Numbers, Fractions, Decimals, and Percents

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These are the solutions to Mathematics questions on the topics: Numbers, Fractions, Decimals, and Percents.
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Percent Applications

$ \underline{\text{Change and Percent of Change}} \\[3ex] (1.)\:\: change = new - initial \\[5ex] (2.)\:\: \%\;\;of\;\;change = \dfrac{change}{initial} * 100 \\[5ex] (3.)\;\; \text{If A is p% more than B, then A = (100 + p)% of B} \\[3ex] (4.)\;\; \text{If A is p% less than B, then A = (100 – p)% of B} \\[5ex] \underline{\text{Percent:Proportion}} \\[3ex] (5.)\;\; \dfrac{is}{of} = \dfrac{\%}{100} \\[5ex] (6.)\;\;\underline{\text{Percent:Equation}} \\[3ex] is \rightarrow equal\;\;to \\[3ex] of\;\; \rightarrow multiply \\[3ex] what \;\; \rightarrow variable \\[5ex] \underline{\text{Wholesale and Retail}} \\[3ex] (7.)\:\: \text{Sale Price } = \text{Initial Price } - \text{Discount} \\[5ex] (8.)\:\: \%\;Discount = \dfrac{Discount}{\text{Initial Price}} * 100 \\[5ex] (9.)\;\; \text{Profit } = \text{Selling Price } - \text{Cost Price} \\[3ex] (10.)\;\; \%\;Profit = \dfrac{Profit}{\text{Cost Price}} * 100 \\[5ex] (11.)\;\; \text{Loss } = \text{Cost Price } - \text{Selling Price} \\[3ex] (12.)\;\; \%\;Loss = \dfrac{Loss}{\text{Cost Price}} * 100 $

(1.) Number Theory Suppose that X and Y are angles with $\tan X = \dfrac{1}{m}$ and $\tan Y = \dfrac{a}{n}$ for some positive integers a, m, and n.
Determine the number of positive integers $a \le 50$ for which there are exactly 6 pairs of positive integers (m, n) with $X + Y = 45^\circ$.

$ \underline{\text{Hint:}} \tan (\alpha \pm \beta) = \dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}...\text{Sum and Difference Formulas} \\[5ex] $

$ \text{Let us solve for }a \\[3ex] \tan (X + Y) = \dfrac{\tan X + \tan Y}{1 - \tan X \tan Y}...\text{Sum Formulas} \\[5ex] X + Y = 45^\circ \\[3ex] \tan X = \dfrac{1}{m} \\[5ex] \tan Y = \dfrac{a}{n} \\[5ex] \tan 45^\circ = \left(\dfrac{1}{m} + \dfrac{a}{n}\right) \div \left[1 - \left(\dfrac{1}{m}\right) * \left(\dfrac{a}{n}\right)\right] \\[5ex] 1 = \left(\dfrac{n + am}{mn}\right) \div \left(1 - \dfrac{a}{mn}\right) \\[5ex] 1 = \left(\dfrac{n + am}{mn}\right) \div \left(\dfrac{mn - a}{mn}\right) \\[5ex] 1 = \left(\dfrac{n + am}{mn}\right) * \left(\dfrac{mn}{mn - a}\right) \\[5ex] 1 = \dfrac{n + am}{mn - a} \\[5ex] n + am = mn - a \\[3ex] am + a = mn - n \\[3ex] a(m + 1) = n(m - 1) \\[3ex] a = \dfrac{n(m - 1)}{m + 1} ...eqn.(1) \\[5ex] \implies \\[3ex] \dfrac{n(m - 1)}{m + 1} \le 50 \\[5ex] $ Solving this inequality: two variables in one inequality is time-consuming...we do not know the value of either m or n
Even though we know that a, m, and n are positive integers, the guesses will take time.
So, let's see whether we can find a theorem that works in Number Theory to help us solve the question.

$ \text{From:} \\[3ex] n + am = mn - a \\[3ex] a = mn - am - n \\[3ex] a = m(n - a) - n \\[3ex] \text{To enable us to get a product of factors: add a to both sides} \\[3ex] a + a = m(n - a) - n + a \\[3ex] 2a = m(n - a) - 1(n - a) \\[3ex] 2a = (n - a)(m - 1) \\[3ex] $