# 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?

## 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.

Mar 4, 2016

For the first problem, note that a number is divisible by $6$ if and only if it is divisible by both $2$ and by $3$. As any three consecutive integers will contain at least one multiple of $2$ and exactly one multiple of $3$, the product of any three consecutive integers will be divisible by both, and thus by $6$.

For the second problem, we can solve this using modular arithmetic. The basic idea behind modular arithmetic is that rather than look at the specific value of a given integer, we look at its remainder when divided by a given modulus. This is just like how when we use an analog clock, we will arrive at the same time if we wait $12$ hours or $24$ hours. We are operating modulo $12$, and rather than saying that $12$ and $24$ are equal we say that they are congruent modulo 12 (and we use the symbol $\equiv$ to denote this). Modular arithmetic is a very useful tool and is worth reading up on.

As a number is divisible by $7$ if and only if it is congruent to $0$ modulo $7$, we can calculate the sum modulo $7$ to demonstrate the result.

As ${1}^{n} = 1$ for all $n$, we have ${1}^{2015} \equiv 1 \text{ (mod 7)}$

As ${2}^{3} = 8$ and $8 \equiv 1 \text{ (mod 7)}$ we have
${2}^{2015} \equiv {2}^{2} {\left({2}^{3}\right)}^{671} \equiv 4 \cdot {1}^{671} \equiv 4 \text{ (mod 7)}$

As ${3}^{3} = 27$ and $27 \equiv - 1 \text{ (mod 7)}$ we have
${3}^{2015} \equiv {3}^{2} {\left({3}^{3}\right)}^{671} \equiv 9 {\left(- 1\right)}^{671} \equiv - 9 \equiv - 2 \text{ (mod 7)}$

As ${4}^{3} = 64$ and $64 \equiv 1 \text{ (mod 7)}$ we have
4^2015-=4^2(4^3)^2015-=16(1^671-=16-=2" (mod 7)"

As ${5}^{3} = 125$ and $125 \equiv - 1 \text{ (mod 7)}$ we have
${5}^{2015} \equiv {5}^{2} {\left({5}^{3}\right)}^{671} \equiv 25 {\left(- 1\right)}^{671} \equiv - 25 \equiv 3 \text{ (mod 7)}$

As $6 \equiv - 1 \text{ (mod 7)}$ we have
${6}^{2015} \equiv {\left(- 1\right)}^{2015} \equiv - 1 \text{ (mod 7)}$

Then, substituting our newly found values into the given expression, we have

${1}^{2015} + {2}^{2015} + {3}^{2015} + {4}^{2015} + {5}^{2015} + {6}^{2016} \equiv$

$\equiv 1 + 4 - 2 + 2 + 3 - 1 \text{ (mod 7)}$

$\equiv 7 \text{ (mod 7)}$

$\equiv 0 \text{ (mod 7)}$

Therefore, as the expression is congruent to $0$ modulo $7$, it is divisible by $7$.

Mar 20, 2016

Alternative way

#### Explanation:

For the second part of the problem this can be proved for first $2 n$ terms of natural number having odd power where$\text{ } n = 1 , 2 , 3 , 4 , 5 , 6 , \ldots . .$

We know that a^n+b^n"" is always divisible by$\left(a + b\right)$ when n is odd. It can be easily verified by putting $- b$ in place of a when the value of ${a}^{n} + {b}^{n}$ becomes zero
So if we rearrange the given problem as below

$\left({1}^{2015} + {6}^{2015}\right) + \left({2}^{2015} + {5}^{2015}\right) + \left({3}^{2015} + {4}^{2015}\right)$
we see that here sum of the bases in every pair with in parentheses
like (1+6);(2+5);(3+4) are all 7. So as per the rule of divisibility the given sum is divisible by 7

similarly the following series of 12 terms is also divisible by 13

${1}^{2015} + {2}^{2015} + {3}^{2015} + {4}^{2015} + {5}^{2015} + {6}^{2015} + {7}^{2015} + {8}^{2015} + {9}^{2015} + {10}^{2015} + {11}^{2015} + {12}^{2015}$