Say whether the following is true or false and support your answer by a proof: The sum of any five consecutive integers is divisible by 5 (without remainder)?

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Answer 1 · 2017-08-25T13:18:32

See a solution process below:

The sum of any 5 consecutive integers is, in fact, evenly divisible by 5!

To show this let's call the first integer: #n#

Then, the next four integers will be:

#n + 1#, #n + 2#, #n + 3# and #n + 4#

Adding these five integers together gives:

#n + n + 1 + n + 2 + n + 3 + n + 4 =>#

#n + n + n + n + n + 1 + 2 + 3 + 4 =>#

#1n + 1n + 1n + 1n + 1n + 1 + 2 + 3 + 4 =>#

#(1 + 1 + 1 + 1 + 1)n + (1 + 2 + 3 + 4) =>#

#5n + 10 =>#

#5n + (5 xx 2) =>#

#5(n + 2)#

If we divide this sum of any 5 consecutive integers by #color(red)(5)# we get:

#(5(n + 2))/color(red)(5) =>#

#(color(red)(cancel(color(black)(5)))(n + 2))/cancel(color(red)(5)) =>#

#n + 2#

Because #n# was originally defined as an integer #n + 2# is also an integer.

Therefore, the sum of any five consecutive integers is evenly divisible by #5# and the result is an integer with no remainder.