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?

Question

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.

Ethan Adkins · Accepted Answer

Alternative way

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 Likewise, the next set of twelve terms divides by thirteen. #1^2015+2^2015+3^2015+4^2015+5^2015+6^2015+7^2015+8^2015+9^2015+10^2015+11^2015+12^2015#

Julia Adams · Answer

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 #-=# 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)"# Next, by entering our recently discovered values into the provided 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#.

Savannah Anderson · Answer

undefined 1. 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.

2. 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.

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.