0\)
7. Since \(0\) is a multiple of \(11\), 5335 is divisible by 11.

(b) 9020814

1. Write places from the right: 4 (1st), 1 (2nd), 8 (3rd), 0 (4th), 2 (5th), 0 (6th), 9 (7th).
2. Odd places: 1st, 3rd, 5th, 7th → 4, 8, 2, 9.
3. Even places: 2nd, 4th, 6th → 1, 0, 0.
4. Odd-place sum:
   \(4 + 8 = 12\)
   \(12 + 2 = 14\)
   \(14 + 9 = 23\)
5. Even-place sum:
   \(1 + 0 = 1\)
   \(1 + 0 = 1\)
6. Difference:
   \(23 - 1 = 22\)
7. Since \(22\) is a multiple of \(11\), 9020814 is divisible by 11.