Math, asked by ashwanipyl01, 1 month ago

14. Test the divisibility of: (i) 10000001 by 11 (iv) 10001001 by 3 (ii) 19083625 by 11 (v) 10203574 by 4 nrime numbers? (iii) 2134563 by 9 (vi) 12030624 by 8​

Answers

Answered by dreampride94
0

Answer:

Sum of the digits at odd places = 8 + 3 + 6 + 0 Sum of the digits at even places = 0 + 9 + 1 + 7 = 17 Difference = 17 - 17 = 0 As the difference between the sum of the digits at odd places and the sum of the digits at even places is O, therefore, 70169308 is divisible by 11.

(e) 10000001

Sum Of the digits at Odd places = 1 Sum of the digits at even places 1 Difference = 1 - 1 = 0 As the difference between the sum of the digits at odd places and the sum of the digits at even places is O, therefore, 10000001 is divisible by 11.

(f) 901153

Sum of the digits at odd places = 3 + 1 + 0 = 4 Sum Of the digits at even places = 5 + 1 + 9 = 15 Difference = 15 - 4 = 11 The difference between the sum of the digits at odd places and the sum of the digits at even places is 11, which is divisible by 11. Therefore, 901153 is divisible by 11.

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