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.

2 Answers
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 #-=# 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#.

Mar 20, 2016

Answer:

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#