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Mathieu d.
Excellent (grades/ SAT scores), tutored independently since high school, mentor for linear algebra
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Linear Algebra
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Question:

You have $$11$$ coins, totaling $$44$$ cents. You know that each coin is either a penny ($$1$$ cent), nickel ($$5$$ cents), or a dime ($$10$$ cents). How many of each type of coin do you have?

Mathieu d.

Let's have $$x_1$$ represent the number of dimes, $$x_2$$ represent the number of nickels, and $$x_3$$ represent the number of pennies. We know that in total we have $$11$$ coins, so we can write: $(x_1 + x_2 + x_3 = 11$) We also know that if we sum up the values of the coins, the total will be $$44$$ cents. We can express this with $(10x_1 + 5x_2 + x_3 = 44$) because each $$x_1$$ is worth $$10$$ cents, each $$x_2$$ is worth 5 cents, and each $$x_3$$ is worth 1 cent. We have now expressed this problem as a system of equations: $(x_1 + x_2 + x_3 = 11 \\ 10x_1 + 5x_2 + x_3 = 44$) This system of equations can be represented as an augmented matrix: $( \left( \begin{array}{ccc|c}1 & 1 & 1 & 11 \\10 & 5 & 1 & 44 \end{array} \right)$) We can use Gaussian elimination to find the reduced row echelon form of the matrix, which is $( \left( \begin{array}{ccc|c}1 & 0 & -\frac{4}{5} & -\frac{11}{5} \\0 & 1 & \frac{9}{5} & \frac{66}{5} \end{array} \right)$) Turning this matrix back into separate equations, we can substitute the free variable $$t$$ in for $$x_3$$ to get (\begin{aligned} x_1 &= \frac{4}{5}t - \frac{11}{5} \\ x_2 &= -\frac{9}{5}t + \frac{66}{5} \\ x_3 &= t \end{aligned}) This appears to give infinitely many solutions (note that none of these equations give any restrictions on the value of $$t$$). However, there are two constraints we need to account for. First, we need to make sure that $$x_1$$, $$x_2$$, and $$x_3$$ are integers. This is because they represent our unknown numbers of coins, and it wouldn't make sense to have a fraction of a coin. Second, we need to make sure that $$x_1$$, $$x_2$$, and $$x_3$$ are at least $$0$$ (because we can't have a negative number of coins) and at most $$11$$ (because we only have $$11$$ coins in total). These two constraints impose some restrictions on $$t$$. Namely, since we have $$x_3 = t$$, we know that $$t$$ must be an integer between $$0$$ and $$11$$. We can try every possible value of $$t$$, from $$0$$ to $$11$$, in those three equations above. The only value that gives integer values between $$0$$ and $$11$$ for $$x_1, x_2, x_3$$, is $$t=4$$. When $$t=4$$ we get $$x_1=1, x_2=6, x_3=4$$. So we must have $$1$$ dime, $$6$$ nickels, and $$4$$ pennies.

Number Theory
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Question:

What is the remainder of $$7^{200127}$$ when divided by $$12$$?

Mathieu d.

The problem above is equivalent to solving $$7^{200127} \mod 12$$. We can rewrite this as $$7^{200124}*7^{3} \mod 12$$, and then again as $$(7^{50031})^4*7^3 \mod 12$$. $$(7, 12)$$ = 1, thus $$(7^{50031}, 12) = 1$$, and $$\phi(12) = 4$$, so we can use Euler's theorem. With a straightforward application of Euler's theorem, our equation simplifies to $$7^3 \mod 12$$. From here, there are a number of ways of completing the problem, including just using long division to find the remainder of $$7^3/12$$. We can also note that $$7^3 = 7^2*7$$, where $$7^2 \equiv 1 \mod 12$$. So our equation can be simplified to $$1*7 \mod 12$$. Therefore, the remainder of $$7^{200127}$$ when divided by $$12$$ is $$7$$.

Calculus
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Question:

$(lim_{n \rightarrow \infty}\sum_{i=1}^n\frac{\pi sin(\frac{\pi i}{n})}{n}$)

Mathieu d.

First, we can rearrange a few terms to get a better idea of what we're looking at. We rewrite the equation as: $(lim_{n \rightarrow \infty}[\frac{\pi}{n}*\sum_{i=1}^n sin(\frac{\pi i}{n})]$) Let's look at this equation intuitively. Ignoring the limit for now, we can see that we are summing $$sin(\frac{\pi}{n})*\frac{\pi}{n} + sin(\frac{2\pi}{n})*\frac{\pi}{n} + \cdots + sin(\frac{n\pi}{n})*\frac{\pi}{n}$$. On a graph of $$sin(x)$$, this sum could be found by drawing a horizontal line from $$x=0$$ to $$x=\pi/n$$ which hits the graph of $$sin(x)$$ at $$x=\pi/n$$. Note that we have created a rectangle with width $$\pi/n$$, and height $$sin(\frac{\pi}{n})$$. Continue this procedure by drawing a horizontal line from $$x=\pi/n$$ to $$x=2pi/n$$ which hits the graph of $$sin(x)$$ at $$x=2\pi/n$$, thus creating a rectangle with with width $$\pi/n$$, and height $$sin(\frac{2\pi}{n})$$. Thus, if we continue this pattern until $$x=\pi$$, we will have created a right-hand estimate of the area under the curve $$sin(x)$$, as $$x$$ goes from $$0$$ to $$\pi$$. If we now consider the limit, we see that we are taking more and more precise estimates as $$n \rightarrow \infty$$. We now see that our sum is actually the definition of the riemann integral: $(\int_0^\pi sin(x) dx = [-cos(x)]_0^\pi = -cos(\pi) - cos(0) =2$)

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