# 1.Show that n(n+1)(n+2) is divisible by 6. 2.Show that #1^2015+2^2015+3^2015+4^2015+5^2015+6^2015# is divisible by 7. How do I solve these?

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I have no lead on the first problem. I know the divisibility rule of 7 but I have to know the whole number for that whereas I can only know the last few digits.

I have no lead on the first problem. I know the divisibility rule of 7 but I have to know the whole number for that whereas I can only know the last few digits.

Alternative way

Likewise, the next set of twelve terms divides by thirteen.

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Next, by entering our recently discovered values into the provided expression, we have

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- To show that n(n+1)(n+2) is divisible by 6, we can use the principle of mathematical induction.

First, we check the base case: For n = 1, we have 1(1+1)(1+2) = 1(2)(3) = 6, which is divisible by 6.

Next, we assume that the statement holds for some arbitrary positive integer k, i.e., k(k+1)(k+2) is divisible by 6.

Now, we need to prove that the statement holds for k+1, i.e., (k+1)(k+2)(k+3) is divisible by 6.

Expanding the expression, we get (k+1)(k+2)(k+3) = (k^2 + 3k + 2)(k+3) = k^3 + 6k^2 + 11k + 6.

We can rewrite this expression as k(k+1)(k+2) + 5(k^2 + 2k + 1).

Since we assumed that k(k+1)(k+2) is divisible by 6, and 5(k^2 + 2k + 1) is divisible by 6 (as 5 is divisible by 6), we can conclude that (k+1)(k+2)(k+3) is divisible by 6.

By the principle of mathematical induction, we have shown that n(n+1)(n+2) is divisible by 6 for all positive integers n.

- To show that 1^2015 + 2^2015 + 3^2015 + 4^2015 + 5^2015 + 6^2015 is divisible by 7, we can use the concept of modular arithmetic.

We can observe that for any positive integer k, k^6 ≡ 1 (mod 7) by Fermat's Little Theorem.

Now, let's consider the given expression: 1^2015 + 2^2015 + 3^2015 + 4^2015 + 5^2015 + 6^2015.

We can rewrite this expression as (1^6)^335 + (2^6)^335 + (3^6)^335 + (4^6)^335 + (5^6)^335 + (6^6)^335.

Using the property mentioned earlier, we can simplify this expression as 1^335 + 1^335 + 1^335 + 1^335 + 1^335 + 1^335.

Since any number raised to the power of 335 is still that number, we have 1 + 1 + 1 + 1 + 1 + 1 = 6.

Therefore, the given expression is divisible by 7.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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