How do you prove that n 3 2n is divisible by 3?

Example Prove by induction that n3 + 2n is divisible by 3 for every non-negative integer n. Solution Let P(n) be the mathematical statement n3 + 2n is divisible by 3. Base Case: When n = 0 we have 03 +0=0=3 × 0. So P(0) is correct.

How can a number n divisible by 3 be represented?

A number is divisible by 3 if sum of its digits is divisible by 3. Illustration: For example n = 1332 Sum of digits = 1 + 3 + 3 + 2 = 9 Since sum is divisible by 3, answer is Yes.

How do you prove that N 3 N is even?

(1) For all integers n, if n is even, then n3 is even. Proof: Let n be an even integer, so that n = 2k for some integer k. Then n3 = (2k)3 = 8k3 = 2(4k3), which is even. n3 = (2k + 1)3 = (2k + 1)(2k + 1)2 = (2k + 1)(4k2 + 4k + 1) = 8k3 + 8k2 + 2k + 4k2 + 4k + 1 = 8k3 + 12k2 + 6k + 1 = 2(4k3 + 6k2 + 3k)+1, which is odd.

How do you prove something is a multiple of 3?

For example if n = 2:n-1 = 1n = 2n+1 = 3If you have a series of 3 consecutive numbers, clearly one of them will be a multiple of 3. Hence if; n3-n = (n-1)(n)(n+1), for all n and one of the numbers n-1, n, n+1 is a multiple of 3, then n3-n is also a multiple of 3.

How do you prove divisible by 3 proofs?

If sum of the digits of a number is divisible by 3, then the number is divisible by 3. This is known to all.

How do you prove that a number is divisible by 3 if the sum of its digits is divisible by 3?

A number is divisible by 3 if the sum of its digits is divisible by 3. For large numbers this rule can be applied again to the result. A.) 504: 5 + 0 + 4 = 9, so it is divisible by 3.

How do you prove a number is a multiple of 3?

If sum of digits in a number is multiple of 3 then number is multiple of 3 e.g., for 612 sum of digits is 9 so it’s a multiple of 3.

How to prove n^3 – n is divisible by 6?

Prove that for any positive interger n, n^3 – n is divisible by 6. there are two methods to solve the problem which are discussed below. 1. Out of three (n – 1), n, (n + 1) one must be even, so a is divisible by 2. 2. (n – 1) , n, (n + 1) are consecutive integers thus as proved a must be divisible by 3.

How do you know if a number is divisible by 2?

∴ n = 2q or 2q + 1, where q is some integer. Then n is divisible by 2. n+1= 2 (q + 1) is divisible by 2. So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2. ∴ n (n – 1) (n + 1) is divisible by 2.

Which equation is divisible by 3?

The statement is vacuously true for n = 1, 2 and obviously true for n = 3. Here P = n*3 ==> divisible by 3 Q = (n-1) n (n+1) ==> this is multiplication of three consecutive number and hence it is divisible by 3 ==> P + Q is divisible by 3

What is the value of 3 ∤ n2?

Note n 2 = 3 ( 3 k 1 2 + 2 k 1) + 1, where 3 k 1 2 + 2 k 1 ∈ N, since N closed under multiplication and addition. So, since we have remainder r = 1 we see that 3 ∤ n 2.