A. Rounding and Truncation
A1-A2.
To determine if Student 1 will receive an A, we must first calculate the percentage. Divide the total points earned by the total points possible. THis gives us a solution of .895 or 89.5%. Next we round up to the nearest whole number. When rounding, we focus on the digit to the right of the digit we are trying to round. If it is 5 or greater, we round up by adding one. If it is less than 5, then we leave the digit alone. 89.5% rounds up to 90%. Student 1 will receive an A.
If the teacher were truncate the grade, then the student would not receive an A. He/she would only have a 89% because we drop all digits to the right of the digit to be truncated.
A3a.
A taxpayer would hope for the tax rate to be truncated because they would owe less taxes. You are basically erasing everything behind the decimal with truncation. Thus the resulting number will always be smaller. This is ideal especially whenever dealing money owed. In this case the taxpayer would have a tax rate of 27%.
A3b.
The government obviously prefers taxpayers to round, due to the possibility of a higher tax rate. If the digit to the right of the digit to be rounded is 5 or greater, then we add a 1 when rounding. So in this example the tax rate would increase to 28%. Let’s say the taxpayer makes $80,000 a year. The government would be due an extra $800. Now if you multiply this by several taxpayers, and it’s evident how much more the government will benefit from simply rounding up to the next percentage point.
In real world situations we use rounding and truncating to help us estimate and perform mental math. One example would be when shopping. I routinely estimate the cost of items by either rounding or subtracting. Let’s say I have four items with prices of $6.99, $1.50, $2.99, and $1.29. Rounding to the nearest dollar would give me a total of $13.00. Truncating would give me $10.00. In this case, rounding would give me a more accurate estimated total.
Outside of computer programming, there are not many situations where truncating would be preferable. But let’s say i’m throwing a party for my son. I have a package of 14 toy cars to share among the 8 guests as a party favor. 14/8 = 1.75. I would certainly truncate this to just 1 car per guest and keep the leftovers. Rounding and promising 2 cars would lead to some disappointed six year olds.

B. Primes and Composites:
B1-B3
The teacher would look for the greatest common divisor of the boys and girls to determine how to divide them into even groups.
To find the greatest common divisor we can list the factors of both groups. For the 20 boys that would be 1x20, 2x10, and 4x5. The 24 girls are 1x24, 2x12, 3x8, and 4x6. 4 is the largest number that can be placed in each group. Thus the 20 boys can form 5 groups with
4 boys per group. The 24 girls can form 6 groups with 4 girls per group. This would leave the class with a total of 11 groups.
B4.
Euclid’s proof, which states: Suppose that p1=2 < p2 = 3 < ... < pr are all of the primes. Let P = p1p2...pr+1 and let p be a prime dividing P; then p can not be any of p1, p2, ..., pr, otherwise p would divide the difference P-p1p2...pr=1, which is impossible. So this prime p is still another prime, and p1, p2, ..., pr would not be all of the primes. can be used to explain the infinitude of primes.
For me it is simpler to look at it like this:
If there was a finite number of primes, then there would exist some "biggest prime number". Call it .
Therefore, you could create a number that was the product of all the prime numbers.
Call it .
So .
Then .
If you were to divide by ANY of the prime numbers, then you would ALWAYS have a remainder of .
That means that either there is a prime number larger than which divides , or else is itself a prime number.
Either way, this contradicts our original statement that is the largest prime.
Therefore, there is an infinite number of prime numbers.

C. Modular Operations:
C1a
With Modular arithmetic we see cyclical behavior. This is most frequently encountered with the 12 hour standard clock. It takes 12 ticks to reset itself on it’s never ending cycle. Along with the operation of clock addition this forms a modulo 12 system.
So we could say 9am plus 8 hours of work equals 5pm, i.e. a ‘9to5’. We could also write this as 9 (mod 12) + 8 (mod 12)= 5 (mod 12). Using mental math, the easiest way to solve this problem was to use the remainder method. So 8+9=17/12=1 R5 The answer is 5. This method works just as easily using negative integers. (-3 +-5) mod 12 = 4. WE can view this as moving counter clockwise.
Multiplying positive or negative integers in modular multiplication is as simple as doing the normal arithmetic first then using the remainder method. For example (9x5)mod 12 =
9(mod 12). We divide the 45/12 and are left with a remainder of 9. Similarly, (-11x3)mod
12 = 3(mod 12). We divide -33/12 and are left with a remainder of -9.
C1b.
10(mod 6) = 4 (mod 6)
C1c.
Adding and multiplying in mod 7 would be like using a clock with 7 parts. It would consist of seven elements, 0 through 6. (6+6) (mod 7) = 5 (mod 7) and (6x6) (mod 7) = 1 (mod 7). Using negative integers, (-10+-2) (mod 7) = 2 (mod 7) and (-10x-2) (mod 7) = 6 (mod 7).

Resources:
Confused About Euclid’s Proof of Infinitude Prime Numbers. mathhelpforum.com