Thread: modular arith

What is the multiplication of 67th and 68th term in the sequence 3, 4, 2, 1, 3, 4, 2, 1, 3, 4…......?

Is it possible to solve with mod arithmetic? So, sequence starts over every 4 numbers, so take mod 4. Then, 67 is equivalent to 3 (mod 4) and 68 is equivalent to 0 (mod 4). So we have 1(3)=3?

This seems off to me.

(nevermind this, didn't properly see the sequence). Yes, this is right in my opinion. It makes sense.

hmm. i think the real answer is 1(2)=2 though. Another way to do it would be to recognize that the 64th term will be 1 and so the 67th = 2 and 68th=1....

Originally Posted by sfspitfire23

hmm. i think the real answer is 1(2)=2 though. Another way to do it would be to recognize that the 64th term will be 1 and so the 67th = 2 and 68th=1....

Yes it is 2*1 = 2.

67 is congruent to 3 (mod 4) means, use the 3rd term, which is 2.
68 is congruent to 0 (mod 4) means, use the 4th term, which is 1.

There is no 0th term so you use the 4th term.