How do you write 78 as a product of prime factors?

1 Answer
Mar 16, 2018

#78 = 2 * 3 * 13#

Explanation:

Given #78# to factorise

  • #78# ends with an even digit, so is divisible by #2# and we find:
    #78 = 2 * 39color(white)(0/0)#

  • #39# ends with an odd digit, so is not divisible by #2#.

  • The digits of #39# add up to a multiple of #3#, namely #3+9 = 12#. So we can tell that #39# is divisible by #3#:
    #39 = 3 * 13color(white)(0/0)#

  • #13# is a prime number, not divisible by any number greater than #1# or less than #13#.

So the prime factorisation of #78# is:

#78 = 2 * 3 * 13#

This can also be represented by a factor tree:

#color(white)(0000)78#
#color(white)(000)"/"color(white)(00)"\"#
#color(white)(00)2color(white)(000)39#
#color(white)(00000)"/"color(white)(00)"\"#
#color(white)(0000)3color(white)(000)13#