Question

Prove that an integer consisting of 3n ones is divisible by 3n (e.g. 111 is divisible by 3, 111,111,111 is divisible by 9, etc.) Prove it by induction. Notice, that while the number with a sum of digits divisible by 3 is itself a multiple of 3 and the number with a sum of digits divisible by 9 is itself a multiple of 9, it is NOT true that the number with a sum of digits divisible by 27 is necessarily a multiple of 27.

Answer 1

For proof of 3 divisibility, abc is a divisible by 3 if the sum of abc (a + b + c) is a multiple of 3.

Integers divisible by 3

The proof for divisibility of 3 implies that an integer is divisible by 3 if the sum of the digits is a multiple of 3.

Proof for the divisibility

111 = 1 + 1 + 1 = 3  (the sum is multiple of 3 = 3 x 1)  (111/3 = 37)

222 = 2 + 2 + 2 = 6 (the sum is multiple of 3 = 3 x 2)  (222/3 = 74)

213 = 2 + 1 + 3 = 6 ( (the sum is multiple of 3 = 3 x 2)  (213/3 = 71)

27 = 2 + 7 = 9  (the sum is multiple of 3 = 3 x 3)  (27/3 = 9)

Thus, abc is a divisible by 3 if the sum of abc (a + b + c) is a multiple of 3.

Learn more about divisibility here: brainly.com/question/9462805

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