Prime Factor Trees
0
If the number ends in a “0”, factor out a 10.
Then break down the 10 into 2 ∙ 5.
630
10
2
1200
63
5
(more)
At the end: 630= 2∙3∙3∙5∙7
5
10
120
2 5
10 12
2 5 (more)
At the end:
1200= 2∙2∙2∙2∙3∙5∙5
375
63
(more)
At the end: 315=3∙3∙5∙7
5
981
75
5
Final:
375=3∙5∙5∙5
354
5
When it’s done: 354=2∙3∙59
3
If it’s an Even Number – it ends in
“2”, “4”, “6”, “8”, factor out a 2.
This also works with “0”.
768
177
(more)
It doesn’t end in a 5, so use
other tricks to break it
down. See below, under “9”.
15
2, 4, 6, 8
2
It doesn’t end in a 0, so use
other tricks to break it
down. This one ends in 5.
See below.
If the number ends in a “5” or a “0”, factor out a 5.
This also works if the number ends in a “0”.
315
5
315
2
384
2
It will
be 28∙3
261
192
2 again,etc.
It’s not an even number, so
use other tricks to break it
down.
It turns out to be 3∙3∙29
More on the reverse side > > > > > > > > > > > > > > > > > > > > > > > >
PrimeFactorTrees.docx
D.R.S. 10/15/2013 1:56 PM
Prime Factor Trees
Sum div by 9
981: 9 + 8 + 1 = 18
18 div by 9 = 2, remainder 0
981
9
6777: 6 + 7 + 7 + 7 = 27
27 div by 9 = 3, remainder 0
6777
109
3
3
And 109 happens to be a
prime number! So it turns
out to be 3∙3∙109
9
3
753
3
to be
continued
below
Sum div by 3
147: 1 + 4 + 7 = 12
12 div by 3 = 4 remainder 0
147
3
49
7
Final: 147 = 3∙7∙7
1617: 1 + 6 + 1 + 7 = 15
15 div by 9 = 1 remainder 6
1617
Since the sum of the digits
doesn’t divide by 9 exactly,
9 is not a factor.
So use other tricks to break
it down. See below.
Add up the digits.
Does 3 divide evenly into the sum?
If so, then factor out a 3.
753: 7 + 5 + 3 = 15
15 div by 3 is 5 remainder 0
6777 continued
9
7
Add up the digits.
Does 9 divide evenly into the sum?
If so, then factor out a 9,
Then break down the 9 into 3 ∙ 3.
753
3 3
3 251
And 251 happens to be a
prime number! So 33∙251
1617: 1 + 6 + 1 + 7 = 15
15 div by 3 is 5 remainder 0
1617
3
539
(more)
It turns out that
1617= 3∙7∙7∙11
7, 11, 13, 17, 19,
23, 29, 31, 37, etc.
Trying 10, 5, 2, 9, and 3 will get you a
long way! After those are used up,
try the other primes, starting with 7
and moving up.
When should I stop? When you try to divide a prime number into the number you’re trying
to factor and you get a quotient that’s smaller than the prime number, you can stop.
Consider the 251 in the middle example under “Sum div by 3”. We try 7 into 251 = 35 r 6. We try 11
into 251 = 22 r 9. We try 13 into 251 = 19 r 4. We try 17 into 251 = 14 r 3. We stop here because
prime 17 is bigger than the quotient 14. Therefore 251 must be a prime number.
PrimeFactorTrees.docx
D.R.S. 10/15/2013 1:56 PM