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When the positive integers w and n are divided by 6, the remainders are 3 and 2, respectively. What is the remainder when the product wn is divided by 6?
The remainder should be 0 but how can I show this? Do we need the division algorithm for this to help prove it? If so, I'm not sure how to do this by proof (I know it works by taking examples substituting positive integers)
k and m are (+) integers ...
w = 6k + 3
n = 6m + 2
wn = (6k + 3)(6m + 2) = 36km + 18m + 12k + 6 = 6(6km + 3m + 2k + 1)
so ... what does that last expression tell you?
Ahh right! Thus, wn is divisible by 6 (remainder is 0) since we can let "q" be the quotient of the stuff in parenthesis? Is that right?
When the positive integer S is divided by 12, the remainder is 4. When the positive integer T is divided by 12, the remainder is 5. What is the remainder when the product ST is divided by 6?
I'm stuck at the bottom, how can I show the remainder is 2?
12k + 4 = S
12u + 5 = T
ST = (12k+4)(12u+5)
= 144ku+48u+60k+20
ST/6 = (144ku+48u+60k+20)/6 = 24ku+8u+10k+10/3
ST = 144ku+48u+60k+20 = (144ku+48u+60k+18)+2 = 6(24ku+8u+10k+3) + 2Quote:
Originally Posted by donnagirl
Ahh that's clever algebra skeeter, I wouldn't have seen it but I do now :)
S = 4 mod 12
T = 5 mod 12
ST = 20 mod 12 = 8 mod 12
this means ST = 12k + 8
thus ST/6 = 12k/6 + 8/6 = 2k + 1 + 2/6
so ST = 6(2k+1) + 2 that is:
ST = 2 (mod 6)
(what can i say? i'm lazy)