Prime Factorization Examples
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If a whole number N less than 169 = 132 is not divisible by
2, 3, 5, 7 nor 11 [the primes smaller than 13] then the number is
prime. Otherwise, it - the number N - is a product of 2, 3, 5, 7 or 11
with another factor N' less than 169.
Knowledge of times tables, divisibility rules or a calculator - one
displaying results to two plus decimals, may be used to recognize
multiples of 2, 3, 5, 7 and 11. Two decimals are sufficient because the
fractions one half to one eleventh are all more than 0.01 = one
hundredth.
In learning and applying algebra exactly, one usually needs to compute
with fractions or ratios of whole numbers less than 169 or so. Learning how to efficiently use the above method for recognizing
primes and obtaining prime factorization of of whole numbers less than
169 is sufficient for most ends and purposes purposes in high school and
college mathematics and science courses. The above method is based on the
square method below.
In ceneral, if a whole number N less than the square of a given prime is not divisible by
all the the primes smaller than given one then the number is
prime. Otherwise, it - the number N - is a product of one smaller prime
and another factor N' less than the square of the given prime.
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Problem A: Find the prime decomposition of 158
A long solution follows. It includes all the details. A more
compact version would be in order when a student follow the
pattern.
Long Solution: Squaring the first six primes 2, 3, 5, 7 or 11 and
13 squared give the sequence 4, 9, 25, 49, 121 and 169 of whole
numbers. Here p = 2, 3, 5, 7 and 11 have the property that
p2 < 156
Step 1: List the possible primes factors with p2
< N = 158
2, 3, 5, 7, 11
Find the smallest prime, if any, in the list which is a
divisor of N? Here the smallest prime is 2, as N = 158 = 2 x 78 = 2
x N'
Step 2: List the possible primes factors with p2
< 78 = N'
2, 3, 5, 7
Find the smallest prime, if any, in the list which is a
divisor of N'? Here the smallest prime is 2, as N' = 78
= 2 x 39 = 2 x N''
Step 3: List the possible primes factors with p2
< 39 = N''
2, 3, 5, 7
Find the smallest prime, if any, in the list which is a
divisor of N'? Here the smallest prime is 3 as N'' = 39 = 3 x 13 =
3 x N'''
List the possible primes factors with p2 < 13
= N''
2, 3,
No prime in the list divides 13. So 13 is prime.
Step 4: We conclude N = 2 x N' = 2 x 2 x N'' = 2x2x3xN''' =
2x2x3 x 13
So the prime decomposition of 158 =
2231(13)1
Note to avoid ambiguity, for the square of a whole number like 345 write
(345)2 instead of 3452.
Problem B: Find the prime decomposition of 4581
Solution: Before listing all primes p with p2
< 4581, let us get smaller numbers. Observe
modulo 9, 4581 = 4+ 5 + 8 + 1 = 0.
So 4581 = 9 x 509. So we need to find the prime decomposition (a.k.a
factorization) of 509 < 625 = 252. The list of primes
< 25 is as follows.
2, 3, 5, 7, 9, 11, 13, 17, 19, 23.
Immediately, divisibility rules say 509 is not a multiple of 2, 3, 5, 9
and 11 as the last digit is odd, the sum of digits is nonzero modulo 3
and 9, the last digit is not a 5 and as
modulo 11, 509 = 9 - 0 + 5 = 14 = 3 =\=0
The foregoing gives a reduce set of primes
2, 3, 5, 7, 9, 11, 13, 17, 19, 23.
that could be divisors of 509. With a calculator, we see
509 = 39.15 is not an
integer,
13
so 13 is not a divisor.
2, 3, 5, 7, 9, 11, 13, 17, 19, 23.
Next we try 17,
509 = 29. 94 is not an integer,
17
so 17 is not a divisor.
2, 3, 5, 7, 9, 11, 13, 17, 19, 23.
Next we try 19,
509 = 26.78 is not an integer,
19
so 19 is not a divisor.
2, 3, 5, 7, 9, 11, 13, 17, 19, 23.
Lastly we try 23,
509 = 22.13 is not an integer,
23
so 23 is not a divisor.
2, 3, 5, 7, 9, 11, 13, 17, 19, 23.
Therefore 506 is prime as all primes with square < 509 do not
divide into it a whole number of times.
Solution: the prime decomposition of 4581 is given by 9 x 509 and
9 x 509 = 4581 for reasons shown above.
Problem C: Find the prime decomposition of 11830
Solution: Before listing all primes p with p2
< 4581, let us get smaller numbers. Observe
11830 = 1183 x 10
The 10 gives two prime factors 2 and 5. Now
the square root of 1183 is 34.39 < 35. The list of primes < 35 is
2, 3, 5, 7, 13, 17, 19, 23, 29, 31
Divisibility rules allow us to eliminate 2 and 5 immediately - they are
not divisors of 1183.
2, 3, 5, 7, 13, 17, 19, 23, 29, 31
Now
modulo 3, 1183 = 1 + 1 + 8 + 0 = 10 = 1.
So the remainder on division by 3 is 1 and 3 is not a divisor. Eliminate
3.
2, 3, 5, 7, 13, 17,
19, 23, 29, 31
Now
1183 = 169
7
according to my calculator. To avoid be misled by a possible
rounding to 169, I clear the display and enter 169 into the calculator
and multiply by 7. The result is 7 x 169 = 1183 exactly.
So 7 is a factor. We now try to factor 169.
2, 3, 5, 7, 13, 17,
19, 23, 29, 31
Here 7 might be a factor again. But
169 = 24.14 is not an integer.
7
Eliminate 7 from the list of possible prime divisors of 169.
2, 3, 5,
7, 13, 17, 19, 23, 29, 31
Note 1183 = 7 x 169. So the primes 2, 3 and 5 remain cross-out or
eliminated as we know 2, 3 and 5 are not divisors of 1183 and hence
cannot be divisors of 169 as if they were, they would be divisors of 1183
as well.
So now we try 13. From a calculator or memory of the square of 13
169 = 13 a prime
13
Therefore 169 = 13 x 13
So 1183 = 7 x 169 = 7 x 132 and hence
11830 = 1183 x 10 = 7 x 132 x 10 = 2 x 5 x
7 x 132
The latter product provides the prime decomposition of 11830
Exercise: Identify all multiples of 7 less than 121 that are not
also multiples of smaller primes 2, 3 and 5. The first three are
indicated above.
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