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YOU are better than YOU think. Show yourself how:
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-/[]\- Logic chapters 1 to 5 re- appear not in sequence, as is or longer, in Volume 1A, Pattern Based Reason, Bon Appetite. Logic
Mastery Logic mastery makes the hard, easier. Logic mastery leads to better, stronger and richer comprehension. Logic mastery improves reading and writing. Logic mastery ease learning difficulties. Logic mastery gives a headstart. In sum, logic mastery will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck. After logic, (a) continue reading Three Skills for Algebra, chapters 8 to 14 and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes & More Math, chapters 2 to 6;
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-/[]\- What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts. Try the Twiddla
Whiteboard. In principle, it allows
to people to draw and chat together online on a copy of this webpage or a clean
sheet. The chat may be via text or audio. Visit www.twiddla.com
to set up whiteboards to work with the webpage of your choice. |
Prime Factorization Examples.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
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'
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''
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''
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
The foregoing gives a reduce set of primes
that could be divisors of 509. With a calculator, we see
so 13 is not a divisor.
Next we try 17, 509 = 29. 94 is not an integer, so 17 is not a divisor.
Next we try 19, 509 = 26.78 is not an integer, so 19 is not a divisor.
Lastly we try 23, 509 = 22.13 is not an integer, so 23 is not a divisor.
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
Divisibility rules allow us to eliminate 2 and 5 immediately - they are not divisors of 1183.
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.
Now
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.
Here 7 might be a factor again. But
Eliminate 7 from the list of possible prime divisors of 169.
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 Therefore 169 = 13 x 13 So 1183 = 7 x 169 = 7 x 132 and hence
The latter product provides the prime decomposition of 11830
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