practice problems pending

Q1. Determine all solutions in the positive integers of 18x+5y=48

S1.

y = (48-18x)/5

48-18x = 0 mod 5

3 = 3x mod 5

x = 1 mod 5

x = 1 => y = 6

x >= 6 not possible else y will become negative.

Q2. 5x+3y=52.
S2.

x = (52 - 3y)/5 => 2 = 3y mod 5 => y = 4 mod 5
y = 4,9,14 => x = 8, 5, 2

Q3. Find all residues r with 0≤r≤7 such that x^2 ≡r(mod8) has a solution.
S3.
0,1,4

Q5.

A single bench section at a school event can hold either 7 adults or 11 children. When N bench sections are connected end to end, an equal number of adults and children together will occupy all the bench space. What is the least possible positive integer value of N?
S5.

a + c = N
7a = 11c
7,11 are primes => smallest (a,c) = (11,7) => smallest N = 18.|


Q6.
Find the general solutions of 123x+360y=99
S6.

41x = 33 - 120y
33 = 120y mod 41
-8 = 38y  = -3y mod 41
8 = 3y mod 41
y = 14 => RHS = 1
so y = 14*8 => RHS = 3 * 112 = 3 * 30 mod 41
=> y = 30 mod 41
=> 41x = 33 - 120(30 + 41k) = 33-120*30 - 120.41.k
=> x = (33 - 3600)/41 - 120k = -87 - 120k
so these are the general solutions.

Q7.

When Mr. Smith cashed a check at his bank, the teller mistook the number of cents for the number of dollars and vice versa. Unaware of this, Mr. Smith spent 68 cents and then noticed to his surprise that he had twice the amount of the original check. Determine the smallest value for which the check could have been written.

S7.
o = 100D + C
100C + D - 68 = 2.o = 200D + 2C
98C = 199D + 68
C = (199D + 68)/98
199D + 68 = 0 mod 98
3D = -68 mod 98
3D = 30 mod 98
D = 10 mod 98
C = (199*10 + 58)/68 = 21
Answer = 10 dollars 21 cents

Q8.
A shop packs rice into boxes that hold either 4 kg or 7 kg. Using only whole boxes, the shop wants to pack exactly 100 kg using as few boxes as possible. How many boxes of each size should be used, and what is the minimum total number of boxes?

S8.
4x + 7y = 100
y = (100-4x)/7
100 = 4x mod 7 => 2 = 4x mod 7 => x = 4 mod 7
For total minimum boxes, use min 4kg boxes.
x = 4 => y = 12
total min = 16

Q9. All positive solutions of 158x - 57y = 7
S9.

y = (158x - 7)/57
158x = 7 mod 57 = 44x mod 57
44x = 7 mod 57
-13x = 7 mod 57
13x = 50 mod 57

Now let's make some trials:
13.4 = -5 mod 57
13.4.10 = -50 mod 57 = 7 mod 57

So -13x = 13.40 mod 57 => x = -40 mod 57 = 17 mod 57

x = 57t + 17 for t >= 0
y = (158*(57t + 17) - 7)/57 = 158t + (158*17 - 7)/57 = 158t + 47 for t>= 0

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