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What method do you like best for gcf why?
This answer is in the context of numbers that are likely to crop up in school or graduate mathematics, not in cryptology or advanced research. Complete factorisation of large numbers is tedious and may require a lot of false starts (if the next factor is quite large). So, for large numbers I would start with Euclid's algorithm (see link). All that is required for this process is subtraction. The problem of finding the GCF of two large numbers soon becomes that of finding the GCF of two very much smaller numbers. Although the method can be carried out to the end, I will often choose to switch to factorisation when the numbers are smaller: below 400, say, when all you need to know is the times tables to 20. Given all the divisibility rules, this is not as daunting as it may sound.
What is the GCF of the numbers 125 and 50?
The GCF is 25.
What is the GCF of the numbers 141 and 192?
The GCF is 3.
What is the GCF of the numbers 40 and 64?
The GCF is 8.
Is this true or false ' if 2 numbers are each divisible by another number then their difference is also divisible by that number'?
True, and this property is useful for finding the greatest common factor (GCF) of two (or more) large numbers.If A > B, then GCF(A , B) = GCF(A - B, B) where A - B is smaller than A.Repeat, each time subtracting the smaller number from the bigger.Keep going until both numbers in the parentheses are the same: that number is the GCF of A and B.GCF by subtraction rather than factorising or division. Unfortunately, it can be quite slow. You could speed it up by doing A - 2B or A - 3B etc rather than A - B.True, and this property is useful for finding the greatest common factor (GCF) of two (or more) large numbers.If A > B, then GCF(A , B) = GCF(A - B, B) where A - B is smaller than A.Repeat, each time subtracting the smaller number from the bigger.Keep going until both numbers in the parentheses are the same: that number is the GCF of A and B.GCF by subtraction rather than factorising or division. Unfortunately, it can be quite slow. You could speed it up by doing A - 2B or A - 3B etc rather than A - B.True, and this property is useful for finding the greatest common factor (GCF) of two (or more) large numbers.If A > B, then GCF(A , B) = GCF(A - B, B) where A - B is smaller than A.Repeat, each time subtracting the smaller number from the bigger.Keep going until both numbers in the parentheses are the same: that number is the GCF of A and B.GCF by subtraction rather than factorising or division. Unfortunately, it can be quite slow. You could speed it up by doing A - 2B or A - 3B etc rather than A - B.True, and this property is useful for finding the greatest common factor (GCF) of two (or more) large numbers.If A > B, then GCF(A , B) = GCF(A - B, B) where A - B is smaller than A.Repeat, each time subtracting the smaller number from the bigger.Keep going until both numbers in the parentheses are the same: that number is the GCF of A and B.GCF by subtraction rather than factorising or division. Unfortunately, it can be quite slow. You could speed it up by doing A - 2B or A - 3B etc rather than A - B.
