27 Jan 10
Originally posted by Palynkaspooky. it's like that thing where, if you add up the individual numerals in a number, and the sum is divisible by three, then the number itself is divisible by three.
BA = 2+1=3
GB = 7+2=9
VR = 22+18=40
Cross multiplying we would get that GB*VR = BA*? <=> ? = 9*40/3=120
An exact 120. Coincidence?
e.g. 1569: 1+5+6+9=21, which is divisible by three, and 1569 is divisible by three (1569/3=523).
on the other hand, 1733: 1+7+3+3=14, which is not divisible by three, and 1733 is not divisible by three (1733/3=577.66 repeating).
is there a proof of this?
27 Jan 10
Originally posted by BlackampThere should be a proof, since by this way you estimate the divisibility by three if you check a number if it is prime
spooky. it's like that thing where, if you add up the individual numerals in a number, and the sum is divisible by three, then the number itself is divisible by three.
e.g. 1569: 1+5+6+9=21, which is divisible by three, and 1569 is divisible by three (1569/3=523).
on the other hand, 1733: 1+7+3+3=14, which is not divisible by three, and 1733 is not divisible by three (1733/3=577.66 repeating).
is there a proof of this?
27 Jan 10
1 edit
Originally posted by BlackampIt comes from the decimal notation. The two digit number "xy" = 10x + y = 9x+(x+y). You know that the first element (9x) is divisible by 3 (=3x) and the second one is exactly the sum of its digits!
is there a proof of this?
So it will only be divisible by 3 if the sum of its digits is divisible by 3. You can easily extend this to numbers with more digits (100z = 99z+z, etc.)