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Number theory Difficulty 3.2 AMC 10/12 Prove it Canada

IMG0 There are $M$ integers between $10\,000$ and $100\,000$ that are multiples of $21$ and whose units (ones) digit is $1$ . What is the value of $M$ ? There are $N$ students who attend Strickland S.S., where $500 < N < 600$ . Among these $N$ students, $\frac{2}{5}$ are in the physics club and $\frac{1}{4}$ are in the math club. In the physics club, there are $2$
times as many students who are not in the math club as there are
students who are in the math club. Determine the number of students who
are not in either club.

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Solution

Suppose that the length of the track is $2L \text{ m}$ , that Arun’s constant speed is $a\text{ m/s}$ , and that Bella’s constant speed is $\$b\text{}$
m/s}. When Arun and Bella run over the same interval of time, the ratio of the distances that they run is equal to the ratio of their speeds. Consider the interval of time from the start to when they first meet. In the diagram,

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Figure for this problem A $is Arun’s starting point,$ B $is Bella’s starting point, and$ P is this first meeting point. [[IMAGE0]] Since Arun has run

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Figure for this problem $100\text{}$ m} and together they have covered half of the length of the track, then Bella has run

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Figure for this problem (L - $100)\text{}$
m} $. Thus,$ $\dfrac{a}{b} = \dfrac{100}{L-100}$ . From their first meeting point

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Figure for this problem P to their second meeting point, which we label

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Figure for this problem Q $, Bella runs$ $150\text{}$ m}. [[IMAGE1]] Over this time, Arun runs from

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Figure for this problem P $to$ B $to$ Q $. Since Bella runs$ 150 - 100 = $50\text{}$
m} $past$ A $, then$ QB = (L - $50)\text{}$ m} $(because$ AB = $L\text{}$ m} $and$ AQ = $50\text{}$ m} $) and so Arun runs$ (L - $100)\text{}$ m} + (L - $50)\text{}$ m} $which is equal to$ (2L - $150)\text{}$
m}. Thus, over this second interval of time,

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Figure for this problem $\dfrac{a}{b} = \dfrac{2L-150}{150}$ $. Equating expressions for$ $\dfrac{a}{b}$ $and solving,$ $\text{100 L-100 = 2L-150 150 100 150 = (L-100)(2L-150) 15 ,000 = 2L 2 - 350L + 15 ,000 350L = 2L 2}$ $Since$ L $\neq$ 0 $, then$ 2L = 350, and so the total length of the track is

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Figure for this problem $350\text{}$ m}. Checking, if the length of the track is

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Figure for this problem $350\text{}$ m}, then half of the length is

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Figure for this problem $75\text{}$ m}. This means that from the start to

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Figure for this problem P $, Arun runs$ $100\text{}$ m} $and Bella runs$ $75\text{}$ m} $. Also, from$ P $to$ Q $, Bella runs$ $150\text{}$ m} $and Arun runs$ $200\text{}$ m} $. Note that$ $\dfrac{100}{75} = \dfrac{200}{150}$ so these numbers are consistent with the given information. Using exponent laws, the following equations are equivalent:

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Figure for this problem $\text{4 1 + 3 = 2 2 - 8 2 (2 2) 1 + 3 = 2 2 - (2 3) 2 2 2 + 2 3 = 2 2 - 2 3 2 2 2 + 2 3 = 2 2 - + 3 2 2 + 2 3 = 2 - + 3 2 2 3 - 3 2 + = 0 (2 2 - 3 + 1) = 0 (2 - 1)( - 1) = 0}$ $and so$ $\cos \theta$ = 0 $or$ $\cos \theta$ = 1 $or$ $\cos \theta = \frac{1}{2}$ $. Since$ $0\degree \leq \theta \leq 360\degree$ $, the solutions are$ $\theta = 90\degree, 270\degree, 0\degree, 360\degree, 60\degree, 300\degree\$.$

Listing these in increasing order, the solutions to the original
equation are $\theta = 0\degree, 60\degree, 90\degree, 270\degree, 300\degree, 360\degree$

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.