How do you write 98 as a product of prime factors?

1 Answer
George C.
Nov 19, 2016

#98 = 2*7*7#

Explanation:

Since #98# ends with an even digit, we can tell that it is divisible by #2#:

#98/2 = 49#

Try each possible prime factor in turn:

  • #49# ends in an odd digit, so is not divisible by #2#.

  • The sum of its digits is #4+9 = 13#, which is not divisible by #3#, so #49# is not divisible by #3# either.

  • #49# does not ends with #5# or #0#, so is not divisible by #5#.

  • #49# is divisible by #7#. We find:

    #49/7 = 7#

So:

#98 = 2*7*7#

We can also draw this as a factor tree:

#color(white)(00000)98#
#color(white)(0000)"/"color(white)(00)"\"#
#color(white)(000)2color(white)(000)49#
#color(white)(000000)"/"color(white)(00)"\"#
#color(white)(00000)7color(white)(0000)7#

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