Which of the following numbers are divisible by 11?
(i) 29435417
(ii) 57463828
(iii) 57463824
(iv) 29435416
- A(iii) and (iv)
- B(ii) and (iii)
- C(i) and (ii)
- D(i) and (iii)
Solution & Step-by-step Explanation
A number is divisible by 11 if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is either 0 or a multiple of 11.
Let's test each number:
(i) 29435417
Sum of digits at odd places (from right): 7+4+3+9=23
Sum of digits at even places (from right): 1+5+4+2=12
Difference = 23−12=11 (Divisible by 11)
(ii) 57463828
Sum of digits at odd places (from right): 8+8+6+7=29
Sum of digits at even places (from right): 2+3+4+5=14
Difference = 29−14=15 (Not divisible by 11)
(iii) 57463824
Sum of digits at odd places (from right): 4+8+6+7=25
Sum of digits at even places (from right): 2+3+4+5=14
Difference = 25−14=11 (Divisible by 11)
(iv) 29435416
Sum of digits at odd places (from right): 6+4+3+9=22
Sum of digits at even places (from right): 1+5+4+2=12
Difference = 22−12=10 (Not divisible by 11)
Thus, numbers (i) and (iii) are divisible by 11.
Let's test each number:
(i) 29435417
Sum of digits at odd places (from right): 7+4+3+9=23
Sum of digits at even places (from right): 1+5+4+2=12
Difference = 23−12=11 (Divisible by 11)
(ii) 57463828
Sum of digits at odd places (from right): 8+8+6+7=29
Sum of digits at even places (from right): 2+3+4+5=14
Difference = 29−14=15 (Not divisible by 11)
(iii) 57463824
Sum of digits at odd places (from right): 4+8+6+7=25
Sum of digits at even places (from right): 2+3+4+5=14
Difference = 25−14=11 (Divisible by 11)
(iv) 29435416
Sum of digits at odd places (from right): 6+4+3+9=22
Sum of digits at even places (from right): 1+5+4+2=12
Difference = 22−12=10 (Not divisible by 11)
Thus, numbers (i) and (iii) are divisible by 11.