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Algebra Difficulty 3.2 AMC 10/12 Prove it Canada

IMG0 In the diagram, rectangle $ABCD$ is divided into four smaller rectangles by the lines $PQ$ and $RS$ , which intersect at $X$ .The areas of these smaller rectangles are, in some order, $2$ , $6$ , $3$ , and $a$ . What are the three possible values of $a$ ? Suppose that the parabola with equation $y = x^2 - 4tx + 5t^2 - 6t$ has two distinct $x$ -intercepts. Determine the value of $t$ for which the distance between these $x$ -intercepts
is as large as possible.

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Solution

Every multiple of $21$ is of the form $21k$ for some integer $k$ . For such a multiple to be between 10 000 and 100 000, we need $10\,000 < 21k < 100\,000$ or $\frac{10\,000}{21} < k < \frac{100\,000}{21}$ .

Since $\frac{10\,000}{21} \approx 476.2$ and $\frac{100\,000}{21} \approx 4761.9$ and $k$ is an integer, then $477 \leq k \leq 4761$ . (Note that $k$ is greater than $476.2$ and is an integer, so must be at least $477$ ; similarly, $k$ is at most $4761$ .) We also want the units digit of $21k$ to be $1$ . This means that the units digit of $k$ itself is $1$ , since the units digit of the product of $21$ and $k$ is equal to units digit of $k$ because the units digit of $21$ is $1$ . Therefore, the possible values of $k$ are $481, 491, 501, $\ldots,$ 4751,
4761 $. There are$ 429 such values. To see this, we can see that counting the integers in this list is the same as counting the integers in the list

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Figure for this problem 48, 49,
50, $\ldots,$ 475, 476. This list is equivalent to removing the integers from

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Figure for this problem 1 $to$ 47 from the list of integers from

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Figure for this problem 1 $to$ 476 $, giving$ 476 - 47 = 429 $integers. Thus,$ M = 429 $. Solution 1: We can partition the$ N $students at Strickland S.S. into four groups:$ a students who are in the physics club and are in the math club

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Figure for this problem b students who are in the physics club and are not in the math club

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Figure for this problem c students who are not in the physics club but are in the math club

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Figure for this problem d students who are not in the physics club and are not in the math club In Math Club Not in Math Club In Physics Club

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Figure for this problem ab $Not in Physics Club$ cd From the given information, there are

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Figure for this problem $\frac{2}{5}N$ students in the physics club. In other words,

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Figure for this problem a+b =
$\frac{2}{5}N$ . Among the students in the physics club, twice as many are not in the math club as are in the math club. This means that

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Figure for this problem b = $\frac{2}{3} \cdot \frac{2}{5}N = \frac{4}{15}N$ $and$ a = $\frac{1}{3}\cdot \frac{2}{5}N = \frac{2}{15}N\$.$

From the given information, there are $\frac{1}{4}N$ students in the math club. In other words, $a + c =
$\frac{1}{4}N$ $. Since$ a = $\frac{2}{15}N$ $, then$ c = $\frac{1}{4}N - \frac{2}{15}N = \frac{7}{60}N\$.$

Since $a + b + c + d = N$ , then $d = N - a - b - c = N - \frac{4}{15}N - \frac{2}{15}N - \frac{7}{60}N = \frac{29}{60}N$ .

Lastly, we know that $500 < N < 600$ .

Since each of $a$ , $b$ , $c$ , and $d$ is an integer, then $N$ must be divisible by 60. Therefore, $N = 540$ and so the number of students not in either club is $d =
$\frac{29}{60} \cdot$ 540 = 261 $. Solution 2: Since there are$ N $students at Strickland S.S., then$ $\frac{2}{5}N$ are in the physics club and

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Figure for this problem $\frac{1}{4}N$ are in the math club. Since each of

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Figure for this problem $\frac{2}{5}N$ $and$ $\frac{1}{4}N$ $must be an integer, then$ N $must be divisible by$ 5 $and must be divisible by$ 4 $. Since$ 5 $and$ 4 share no common divisor larger than

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Figure for this problem 1 $, then$ N $must be divisible by$ 5 $\cdot$ 4 = 20 $. Thus, we let$ N = 20m for some positive integer

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Figure for this problem m $. In this case,$ $\frac{2}{5}N$ = 8m students are in the physics club and

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Figure for this problem $\frac{1}{4}N$ = 5m students are in the math club. Now, among the

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Figure for this problem 8m students in the physics club, twice as many are not in the math club as are in the math club. In other words,

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Figure for this problem $\frac{1}{3}$ $of the$ 8m students in the physics club are in the math club. This means that

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Figure for this problem m $must be divisible by$ 3 $, since$ 3 $is a prime number and$ 8 $is not divisible by$ 3 $. Therefore,$ m = 3k for some positive integer

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Figure for this problem k $, which means that$ N = 20m = 60k $and$ $\frac{2}{5}N$ = 8m = 24k $and$ $\frac{1}{4}N$ = 5m = 15k $. Since$ 500 < N < 600 $and$ N $is a multiple of 60, then$ N = 540 $, which means that$ k = 9. Thus, the number of students in the physics club is

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Figure for this problem 24k = 216 $, of whom$ $\frac{1}{3} \cdot$ 216 = 72 are in the math club and

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Figure for this problem $\frac{2}{3} \cdot$ 216=
144 are not in the math club. Also, the number of students in the math club is

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Figure for this problem 15k = 135. Finally, we know that there are

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Figure for this problem 540 $students at the school,$ 72 of whom are in both the physics club and the math club,

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Figure for this problem 144 of whom are in the physics club and not in the math club, and

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Figure for this problem 135 - 72 = 63 are in the math club and not in the physics club. Therefore, the number of students in neither club is

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Figure for this problem 540 - 72 - 144 - 63 = 261$.

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