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

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1.Show that n(n+1)(n+2) is divisible by 6. 2.Show that $1^{2015} + 2^{2015} + 3^{2015} + 4^{2015} + 5^{2015} + 6^{2015}$ $1^2015+2^2015+3^2015+4^2015+5^2015+6^2015$ 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.

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Answer 1 · 2016-03-04T19:50:56

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

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 $12$ $12$ hours or $24$ $24$ hours. We are operating modulo $12$ $12$ , and rather than saying that $12$ $12$ and $24$ $24$ 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.](http://betterexplained.com/articles/fun-with-modular-arithmetic/)

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

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

As $2^{3} = 8$ $2^3 = 8$ and $8 \equiv 1 (mod 7)$ $8-=1" (mod 7)"$ we have
$2^{2015} \equiv 2^{2} {(2^{3})}^{671} \equiv 4 \cdot 1^{671} \equiv 4 (mod 7)$ $2^2015 -= 2^2(2^3)^671-=4*1^671-=4" (mod 7)"$

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

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

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

As $6 \equiv - 1 (mod 7)$ $6-=-1" (mod 7)"$ we have
$6^{2015} \equiv {(- 1)}^{2015} \equiv - 1 (mod 7)$ $6^2015 -=(-1)^2015-=-1" (mod 7)"$

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

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

$\equiv 1 + 4 - 2 + 2 + 3 - 1 (mod 7)$ $-=1+4-2+2+3-1" (mod 7)"$

$\equiv 7 (mod 7)$ $-=7" (mod 7)"$

$\equiv 0 (mod 7)$ $-=0" (mod 7)"$

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

Answer 2 · 2016-03-20T15:42:26

Alternative way

Explanation:

For the second part of the problem this can be proved for first $2 n$ $2n$ 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

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

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1.Show that n(n+1)(n+2) is divisible by 6. 2.Show that $1^{2015} + 2^{2015} + 3^{2015} + 4^{2015} + 5^{2015} + 6^{2015}$ $1^2015+2^2015+3^2015+4^2015+5^2015+6^2015$ 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.

2 Answers

Explanation:

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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+6^201512015+22015+32015+42015+52015+620151^2015+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.

2 Answers

Explanation:

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