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

Question

Mila Anderson · Accepted Answer

#98 = 2*7*7#

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#

Jeremiah Andrews · Answer

undefined To write 98 as a product of prime factors, you can start by dividing it by the smallest prime number, which is 2. Continue dividing until you can't divide anymore.

98 ÷ 2 = 49

49 is not divisible by 2, so you move to the next prime number, which is 3. But 49 is not divisible by 3 either.

Next, try dividing by 5.

49 ÷ 7 = 7

7 is a prime number, so the prime factorization of 98 is:

2 × 7 × 7