How can you use prime factorization to determine if 856 is evenly divisible by 7?

1 Answer
May 1, 2016

#856# is not evenly divisible by #7#

Explanation:

You can reduce the size of the problem if you can separate out other prime factors.

#color(white)(00000)856#
#color(white)(00000)"/"color(white)(0)"\"#
#color(white)(0000)2color(white)(00)428#
#color(white)(0000000)"/"color(white)(0)"\"#
#color(white)(000000)2color(white)(00)214#
#color(white)(000000000)"/"color(white)(0)"\"#
#color(white)(00000000)2color(white)(00)107#

If we knew that #107# is prime then we would be done.

  • It is not divisible by #2# since the last digit is not even.
  • It is not divisible by #3# since the sum of the digits #1+0+7=8# is not divisible by #3#.
  • It is not divisible by #5# since the last digit is neither #5# nor #0#.

The next factor to try would be #7#, which is the one we're wanting to test anyway.

I'm not sure this factorisation has helped us much, but let us at least split it down to make the arithmetic a little easier:

#107 / 7 = (70+35+2) / 7 = 70/7+35/7+2/7 = 10+5+2/7 = 15 2/7#

So #107# is not divisible by #7# and neither is #856#.