Question 7036: Can someone help me with these I just dont understand.
without tables perform the following operations and reduce answer to a whole number less than the modulus.
1. 3+2 (mod5)
2. 15-2(mod5)
3. (3+2)+5(mod6)
4. (12)(3) (mod5)
5. 2(4+11)(mod7)
6. 4cubed(mod12)
7. (624)(589)(mod9)
Thank you all so much!!
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! m = r (mod n) means m =nq+r (mod n) where 0<=r < n. q as quotient, r as
remainders. And the mod prerseves addition & multiplication.
1. 3+2 (mod 5) = 5 (mod 5) = 0 (mod 5)
2. 15-2(mod 5) = 13 (mod 5) = 3 (mod 5)
3. (3+2)+5(mod 6) = 10 (mod 6) = 4 (mod 6)
4. (12)(3) (mod 5) = 12 (mod 5) * 3 (mod 5) = 2 (mod 5) * 3 (mod 5)
= 6 (mod 5)= 1 (mod 5).
5. 2(4+11)(mod7) = 30 (mod 7) = 2 (mod 7)
6. 4cubed(mod12) [Never type cubed or squared any more]
= 4^3 (mod 12) = 16* 4 (mod 12) = 16 (mod 12) * 4 (mod 12)
= 4 (mod 12) * 4 (mod 12) = 16 (mod 12) = 4 (mod 12)
7. (624)(589)(mod 9)
= 624 (mod 9) * 589 (mod 9)
= (69* 9 + 3) (mod 9) * (65* 9 + 4) (mod 9)
= 3 (mod 9) * 4 (mod 9)
= 12 (mod 9)
= 3 (mod 9)
Try to use your eyes to calculate small numbers.
No calculators , even for the big numbers as (7):
by reducing the two big numbers separately, you can get the answer right
away.
If some steps you don't understand, use your brains to think about it.
Don't suppose mod is hard.
Kenny
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