How do you show that the product of #3# consecutive integers is always divisible by #6# ?

1 Answer
Apr 30, 2016

Answer:

See explanation...

Explanation:

Consider three consecutive integers: #n#, #n+1# and #n+2#.

#n = 3q+r#

for some integer quotient #q# and remainder #r in {0, 1, 2}#

If #r=0# then #n# is divisible by #3#

If #r=1# then #n+2# is divisible by #3#

If #r=2# then #n+1# is divisible by #3#

So at least (in fact exactly) one of #n#, #n+1# and #n+2# will be divisible by #3#.

Hence #n(n+1)(n+2)# will be divisible by #3# too.

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Similarly for #2#...

#n = 2q_1+r_1#

for some integer quotient #q_1# and remainder #r_1 in {0, 1}#

If #r_1=0# then #n# is divisible by #2#

If #r_1=1# then #n+1# is divisible by #2#

So at least one of #n# and #n+1# will be divisible by #2#.

Hence #n(n+1)# (and #n(n+1)(n+2)#) will be divisible by #2# too.