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Which number under 100 has the most primed factors

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Hi johnhboland
Prime numbers are 2,3,5,7,11,13,17,23 .........
Take different combinations from the above list and multiply making sure you don't exceed 100

2x3x5 =30
2x3x7 =42
2x3=11 =66
2x3x13 =78
2x5x7 =70
I would say the answers are 30, 42, 66, 78, and 100 all have 3 prime factors
If repetition of the same factor is allowed:
answers would be 64 and 96 both having 6 prime factors
Have a good day
luciana44


Posted on Nov 22, 2009

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1 Answer

What is a prime factor


A factor that is a prime number. Any of the prime numbers that, when multiplied, give the original number. Example: Theprime factors of 15 are 3 and 5 (3×5=15, and 3 and 5 areprime numbers).

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Number 2 is chosen to begin a ladder diagram to find the prime factorization of 66. What other numbers could have been used to start the ladder diagram of 66?


By changing the number for your prime factorization, you also have a different diagram.

Prime factorization for 66

66
/\
2*33
/\
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Even if we change the prime factorization order, we still have the same numbers.

66 = 2 * 3 * 11

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Is 8 a prime number? And what are the factors for 20? And is 20 a prime number?


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My casio fx-83GT Plus calculator wont do HCF for me


There is nothing you can do to make the calculator find the HCF for you. No point complaining about that. But if you are interested in doing it by hand (using the calculator to do the divisions for you) here how it is done.
  1. Decompose the first number in prime factors. If a prime factor is repeated use the exponent notation: That helps.
  2. Decompose the second number in prime factors too, using the exponent notation.
  3. Now look at the two decompositions. If a prime factor is present in both decompositions it must be in the HCD /HCF, with the smallest of its two exponents.
  4. Do that for all prime factors
Example: Here are the decompositions of two numbers
(2^5)*3*(5^4)(7^3)*11 and (2^3)*(5^6)*(11^2)*7
The prime factors that are present in both decompositions are
2, 5, 7, and 11
From the two decompositions select the smallest power of each common prime factor. They are represented in bold fonts.
2^3, 5^4, 7, 11
The highest common divisor/Highest common factor is
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Can you do the highest common factor on an fx-83GT PLUS casio calculator


The calculator has no application that will find the highest common divisor (or HCF) but that should not be difficult to do by hand.
Decompose the first number in prime factors. If a prime factor is repeated use the exponent notation: That helps
Decompose the second number in prime factors too, using the exponent notation.
Now look at the two decompositions. If a prime factor is present in both decomposition it must be in the HCD /HCF, with the smallest of the two exponents. Do that for all prime factors
Example;
(2^5)*3*(5^4)(7^3)*11
(2^3)*(5^6)*(11^2)*7
The prime factors that are present in both decompositions are
2, 5, 7, and 11
From the two decompositions select the smallest power of each
2^3, 5^4, 7, 11
The highest common divisor/Highest common factor is
(2^3)*(5^4)*7*11

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Factors of 100


1, 2, 4, 5, 10, 20, 25, 50,100
Prime decomposition of 100
100=1*(2^2)*(5^2)

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What are the factors for 100


1, 2, 4, 5, 10, 20, 25, 50,100
Prime decomposition of 100
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What is prime factorization?


"Prime Factorization" is finding which prime numbers multiply together to make the original number.


Example : What are the prime factors of 12 ? It is best to start working from the smallest prime number, which is 2, so let's check:

12 ÷ 2 = 6
Yes, it divided evenly by 2. We have taken the first step!
But 6 is not a prime number, so we need to go further. Let's try 2 again:

6 ÷ 2 = 3
Yes, that worked also. And 3 is a prime number, so we have the answer:

12 = 2 × 2 × 3

As you can see, every factor is a prime number, so the answer must be right.

Note: 12 = 2 × 2 × 3 can also be written using exponents as 12 = 22 × 3

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#include<stdio.h>
#include<conio.h>
void main()
{
int num,i=1,j,k;
clrscr();
printf("\nEnter a number:");
scanf("%d",&num);
while(i<=num)
{
k=0;
if(num%i==0)
{
j=1;
while(j<=i)
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}
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}
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}
getch();
}

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