How do you find the prime factorization of 153?
4 Answers
Explanation:
Note that
prime factors of 153 =
Explanation:
[1] Let
[2] Let
[3] Let
[4] while
[5]
[6]
[7]
[8]
[9]
[10]
[11] end while
[12] add
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Let's see how this works with the given positive integer
#{: (underline(color(black)(n)),underline(color(black)(f)),underline(color(black)(L)),underline("comment")), (153,2,{},), (153,3,{},), (51,3,{3},), (17,3,{3,3},), (17,5,,5^2>=17), (,,{3,3,17},) :}#
Explanation:
The problem with numbers bigger than about 100, is that we often don't know their factors.
The "RULES for DIVISIBILITY" are very useful and prevent us from wasting time by dividing by numbers which are not factors. If we can find just one factor, the others are usually easy to find.
Does 153
Only even numbers are divisible by 2. 153 is odd so it has no even number as a factor.
Does 153
If the sum of the digits is divisible by 3, then 3 is a factor.
(1+5+3 = 9), so 3 is a factor.
Does 153
153 does not end in a 5 or 0, so 5 is not a factor.
We only need to look for factors smaller than
#sqrt153 = 12.4
The only prime numbers left to check would be 7 and 11.
But we know that 3 is a factor, so let's start by dividing by 3.
Explanation:
Starting point
If you add the digits of 153 you have 1+5+3 = 9
As 9 is divisible by 3 exactly then 153 is also divisible by 3 exactly
Build a factor tree:
From this you observe that
as 3 and 17 are prime factors your answer is:
You can obtain a list of prime factors by doing an internet search.