How do you write 78 as a product of prime factors?
1 Answer
Mar 16, 2018
Explanation:
Given
-
#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 = 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#