course Mth 151
april 18 around 11:30
assignment 26-QAq001:
3*9 (mod4) is the remainder when 3*9 is divided by 4. Since 3*9=27 and 27/4 leaves remainder 3, we see that 3*9(mod4)=3.
7*12(mod4) is the remainder when 7*12 is divided by 4. Since 7*12 =84 and 84/4 leaves remainder 0, we see that 7*12(mod4)=0.
11*13(mod4) is the remainder when 11*13 is divided by 4. Since 11* 13=143 and 143/4 leaves remainder 3,we see that 11*13(mod4)=3.
*2
q002:
the table would look like this:
mod 40123
00000
10123
20202
30321
*2
q003:
the table would look like this:
mod 51234
11234
22413
33142
44321
*2
q004:
3x+7=9(mod 4) means that 3x+7-9=0(mod 5) so 3x-2=0 (mod 5). If 3x-2=0(mod 5) then when we divide 3x-2 by 5 we should get remainder 0. So we substitute the different possible values for x into the expression 3x-2 until we get a number of which when divided by 5 gives us remainder 0.
If x=0 then 3x-2=-2, and-2(mod 5)=3
If x=1 then 3x-2=1, and 1(mod 5) =1.
If x=2 then 3x-2=4, and 4(mod 5)=4.
If x=3 then 3x-2=7, and 7(mod 5)=2.
If x=4 then 3x-2=10, and 10(mod 5)=0.
Thus x=4 is a solution to the equation 3x+7=9(mod 5).
*2
q005:
We recall that 3x + 7 = 9 (mod 5) is equivalent to 3x - 2 = 0 (mod 5). We evaluate 3x - 2 (mod 5) for x = 5, 6, 7, 8 and 9 and we find that the results are 3, 1, 4, 2, and 0. So x = 9 is our next solution.
We might also note that the series of results 3, 1, 4, 2, 0 is the same as the series we got for x = 0, 1, 2, 3, 4. Our results therefore seem to indicate a repeating pattern in which the remainder 0 occurs every fifth number starting with 4. This is in fact what happens, and you might wish to think about why this happens. However, you should in a case remember that this is what happens.
In general when we have an equation of the form A x + B = C (mod n), integer solutions happen at intervals of n. for some values of A, B and C integer solutions can also occur at shorter intervals, but they always do occur at intervals of n.
*1
q006:
the first five solutions are x=4,9,14,19 and 24
*2
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