Divisible by 3?
As I haven't seen it be posted before: prove that if an integer is divisible by 3, that the sum of it's digits are as well.
Also show that the remainder after division by 3 of the sum of the digits is the same as the remainder after division by 3 of the original integer.
Also show that the remainder after division by 3 of the sum of the digits is the same as the remainder after division by 3 of the original integer.
Labels: mathemagic
posted by Chris at
1:58 PM
4 Comments:
Divisible by 3
Rule1) A*B mod C =(A mod C * B mod C) mod C
Rule2) (A+B) mod C = A mod C + B mod C
10 mod 3=1 thus from Rule 1 , 10^n mod 3 =1
so the mod of a*10^n mod 3 =a mod 3 e.g. for 50, 50 mod 3 = 5 mod 3 =2, 50/3=16R2
and from Rule 2 for the mod 3 of a n digit number,(call coefficients a,b,c,d for simplicity) it is (a*1)mod 3 +(b*10) mod 3 + (c*100)mod 3 +......(d*10^n-1)mod 3 , which will simply be the sum of its coefficients (a+b+c+....d)
e.g. for 200,097 the sum is 2+9+7=18, 18 mod 3=0 , 200097/3=66699
Thus the sum of its coefficients must be divisible by 3 for the number to be divisible by 3.
Hopefully this is a clear explanation.
Cam
Hi Cam, With practice, you could get to be quite good at this stuff ;)
You didn't answer the second part; which is the part I find more interesting.
I thought the second part was implied by showing that the mod 3 of the number is = to the mod 3 of the sum of digits....
equal mods means equal remainders
Cam
Hi Cam. You're right, sorry. I wasn't paying close enough attention. Prolly because the example you gave had a remainder of 0.
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