Techniques for Factoring Numbers in Your Head
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Re: Techniques for Factoring Numbers in Your Head
#2Re: Techniques for Factoring Numbers in Your Head
#3100 is divisible by 2. 2x50=100
No need to wonder if 51-99 are factors. They aren't. 50 is your largest factor.
Re: Techniques for Factoring Numbers in Your Head
#4Tip: you only need to factor it to the midpoint. Factors occur in pairs. Identifying the smallest factors will automatically identify the largest. 100 is divisible by 2. 2x50=100 No need to wonder if 51-99 are factors. They aren't. 50 is your largest factor.
Re: Techniques for Factoring Numbers in Your Head
#5Tip: you only need to factor it to the midpoint. Factors occur in pairs. Identifying the smallest factors will automatically identify the largest. 100 is divisible by 2. 2x50=100 No need to wonder if 51-99 are factors. They aren't. 50 is your largest factor.
Indeed - you can even extend this idea to only identify factors less than or equal to the square root - all factors greater than this have a matching factor less than the square root.
50 = 2x5x5 So 100 = 2x2x5x5
You can build your other factors of 4,10, 20 and 25 by multiplying the subsets.
Re: Techniques for Factoring Numbers in Your Head
#6Re: Techniques for Factoring Numbers in Your Head
#7Doerfler's _ Dead Reckoning: Calculating Without Instruments _ (pages 47-73) deals extensively with this issue. https://archive.org/stream/deadreckoningcal00doer#page/46/mo...
Re: Techniques for Factoring Numbers in Your Head
#8If b evenly divides a×r_i, where r_i is the current value, divide by b. If not, add or subtract p until the result is evenly divisible by b, then divide by b.
And step 6 should loop back to step 4 and define r_i to be the "result".
This method is very interesting and systematic, but it seems pretty complicated compared to ad hoc reasoning. E.g. 34 is a multiple of 17, so 680 must be too; 986-680 = 306, which is 34 away from 340, a multiple of 17, so 17 is a factor of 986.
Re: Techniques for Factoring Numbers in Your Head
#9The division by subtraction is fairly well known, of course, but it reminds me of the surprising square root by subtraction that I only learned about recently. It's probably too much to do mentally, but easy with pencil & paper.
To compute the digits of the square root of n, set a = 5n, b = 5. Repeat: if a≥b, set a=a-b, add 10 to b. Then if aI had to code it up myself just to believe it would work. http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec02.pdf