course Mth 151
april 18 around 11:50
assignment 26-QUERYq5.4.12:
x mod 10 is the remainder when x is divided by 10.
So [(10+7)*(5+3)] mod 10=(17*8) mod 10=136 mod 10=6, since 136/10 leaves remainder 6.
*2
q5.4.20
60/20 = 3 because 3*20=60. if 2/3=x, then 3x=2. It turns out that 3*4=2, so it follows that 2/3=4
*2
q5.4.42:
(3-27) mod 5=-24 mod 5.
You would go all the way around around backwards 5 complete times to get -25, then move forward 1 to get to -24. That would put you at 1 on the actual clock.
*2
q5.4.20:
7 (mod 4) is 3.
Since (5x-3) mod4=7 mod 4,(5x-3) mod 4 must be 3.
For x=1,2,3,4,..., the expression 5x-3 takes values 2,7,12,17,22,27,32,37,....
These numbers, when divided by 4, give remainders 2,3,0,1,2,3,0,1,2,3,0,1,2,3,0,....
Thus every fourth number, mod 4, is equal to 3.
This starts with the second number, which occurs when x = 2.
Every fourth number, starting with 2, gives us the sequence 2,6,10,14, ...
2 is the first solution, 4 is the difference between solutions.
Thus x can be any element in the set {2,6,10,14,...}.
The general term of this sequence is 2+4n. So we can also say that x=2+4n, where n=0,1,2,3,...
*2
q5.4.30:
the table would look like this:
0123456
00123456
11234560
22345610
33456012
44560123
55601234
66012345
and the properties would be identity, associative, communiative, and is closed.
*2
q5.4.33:
the table would look like this:
0123
00000
10123
20202
30321
and the properties would be identity, communative, and associative. it is also closed
*2
q5.4.70:
The calculation is 2002+[[2002-1/4]]-[[2002-1/100]]+[[2002-1/400]].
[[Q]] means the greatest integer contained in Q.
So we get 2002+[[500.25]]-[[20.01]]+[[5.0025]]=2002+500-20+5=2487.
Now 2487 mod 7 is 2.
Sunday is 0, Monday is 1 so Tuesday is 2
*2
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