Trial question (1 Viewer)

[freaking_out]

hey, i just finished my 4u independent trial paper...i was wondering how people did this question:

if (a+b+c) is divisible by 3 then prove that the number "abc" (not product, but actually the number abc, c is the unit, b the tens etc..) is also divisible by 3.

i remember in maths comp. this is a trick (i.e to add the digits to c if its divisbile), but never knew the proof.


also that crappy circular geometry question- i just left it-too hard for m:mad1:

 

[ND]

a+b+c=3k (where k is an integer)
100a+10b+c = 3k+99a+9b
= 3(k+33a+3b)

Which question was that in?

 

[freaking_out]

Originally posted by ND
a+b+c=3k (where k is an integer)
100a+10b+c = 3k+99a+9b
= 3(k+33a+3b)

Which question was that in?

faarr in!!!!!! that looks so easy now!!!!!!!!!!! i can't believe it that i left this question!!!

anyway, yeah that was in my test and we do the independent.

 

[rx72c]

Hold on i dont get it, how did you show that abc is also divisible by 3
Can someone explain
IM LOst

 

[ND]

Well the number abc can be expressed in teh form 100a+10b+c, and because:
100a+10b+c = 3(k+33a+3b)
(100a+10b+c)/3 = k+33a+3b, which is a whole number.

 

[rx72c]

ok may be im dumb or somethin
could you just outline how abc can be expressed as 100a +10b +c

 

[ND]

Ok i'll give you a specific example:

Consider the number 24. Now 24 is obviously divisible by 3, let's express 24 as 6+9+9. Now a+b+c is to abc what 6+9+9 is to 699, and 100*6 + 10*9 + 9 = 699.

edit: i don't know why i included that stuff about 24, forget that bit.