varoom
Oct 8th, 2005, 11:48 am
Reading another post, I got to thinking about these last night and thought I'd post them here (either that or go downstairs and have my wife put to me work!). The whole idea was to be able to solve math problems (fast) w/o a calculator (okay, okay! w/o a sliderule :eek: ). These are "divisibilty rules" - some are obvious (like 2) but pretty cool stuff...
Divisibility rules (base 10)
A number is divisible by 2 if the last digit is even.
A number is divisible by 3 if the sum of the digits is divible by 3 (465 = 4+6+5 = 15; 15/3 = 5 therefore 465/3? is true.
A number is divisible by 4 if the last two digits are divisible by 4 (728; 28 is divisible by 4)
5: if the last digit is either a 0 or 5
6: if the number is divible by both 2 and 3.
7: this one I had to play around with before I could recall it:
a) double the units digit
b) subtract that from the remaining digits
c) if the difference is divisible by 7 then original number is divisible by 7. If the result is not readily apparent - continue steps a and b until number is/not divisible by 7. For example: 2884
288 - (2*4) = 280, okay bad choice (to easy) 280 = 7 * 40.
Trying 3562: 356 - 4 = 352 (nope 50 * 7 is 350, so not divisible).
8: if last 3 digits are divisible by 8 (ie. 1064 or 2160)
9: if sum of the digits is divisble by 9 (64,593 is 6+4+5+9+3= 27 so 64,593 is divisible by 9.
10: if the last digit of the number is a 0.
11: another *strange* one! if the sum of the odd digits (by place in the number) subtracted by the even number digits is divisible by 11 :eek: :eek: :eek: ! 49731 is (4 + 7 + 1) - (9 + 3) = 12 - 12 = 0 (0 evaluates to true) so true. 135916 is (3 + 9 + 6) - (1 + 5 + 1) = 18 - 7 = 11, so true.
That's all I remember but maybe 12 is true if # is divisible by 3 and 4. Hmmm, 144 fits. 156 works. 324? yep. Just adding 12 in my head going up to 408 - each one of those worked for 3 and 4 :).
13? Forget about it!
Divisibility rules (base 10)
A number is divisible by 2 if the last digit is even.
A number is divisible by 3 if the sum of the digits is divible by 3 (465 = 4+6+5 = 15; 15/3 = 5 therefore 465/3? is true.
A number is divisible by 4 if the last two digits are divisible by 4 (728; 28 is divisible by 4)
5: if the last digit is either a 0 or 5
6: if the number is divible by both 2 and 3.
7: this one I had to play around with before I could recall it:
a) double the units digit
b) subtract that from the remaining digits
c) if the difference is divisible by 7 then original number is divisible by 7. If the result is not readily apparent - continue steps a and b until number is/not divisible by 7. For example: 2884
288 - (2*4) = 280, okay bad choice (to easy) 280 = 7 * 40.
Trying 3562: 356 - 4 = 352 (nope 50 * 7 is 350, so not divisible).
8: if last 3 digits are divisible by 8 (ie. 1064 or 2160)
9: if sum of the digits is divisble by 9 (64,593 is 6+4+5+9+3= 27 so 64,593 is divisible by 9.
10: if the last digit of the number is a 0.
11: another *strange* one! if the sum of the odd digits (by place in the number) subtracted by the even number digits is divisible by 11 :eek: :eek: :eek: ! 49731 is (4 + 7 + 1) - (9 + 3) = 12 - 12 = 0 (0 evaluates to true) so true. 135916 is (3 + 9 + 6) - (1 + 5 + 1) = 18 - 7 = 11, so true.
That's all I remember but maybe 12 is true if # is divisible by 3 and 4. Hmmm, 144 fits. 156 works. 324? yep. Just adding 12 in my head going up to 408 - each one of those worked for 3 and 4 :).
13? Forget about it!