Question

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.

Answer 1

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`.

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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 `-=` 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-=1" (mod 7)"`

As `2^3 = 8` and `8-=1" (mod 7)"` we have
`2^2015 -= 2^2(2^3)^671-=4*1^671-=4" (mod 7)"`

As `3^3 = 27` and `27 -=-1" (mod 7)"` we have
`3^2015 -=3^2(3^3)^671-=9(-1)^671-=-9-=-2" (mod 7)"`

As `4^3=64` and `64-=1" (mod 7)"` we have
`4^2015-=4^2(4^3)^2015-=16(1^671-=16-=2" (mod 7)"`

As `5^3=125` and `125-=-1" (mod 7)"` we have
`5^2015-=5^2(5^3)^671-=25(-1)^671-=-25-=3" (mod 7)"`

As `6-=-1" (mod 7)"` we have
`6^2015 -=(-1)^2015-=-1" (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-=`

`-=1+4-2+2+3-1" (mod 7)"`

`-=7" (mod 7)"`

`-=0" (mod 7)"`

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

Answer 2

Alternative way

Explanation:

For the second part of the problem this can be proved for first `2n` terms of natural number having odd power where`" "n =1,2,3,4,5,6,.....`

We know that `a^n+b^n""` is always divisible by`(a+b)` 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

`(1^2015+6^2015)+(2^2015+5^2015)+(3^2015+4^2015)`
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`