Problem 1 of 30
How may positive integers less than $1000$ have the property that the sum of the digits of each such number is divisible by $7$ and the number itself is divisible by $3$?Solution:
The largest integer less than $1000$ is $999$ hence the maximum sum of the digits is $9 + 9 + 9 = 27$.
Hence, the numbers that satisfy the given conditions can be one of $7$, $14$ and $21$.
Since the number itself is divisible by $3$, the sum of the digits must also be divisible by $3$.
Out of the $3$ cases above only $21$ is divisible by $3$.
Therefore for all numbers whose sum is $21$ will satisfy the given conditions.
The maximum sum of digits for a one-digit or two-digit number cannot be $21$.
Therefore, the number can only be three-digit numbers.
We can take an un-ordered set of $3$ digits adding up to $21$ and find the possible arrangements with those digits:
$9,\ 9,\ 3$ $...$ possible arrangements $= \dfrac{3!}{2!} = 3$
$9,\ 8,\ 4$ $...$ possible arrangements $= 3! = 6$
$9,\ 7,\ 5$ $...$ possible arrangements $= 3! = 6$
$9,\ 6,\ 6$ $...$ possible arrangements $= \dfrac{3!}{2!} = 3$
$8,\ 8,\ 5$ $...$ possible arrangements $= \dfrac{3!}{2!} = 3$
$8,\ 7,\ 6$ $...$ possible arrangements $= 3! = 6$
$7,\ 7,\ 7$ $...$ possible arrangements $= \dfrac{3!}{3!} = 1$
The total number of arrangements (each arrangement resulting in a distinct three-digit number) is $3 + 6 + 6 + 3 + 3 + 6 + 1 = 28$.
Problem 2 of 30
Suppose $a,\ b$ are positive real numbers such that $a\sqrt{a} + b\sqrt{b} = 183$, $a\sqrt{b} + b\sqrt{a} = 182$.Find $\dfrac{9}{5}(a + b)$.
Solution:
To make this problem easy to solve, we can eliminate the fraction powers by substituting $\sqrt{a} = x$ and $\sqrt{b} = y$
We get the given equations as:
$x^2x + y^2y = 183 \Rightarrow$ $x^3 + y^3 = 183$ $...eqn(i)$
$x^2y + y^2x = 182$ $...eqn(ii)$
Since our goal is to find $a + b$, with these substitutions we should find $x^2 + y^2$
Multiplying $eqn(ii)$ by $3$ and adding it to $eqn(i)$, we get:
$x^3 + 3x^2y + 3y^2x + y^3 = 183 + 3\times 182$
$\Rightarrow (x + y)^3 = 3(61 + 182) = 3\times 243 = 3 \times 9 \times 27$
$\Rightarrow x + y = 3 \times 3 = 9$
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Using $eqn(ii)$
$x^2y + y^2x = 182$
$\Rightarrow xy(x + y) = 182$
$\Rightarrow xy \times 9 = 182$
$\Rightarrow xy = \dfrac{182}{9}$
$x^2 + y^2 = (x + y)^2 - 2xy$
$= 9^2 - \dfrac{2 \times 182}{9}$
$= 81 - \dfrac{364}{9}$
$= \dfrac{729 - 364}{9}$
$= \dfrac{365}{9}$
$\therefore \dfrac{9}{5}(a + b) = \dfrac{9}{5}(x^2 + y^2)$
$= \dfrac{9}{5} \times \dfrac{365}{9}$
$= 73$
Problem 3 of 30
A contractor has two teams of workers: team $A$ and team $B$. Team $A$ can complete a job in $12$ days and team $B$ can do the same job in $36$ days. Team $A$ starts working on the job and team $B$ joins team $A$ after four days. The team $A$ withdraws after two more days. For how many more days should team $B$ work to complete the job?Solution:
Team $A$ worked for a total of $6$ days.
Since team $A$ can complete the work in $12$ days, in $6$ days they can finish $\dfrac{1}{2}$ of the work, which means team $B$ has to do the remaining $\dfrac{1}{2}$ of the work, which is equivalent to $18$ days' of work for team $B$.
Team $B$ worked for two days before team $A$ left. Therefore, the remaining work can be completed by team $B$ in $16$ days.
Problem 4 of 30
Let $a,\ b$ be integers such that all the roots of the equation $(x^2 + ax + 20)(x^2 + 17x + b) = 0$ are negative integers. What is the smallest possible value of $a + b$?Solution:
Here for both the quadratic equations, the coefficients of $x^2$ are $1$.
For a generic equation of the form $ax^2 + bx + c$ to have integer roots $b^2 - 4ac$ must be a perfect square, and for them to have both negative roots, $b$ must be of the same sign as $c$
Therefore, for the first quadratic equation $x^2 + ax + 20$ to have positive integer roots:
$a^2 - 80$ must be a perfect square and $a$ must be positive.
The minimum value of $a$ is $9$
Therefore $a = 9$
Similarly, for the equation $x^2 + 17x + b$
$17^2 - 4b = 289 - 4b$ should be a perfect square, and $b$ must be a positive number.
We can write $289 - 4b = n^2$ where $n$ is a positive integer.
$\therefore 4b = 289 - n^2$
For $b$ to be an integer $289 - n^2$ must be an even number, therefore $n$ must be an odd number.
The odd perfect square, less than $289$ is $225$, and $225 = 289 - 4 \times 16$
Therefore, the minimum value of $b$ is $16$
The minimum value of $a + b = 9 + 16 = 25$
Problem 5 of 30
Let $u,\ v,\ w$ be real numbers in geometric progression such that $u \gt v \gt w$. Suppose $u^{40} = v^n = w^{60}$, find the value of $n$.Solution:
Let
$w = a$
$v = ar$
$u = ar^2$
Since $u^{40} = w^{60}$
$(ar^2)^{40} = a^{60}$
$\Rightarrow a^{40}r^{80} = a^{60}$
$\Rightarrow r^{80} = a^{20}$
$\Rightarrow a = r^4$
Therefore,
$v = r^5$
Since $v^n = w^{60}$
$r^{5n} = (r^4)^{60}$
$\Rightarrow 5n = 240$
$\Rightarrow n = 48$
Problem 6 of 30
Let the sum $\displaystyle\sum\limits_{n=1}^{9}\dfrac{1}{n(n+1)(n+2)}$ written in its lowest terms be $\dfrac{p}{q}$. Find the value of $q - p$.Solution:
We can expand the given series as:
$\dfrac{1}{1.2.3} + \dfrac{1}{2.3.4} + ... + \dfrac{1}{n(n+1)(n+2)}$
We will solve this problem using the telescoping method explained .
We can break up the $t\xasuper{th}$ term as:
$\dfrac{1}{t(t+1)(t+2)} = \dfrac{1}{2}\left\{\dfrac{1}{t(t+1)} - \dfrac{1}{(t+1)(t+2)}\right\}$
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Therefore, our series becomes:
$\dfrac{1}{2}\left\{\dfrac{1}{1.2} - \dfrac{1}{2.3}\right\} + \dfrac{1}{2}\left\{\dfrac{1}{2.3} - \dfrac{1}{3.4}\right\} ... + \dfrac{1}{2}\left\{\dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)}\right\}$
$= \dfrac{1}{2}\left[\left\{\dfrac{1}{1.2} - \dfrac{1}{2.3}\right\} + \left\{\dfrac{1}{2.3} - \dfrac{1}{3.4}\right\}... + \left\{\dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)}\right\}\right]$
$= \dfrac{1}{2}\left[\dfrac{1}{1.2} - \dfrac{1}{2.3} + \dfrac{1}{2.3} - \dfrac{1}{3.4} + ... + \dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)}\right]$
In the above series all terms, except the first and the last term will cancel out with the next term, and we will be left with:
$\dfrac{1}{2}\left[\dfrac{1}{1.2} - \dfrac{1}{(n+1)(n+2)}\right]$
Substituting $n = 9$ we get the sum as:
$\dfrac{1}{2}\left[\dfrac{1}{1.2} - \dfrac{1}{(10)(11)}\right]$
$= \dfrac{1}{2}\left[ \dfrac{55 - 1}{110} \right]$
$= \dfrac{27}{110}$
Therefore, $q - p = 110 - 27 = 83$
Problem 7 of 30
Find the number of positive integers $n$, such that $\sqrt{n} + \sqrt{n + 1} \lt 11$. Solution:
Given that $\sqrt{n} + \sqrt{n+1} \lt 11$
$\Rightarrow \sqrt{n+1} \lt 11 - \sqrt{n}$
$\Rightarrow n + 1 \lt 121 + n - 22\sqrt{n}$
$\Rightarrow 1 \lt 121 - 22\sqrt{n}$
$\Rightarrow 22\sqrt{n} \lt 120$
$\Rightarrow \sqrt{n} \lt \dfrac{60}{11}$
$\Rightarrow \sqrt{n} \lt 5 + \dfrac{5}{11}$
$\Rightarrow n \lt \left(5 + \dfrac{5}{11}\right)^2$
$\Rightarrow n \lt 25 + \dfrac{50}{11} + \dfrac{25}{121}$
$\Rightarrow n \lt 25 + 4 + \dfrac{6}{11} + \dfrac{25}{121}$
$\Rightarrow n \lt 29 + \dfrac{91}{121}$
The largest integer value of $n$ is $29$ and the smallest value is $1$ (since it is a positive integer).
Therefore, there are $29$ values of $n$ that satisfies the given inequality.
Problem 8 of 30
A pen costs $\unicode{0x20B9}\ 11$ and a notebook costs $\unicode{0x20B9}\ 13$. Find the number of ways in which a person can spend exactly $\unicode{0x20B9}\ 1000$ to buy pens and notebooks.(Note: For international students $\unicode{0x20B9}$ is the symbol for the currency Indian Rupees)
Solution:
Let $p$ be the number of pens and $n$ be the number of notebooks.
$11p + 13n = 1000$
Since $p$ and $n$ can only be integers, we will use the method of solving linear Diophantine equation described , to solve this problem.
We will use the modular arithmetic approach to find the initial solution.
We know that:
$1000 \equiv 10\ (mod\ 11)$
and
$13 \equiv 2\ (mod\ 11)$
$5 \times 13 \equiv 10\ (mod\ 11)$
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Therefore,
$1000 - 5 \times 13 \equiv 0\ (mod\ 11)$
$\Rightarrow 935 \equiv 0\ (mod\ 11)$
$\therefore p_0 = 85$ and $n_0 = 5$
The parametric equations for these variables can be written as:
$p = -13t + p_0 \Rightarrow p = 85 - 13t$
$n = 11t + n_0 \Rightarrow n = 11t + 5$
Since $n$ is positive:
$11t + 5 \ge 0$
$\Rightarrow 11t \ge -5$
$\Rightarrow t \ge -\dfrac{5}{11}$
Since $p$ is positive:
Putting $85 - 13t \ge 0$
$85 \ge 13t$
$t \le \dfrac{85}{13}$
$t \le 6\dfrac{7}{13}$
Therefore $t$ ranges from $0$ to $6$, therefore there are $7$ possible value of $t$ which will be positive $p$ and $n$ which satisfies the given equation.
Problem 9 of 30
There are five cities $A,\ B,\ C,\ D,\ E$ on a certain island. Each city is connected to every other city by road. In how many ways can a person, starting from city $A$ come back to $A$ after visiting some cities without visiting a city more than once and without taking the same road more than once? (The order in which he visits the cities also matters: e.g., the routes $A \longrightarrow B \longrightarrow C \longrightarrow A$ and $A \longrightarrow C \longrightarrow B \longrightarrow A$ are different.)Solution:
It is clear that the person has to visit at least two other cities, to be able to take a different road back to city $A$.
Suppose he visits $2$ other cities, the first city can be chosen in $4$ ways, and the second city can be chosen in $3$ ways.
Hence there are $4 \times 3 = 12$ ways to do this.
Similarly, if he visits $3$ other cities, the cities can be chosen in:
$4 \times 3 \times 2 = 24$ ways.
If he visits all other cities, the cities can be chosen in:
$4\times 3 \times 2 \times 1 = 24$ ways.
Therefore, there are $12 + 24 + 24 = 60$ ways.
Problem 10 of 30
There are eight rooms on the first floor of a hotel, with four rooms on each side of the corridor, symmetrically situated (that is each room is exactly opposite to one other room). Four guests have to be accommodated in the four of the eight rooms (that is, one in each) such that no two guests are in adjacent rooms or in opposite rooms. In how many ways can the guests be accommodated?Solution:
Let us draw the schematic of the rooms as below, with the rooms number $1$ to $8$, as shown below:
First we can choose the rooms which will be used for the guests. This can be done in exactly two ways:
We can either choose rooms $1,\ 3,\ 5,\ 7$ or we can choose $2,\ 4,\ 6,\ 8$
Now for each of the above choices, there are $4!$ ways of assigning $4$ guests to these rooms.
Therefore, there are a total of $2 \times 4! = 48$ ways of assigning the guests.
Problem 11 of 30
Let $f(x) = sin\dfrac{x}{3} + cos\dfrac{3x}{10}$ for real $x$. Find the least natural number $n$ such that $f(n\pi + x) = f(x)$ for all real $x$.Solution:
The given function is a composite trigonometric function. We know that a periodic function repeats itself after every period.
That is $f(n\pi + x) = f(x)$ for all values of $x$, where $n\pi$ is the period of the function.
To solve this problem we need to find the period of the given composite function using the method described .
Let us first write the given function in the required form, which is:
$f(x) = sin\left(\dfrac{x}{3}\right) + cos\left(\dfrac{3x}{10}\right) = sin\left(\dfrac{x}{3}\right) + cos\left(\dfrac{x}{\frac{10}{3}}\right)$
Using the method described , we know $lcm\left(3,\ \dfrac{10}{3}\right) = \dfrac{lcm(3,\ 10)}{gcd(1,\ 3)} = 30$.
The period of $f(x) = 30\times 2\pi = 60\pi$
Therefore, $n\pi = 60\pi$
$\Rightarrow n = 60$
Problem 12 of 30
In a class, the total numbers of boys and girls are in the ratio $4:3$. On one day it was found that $8$ boys and $14$ girls were absent from the class, and that the number of boys was the square of the number of girls. What is the total number of students in the class?Solution:
Let the number of boys be $4k$ and the number of girls be $3k$.
Based on the given conditions:
$4k - 8 = (3k - 14)^2$
$\Rightarrow 4k - 8 = 9k^2 - 84k + 196$
$\Rightarrow 9k^2 - 88k + 204 = 0$
$\Rightarrow 9k^2 - 54k - 34k + 204 = 0$
$\Rightarrow 9k(k - 6) - 34(k - 6) = 0$
$\Rightarrow (9k - 34)(k - 6) = 0$
$\Rightarrow k = \dfrac{34}{9}$ or $k = 6$
The value of $k$ cannot be $\dfrac{34}{9}$ since $4k$ and $3k$ have to be integers.
Therefore $k = 6$
Therefore, number of student $= 4k + 3k = 7k = 42$
Problem 13 of 30
In a rectangle $ABCD$, $E$ is the midpoint of $AB$; $F$ is a point on $AC$ such that $BF$ is perpendicular to $AC$; and $FE$ perpendicular $BD$. Suppose $BC = 8\sqrt{3}$, find $AB$.Solution:
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Let us call the intersection point of the two diagonals $AC$ and $BD$ as $O$ and the intersection point of $EF$ and $BD$ as $P$.
We draw a line parallel to $AB$, passing through $O$, and meeting $AD$ and $BC$ at $M$ and $N$ respectively, as shown in the following diagram.
Let $\angle CAB = \theta$ as marked in the above diagram.
Considering $\triangle AEB$,
$\angle AEB = 90^\circ$, and $F$ is the midpoint of $AB$.
$EF$ is the median to the hypotenuse $AB$ of a right angled triangle $AEB$.
Therefore, $EF = \dfrac{1}{2}AB = FB$
Therefore, $EF = FA = FB$
Therefore, $\triangle FAE$ is isosceles, and $\angle AEF = \angle FAE = \theta$
$\therefore \angle EFB = 2\theta$ (sum of two opposite interior angles)
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Since $ABCD$ is a rectangle, diagonals $AC$ and $BD$ are equal, and they bisect each other.
$\therefore OA = OB$, and $\triangle OAB$ is isosceles.
$\therefore \angle OAB = \angle OBA = \theta$
$\because \triangle FEB$ is isosceles with $FE = FB$,
$\angle FEB = \dfrac{180 - \angle EFB}{2} = \dfrac{180 - 2\theta}{2} = 90 - \theta$
Considering right angled triangle $ABC$,
$\angle ACB = 90 - \angle CAB = 90 - \theta$
Again considering right angled triangle $BEC$,
$\angle EBC = 90 - \angle ECB = 90 - \angle ACB = 90 - (90 - \theta) = \theta$
$\therefore \angle OBE = 90 - (\angle OBA + \angle EBC) = 90 - (\theta + \theta) = 90 - 2\theta$
Considering right angled triangle $PEB$,
$\angle PEB + \angle PBE = 90^\circ$
$\therefore (90 - \theta) + (90 - 2\theta) = 90^\circ$
$\Rightarrow 3\theta = 90^\circ$
$\Rightarrow \theta = 30^\circ$
$\therefore \angle BOC = \angle OAB + \angle OBA = 30^\circ + 30^\circ = 60^\circ$
Since $OC = OB$ (half of the diagonals)
$\therefore \triangle OBC$ is isosceles with $\angle OBC = \angle OCB$
$therefore \angle OCB = \angle OBC = \dfrac{180 - \angle BOC}{2} = \dfrac{180 - 60}{2} = 60^\circ$
Since all the angles of $\triangle OBC$ are $60^\circ$, it is an equilateral triangle with each side $= 8\sqrt{3}$
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Since we have drawn $MN$ parallel to $O$, $ABMN$ is a rectangle, therefore $MN = AB$ and $MN \perp BC$
$ON$ is the altitude of $\triangle OBC$
Therefore, $ON = \dfrac{\sqrt{3}}{2} \times 8\sqrt{3} = 12$
Therefore, $AB = MN = 2ON = 2 \times 12 = 24$
Problem 14 of 30
Suppose $x$ is a positive real number such that $\{x\}$, $\lfloor x\rfloor$ and $x$ are in geometric progression. Find the least positive integer $n$ such that $x^n \gt 100$. (Here $\lfloor x\rfloor$ denotes the integer part of $x$ and $\{x\}$ denotes $x - \lfloor x\rfloor$.)Solution:
Since $x$ is positive, both $\lfloor x \rfloor$ and $\{ x \}$ are also positive and $\{ x \} \lt 1$.
Since the three numbers are in GP, let:
$\{ x \} = a$, $\lfloor x \rfloor = ar$ and $x = ar^2$
Since all the three numbers are positive $a$ and $r$ are also positive.
Based on the given definitions of $\lfloor x \rfloor$, and $\{ x \}$ we know:
$x = \lfloor x \rfloor + \{ x \}$
$\Rightarrow ar^2 = ar + a$
$\Rightarrow r^2 = r + 1$
$\Rightarrow r^2 - r - 1 = 0$
$r = \dfrac{1 \pm \sqrt{5}}{2}$
$1 \lt \sqrt{5} \Rightarrow 1 - \sqrt{5} \lt 0$
$r = \dfrac{1 - \sqrt{5}}{2}$ will be a negative value, and we will ignore it.
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$\therefore r = \dfrac{1 + \sqrt{5}}{2}$
Observe that $ar$ is a positive integer and $r$ is a rational number.
Therefore, $a$ has to be a number of the form $p \times \dfrac{2}{1 + \sqrt{5}}$ where $p$ is a positive integer, only then $ar$ will be a positive integer.
Since $a \lt 1$ the only possible value of $p = 1$, because if we put $p = 2$,
we get $a = \dfrac{4}{1 + \sqrt{5}}$ which is greater than $1$
Therefore, our GP series is:
$\dfrac{2}{1 + \sqrt{5}}$, $1$, $\dfrac{1 + \sqrt{5}}{2}$ and
$x = \dfrac{1 + \sqrt{5}}{2} \approx 1.6$
Let $1.6^n \gt 100$
$\Rightarrow \left(\dfrac{16}{10}\right)^n \gt 100$
$\Rightarrow log\left(\dfrac{16}{10}\right)^n \gt log(100)$
$\Rightarrow n log\left(\dfrac{16}{10}\right) \gt log(10^2)$
$\Rightarrow n (log(16) - log(10)) \gt 2log(10)$
$\Rightarrow n (log2^4 - 1) \gt 2$
$\Rightarrow n (4log2 - 1) \gt 2$
$\Rightarrow n (4\times 0.301 - 1 \gt 2$
$\Rightarrow n (1.204 - 1) \gt 2$
$\Rightarrow n \times 0.204 \gt 2$
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$\Rightarrow n \gt \dfrac{2}{0.204}$
$\Rightarrow n \gt \dfrac{2.04 - 0.04}{0.204}$
$\Rightarrow n \gt 10 - \dfrac{0.04}{0.204}$
$\Rightarrow n \gt 9 + 1 - \dfrac{40}{204}$
$\Rightarrow n \gt 9 + \dfrac{164}{204}$
The smallest integer value of $n$ satisfying this inequality is $10$
Problem 15 of 30
Integers $1,\ 2,\ 3,\ ...,\ n$ where $n \gt 2$, are written on a board. Two numbers $m,\ k$ such that $1 \lt m \lt k$, $1 \lt k \lt n$ are removed and the average of the remaining numbers is found to be $17$. What is the maximum sum of the two removed numbers?Solution:
Let the maximum sum of $m$ and $k$ be $s$.
The sum of the numbers before removing $s$ is $\dfrac{n(n+1)}{2}$.
The sum of numbers after removing $s$ is $17(n-2)$
The smallest set of numbers that can be removed is greater than $\{1,\ 2\}$
After removing $s$, the remaining sum must be lesser than the sum we would get after removing $3$.
Therefore,
$17(n-2) \lt \dfrac{n(n+1)}{2} - 3$
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Similarly, the largest sum that can be removed must be less than $(n - 1) + n$ that is $2n - 1$.
Thus the remaining sum must be greater than $\dfrac{n(n+1)}{2} - 3 - (2n - 1)$
That is,
$17(n-2) \gt \dfrac{n(n+1)}{2} - (2n - 1)$
Combining the above two inequalities, we get:
$\dfrac{n(n+1)}{2} - (2n - 1) \lt 17(n-2) \lt \dfrac{n(n+1)}{2} - 3$
$\Rightarrow \dfrac{n(n+1)}{2} - 2n + 1 \lt 17(n-2) \lt \dfrac{n(n+1)}{2} - 3$
$\Rightarrow \dfrac{n(n+1)}{2} - 2n + 1 \lt 17(n-2) \lt \dfrac{n(n+1)}{2} - 3$
$\Rightarrow \dfrac{n^2 - 3n + 2}{2} \lt 17(n - 2) \lt \dfrac{n^2 + n - 6}{2}$
$\Rightarrow \dfrac{(n-2)(n-1)}{2} \lt 17(n - 2) \lt \dfrac{(n+3)(n-2)}{2}$
Since $n \gt 2$ $n - 2$ is positive and we can divide all parts of the above inequalities by $(n - 2)$, and multiply everything by $2$ which gives us:
$(n - 1) \lt 34 \lt (n + 3)$
Therefore, we get:
$ 31 \lt n \lt 35$
Since $n$ is an integer, the valid values of $n$ can be $32,\ 33,\ 34$
We know that:
$s = \dfrac{n(n+1)}{2} - 17(n-2)$
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Putting $n = 32$ we get $s = \dfrac{32 \times 33}{2} - 17(30) = 18$
Putting $n = 33$ we get $s = \dfrac{33 \times 34}{2} - 17(31) = 34$
Putting $n = 34$ we get $s = \dfrac{34 \times 35}{2} - 17(32) = 51$
Therefore, the maximum value of $s$ is $51$
Problem 16 of 30
Five distinct two digit numbers are in geometric progression. Find the middle term.Solution:
We will consider the GP series in increasing order, with the first term as the smallest and the common ratio as greater than $1$.
Even if we take the GP series in the reverse order of terms, the middle term will still remain the same.
Since the problem mentions two-digit numbers, we can take the first term $a$ has to be an integer, and its minimum value could be $10$.
If we try to take $r$ as an integer, the minimum possible value is $2$, and the minimum possible value of the fifth term becomes $10 \times 2^4 = 10 \times 16 = 160$. Hence $r$ has to be less the $2$.
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Since $1 \lt r \lt 2$ we can take $r$ as a fraction $\dfrac{p}{q}$ in the fully reduced form, where $q \lt p \lt 2q$
The fifth term is $a \times \left(\dfrac{p}{q}\right)^4$
Therefore, for all the terms to be integers, $a$ has to be divisible by $q^4$.
If we take $q = 2$ the smallest possible value of $a$ is $2^4 = 16$, and if we take $q = 3$ the smallest possible value of $a$ is $3^4 = 81$
If we take $q = 3$, the minimum value of $p$ is $4$ and $r$ becomes $\dfrac{4}{3}$
The series will become:
$81, 108,...$
Hence it is not possible to get all the terms as 2-digit numbers.
If we take $q = 2$, the only possible value of $p$ is $3$, and the terms become,
$16,\ 16\times \dfrac{3}{2},\ 16\times \dfrac{9}{4},\ 16\times \dfrac{27}{8},\ 16\times \dfrac{81}{16}$
$= 16,\ 24,\ 36,\ 54,\ 81$
This series, and its reverse, $81,\ 54,\ 36,\ 24,\ 16$ satisfy all the given conditions.
The middle term, in either case, is $36$.
Problem 17 of 30
Suppose the altitudes of a triangle are $10,\ 12$ and $15$. What is the integer nearest to the value of the semi-perimeter ?Solution:
Note: The problem, as published originally for PRMO, did not specify the criteria $\unicode{0x201C}$integer closest to$\unicode{0x201D}$. Therefore, this would have led to a decimal answer, hence the question was later on discounted. We added the nearest integer condition to make this problem solvable with an integer answer.
We know that the three sides of a triangle are inversely proportional to the corresponding altitudes drawn from these sides.
Therefore, if the three sides of the given triangle are $a$, $b$ and $c$, then
$a : b : c = \dfrac{1}{10}:\dfrac{1}{12}:\dfrac{1}{15} = 6:5:4$
Let the sides be $6k$, $5k$, and $4k$ respectively.
Therefore, the semi-perimeter $= \dfrac{15k}{2}$
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Considering any one side and the corresponding altitude, the area of $\triangle ABC = \dfrac{1}{2} \times 6k \times 10 = 30k$
Using Heron's formula:
$30k = \sqrt{\dfrac{15k}{2}\left(\dfrac{15k}{2} - 6k\right)\left(\dfrac{15k}{2} - 5k\right)\left(\dfrac{15k}{2} - 4k\right)}$
$\Rightarrow 30k = \sqrt{\dfrac{15k}{2}\left(\dfrac{3k}{2}\right)\left(\dfrac{5k}{2}\right)\left(\dfrac{7k}{2}\right)}$
$\Rightarrow 30k = k^2 \sqrt{\dfrac{15 \times 3 \times 5 \times 7}{2 \times 2 \times 2 \times 2}}$
$\Rightarrow 30k = k^2 \times \dfrac{15\sqrt{7}}{4}$
$\Rightarrow k = \dfrac{8}{\sqrt{7}}$
Therefore the semi-perimeter $(s)$ can be obtained using:
$s = \dfrac{15k}{2} = \dfrac{15 \times 8}{2\sqrt{7}}$
$= \dfrac{60 \times \sqrt{7}}{7} = \dfrac{60 \times 2.65}{7} = \dfrac{159}{7} = 22\dfrac{5}{7}$
Therefore, the integer nearest to the semi-perimeter is $23$
Problem 18 of 30
If the real numbers $x,\ y,\ z$ are such that $x^2 + 4y^2 + 16z^2 = 48$ and $xy + 4yz + 2zx = 24$, what is the value of $x^2 + y^2 + z^2$?Solution:
This problem has given two equations, each containing $3$ variables. We know that such an equation cannot be solved unless some special condition is satisfied, like the sum of some squares equal to zero or similar condition.
Let us try to obtain an equation by setting the RHS to $0$.
Subtracting twice the second equation from the first we get:
$x^2 + 4y^2 + 16z^2 - 2xy - 8yz - 4zx = 0$
We can try to express the LHS as the sum of multiple squares, each term involving two of the three unknowns.
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Multiplying both sides by $2$
$2x^2 + 8y^2 + 32z^2 - 4xy - 16yz - 8zx = 0$
$\Rightarrow x^2 -4xy + 4y^2 + 4y^2 - 16yz + 16z^2 + 16z^2 - 8zx + x^2 = 0$
$\Rightarrow (x - 2y)^2 + (2y - 4z)^2 + (4z - x)^2 = 0$
The LHS being the sum of $3$ terms which are squares of real numbers, and being equal to zero, each of the term must be $0$.
Therefore we get:
$x - 2y = 0 \Rightarrow x = 2y \Rightarrow x^2 = 4y^2$
$2y - 4z = 0 \Rightarrow y = 2z$
$4z - x = 0 \Rightarrow 4z = x \Rightarrow 16z^2 = x^2$
Substituting $4y^2 = x^2$ and $16z^2 = x^2$, we get:
$x^2 + x^2 + x^2 = 48$
$x^2 = 16$
$x = 4$
$y = 2$
$z = 1$
Therefore, $x^2 + y^2 + z^2 = 16 + 4 + 1 = 21$
Problem 19 of 30
Suppose $1,\ 2,\ 3$ are the roots of the equation $x^4 + ax^2 + bx = c$. Find the value of $c$.Solution:
Since $1,\ 2,\ 3$ are the roots of the given equation, we can substitute $x$ with these three values, and obtain $3$ linear equations in $a,\ b$ and $c$ which we can easily solve.
Substituting $x = 1$,
$1 + a + b = c$ $...eqn\ (i)$
Substituting $x = 2$
$16 + 4a + 2b = c$ $...eqn\ (ii)$
Substituting $x = 3$
$81 + 9a + 3b = c$ $...eqn\ (iii)$
Subtracting $eqn\ (i)$ from each of $eqn\ (ii)$ and $(iii)$ we get:
$15 + 3a + b = 0$ $...eqn\ (iv)$
and
$80 + 8a + 2b = 0$ $...eqn\ (v)$
$\Rightarrow 40 + 4a + b = 0$
Subtracting $eqn\ (iv)$ from $(v)$ we get:
$25 + a = 0$
$\Rightarrow a = -25$
$\therefore 15 + 3(-25) + b = 0$
$\Rightarrow b = 75 - 15 = 60$
$\therefore c = 1 + a + b = 1 + (-25) + 60 = 36$
Problem 20 of 30
What is the number of triples $(a,\ b,\ c)$ of positive integers such that,$(i)\ a \lt b \lt c \lt 10$ and
$(ii)\ a,\ b,\ c,\ 10$ form the sides of a quadrilateral?
Solution:
We can see that there are two sets of restrictions for choosing the values of $a,\ b,\ c$, which are:
$a \lt b\lt c \lt 10$ and $a + b + c \ge 11$
We can solve this problem both ways, namely:
- We can enumerate all cases considering the restrictions and add them.
- We can find the cases without any restriction and then eliminate the cases which do not satisfy the restrictions.
We will do it the second way, since it involves less counting:
Since $a \lt b \lt c \lt 10$, we can choose three distinct integers from the set $\{1,\ 2,\ 3,\ ...,\ 9\}$, and the ascending arrangement of these integers will give us a set of possible values for $a,\ b,\ c$.
This can be done in $\xacomb{9}{3} = 84$ ways.
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The minimum possible value of $a + b + c$ from the above selections is $1 + 2 + 3 = 6$
Now we need to eliminate the cases that will not satisfy the condition $a + b + c \ge 11$, which are:
The following cases will not satisfy this condition:
$a + b + c = 6$
This has just one case $a = 1,\ b = 2,\ c = 3$
$a + b + c = 7$
This also has just one case:
$\{1,\ 2,\ 4\}$
$a + b + c = 8$
$\{1,\ 2,\ 5\}$, $\{1,\ 3,\ 4\}$
Two cases.
$a + b + c = 9$
$\{1,\ 2,\ 6\}$, $\{1,\ 3,\ 5\}$, $\{2,\ 3,\ 4\}$
Three cases.
$a + b + c = 10$
$\{1,\ 2,\ 7\}$, $\{1,\ 3,\ 6\}$, $\{1,\ 4,\ 5\}$, $\{2,\ 3,\ 5\}$
Four cases.
Therefore, a total of $11$ cases do not satisfy the condition $a + b + c \ge 11$
Therefore, the total number of cases that satisfy the given conditions is $84 - 11 = 73$
Problem 21 of 30
Find the number of ordered triples $(a,\ b,\ c)$ of positive integers such that $abc = 108$.Solution:
The number $108$ can be factored as $3^3 \times 2^2$.
Each of the $3$s are identical to one another, and so are the $2$s.
The three integers $a,\ b,\ c$ are distinct.
Let us start with $a = 1$, $b = 1$ and $c = 1$ and then we can assign one or more of our prime factors to $a,\ b,\ c$
If we are able count the distinct ways to distribute the three $3$'s amongst $a,\ b,\ c$ the each will give us a distinct set of ordered triples.
For each of these distributions we can then again distribute the two $2$s in all possible way to get a distinct set of ordered triples, which will always multiply up to $3^3 \times 2^2$.
Observe that it is possible that in a particular distribution one or two of $a,\ b,\ c$ does not receive either a $2$ and $3$, which means that number will be $1$.
We will use the method described , to count the number of ways each identical prime factor can be distributed amongst the $3$ distinct numbers.
The number of ways three identical $3$s can be distributed amongst three distinct bins is:
$\xacomb{3 + 3 - 1}{3 - 1}$ and for each of these distributions the two $2$s can be distributed amongst $3$ bins is
$\xacomb{2 + 3 - 1}{3 - 1}$.
Therefore, the total number of ways $\xacomb{3 + 3 - 1}{3 - 1} \times \xacomb{2 + 3 - 1}{3 - 1}$
$= \xacomb{5}{2} \times \xacomb{4}{2}$
$= 10 \times 6 = 60$ ways
Problem 22 of 30
Suppose in a plane $10$ pairwise nonparallel lines intersect one another. What is the maximum possible number of non-overlapping polygons (with finite areas) that can be formed?Solution:
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Note:This problem, as published for PRMO, did not contain the word $\text{non-overlapping}$, which made it ambiguous. The results will be different when all possible polygons are counted and where only non-overlapping polygons are counted. This problem was discounted from the final result. We have added the word $\text{non-overlapping}$ to eliminate the ambiguity.
We will use the second approach described in this and extend the same to solve this problem.
We will start with the diagram from this approach, as shown below:
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We have seen that for a set of $n$ lines there are $\xacomb{n}{2}$ regions which have a lowest point and there are $n+1$ regions which are open at the bottom and do not have a lowest point.
The count $\xacomb{n}{2}$ includes the regions open at the top. If we can determine the count of those, then we can easily find the count of regions that have a highest and a lowest point. It can be easily shown that if a region has a highest and a lowest point it has to be bounded, that is, it is of finite area.
To count the number of regions which has a lowest point, we just take the previous image and turn it upside down as shown below:
As before, we can see that every intersection point is the bottom-most point of some bounded region and some regions that are open at the top, which were earlier open at the bottom.
If we again count the number of regions open at the bottom by drawing horizontal line $l_2$ we will see that there are $n + 1$ regions with no lowest point.
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Out of these $n +1$ regions, two regions where the two extreme parts of $l_2$ lie have been counted in the previous case as well.
It can also, be shown that there will only be two such regions which will be counted twice.
Now, the final count of bounded regions will be:
$\xacomb{n}{2} - (n + 1) + 2 = \xacomb{n}{2} - n + 1$
We could have achieved the same result, by just counting the regions where each intersection point form the highest point and the finding the count of regions without a highest point. This was just for the ease of understanding.
For a set of $10$ lines, the maximum possible number of regions is $\xacomb{10}{2} - 10 + 1 = 45 - 10 + 1 = 36$
Problem 23 of 30
Suppose an integer $x$, a natural number $n$ and a prime number $p$ satisfy the equation $7x^2 - 44x + 12 = p^n$. Find the largest value of $p$.Solution:
Given that:
$7x^2 - 44x + 12 = p^n$
$\Rightarrow (7x - 2)(x - 6) = p^n$
Because the right hand side is a power of a prime $p$, then each factor on the left side can only be powers of the same prime $p$
Therefore, $7x - 2 = p^a$
and $x - 6 = p^b$
Since $p$ is a prime both $p^a$ and $p^b$ are positive numbers.
$\therefore x - 6 \gt 0$ and $7x - 2 \gt 0$
Therefore $x \gt 6$
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Therefore, $x - 6$ is divisible by $7x - 6$ if $b \ge a$ or
$7x - 2$ is divisible by $x - 6$ if $a \gt b$
If $b \ge a$
$\Rightarrow x - 6 \ge 7x - 2$
$\Rightarrow 4 \ge 6x$
$\Rightarrow x \le -\dfrac{2}{3}$
This contradicts our observation that $x \gt 6$, since there is no solution for $x$ satisfying both conditions.
If $a \gt b$
$\Rightarrow 7x - 2 \gt x - 6$
then $\dfrac{7x - 2}{x - 6}$ is an integer
$\Rightarrow \dfrac{7x - 42 + 40}{x - 6}$ is an integer
$\Rightarrow \dfrac{7(x - 6) + 40}{x - 6}$ is an integer
$\Rightarrow 7 + \dfrac{40}{x - 6}$ is an integer
Therefore, $x - 6$ is a divisor of $40$
Therefore $(x - 6)\ \epsilon\ \{1,\ 2,\ 4, 5,\ 8,\ 10,\ 20,\ 40\}$
$\Rightarrow x\ \epsilon\ \{7,\ 8,\ 10, 11,\ 14,\ 16,\ 26,\ 46\}$
Putting $x = 7$ we get $x - 6 = 1$ and $7x - 2 = 47$
And $p^n = 47 \Rightarrow p = 47,\ n = 1$
For the remaining possibilities we can see that the largest value of $x$ is $46$,
therefore $x - 6 = 40$ which is not a power of prime.
Likewise no other value in the set can give us a power of a prime greater than $47$
Problem 24 of 30
Let $P$ be an interior point of a triangle $ABC$ whose sidelengths are $39,\ 65,\ 78$. The line through $P$ parallel to $BC$ meets $AB$ in $K$ and $AC$ in $L$. The line through $P$ parallel to $CA$ meets $BC$ in $M$ and $BA$ in $N$. The line through $P$ parallel to $AB$ meets $CA$ in $S$ and $CB$ in $T$. If $KL$, $MN$, $ST$ are of equal lengths, find the integer value closest to this common length.Solution:
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Note: The original problem, as published for PRMO, specified the three sides as $26$, $65$ and $78$, which had no solution for point $P$ inside $\triangle ABC$. We modified the value of a single side from $26$ to $39$ to make this problem solvable.
Let us draw the diagram for the figure as shown below:
We assume that $AB = 39$, $BC = 65$ and $CA = 78$
We assume that the areas of $\triangle PNK$, $\triangle PTM$ and $\triangle SPL$ are $x^2$, $y^2$ and $z^2$ respectively.
Let the common length of $MN$, $KL$ and $ST$ be $l$
Based on the property of triangles explained ,
$ar[\triangle ABC] = (x + y + z)^2$
$ar[\triangle AKL] = (z + x)^2$
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Also, from the theory of similar triangles we know that:
$\dfrac{ar[\triangle AKL]}{ar[\triangle ABC]} = \dfrac{KL^2}{BC^2} = \dfrac{l^2}{BC^2}$
$\Rightarrow \dfrac{l^2}{BC^2} = \dfrac{(z + x)^2}{(x + y + z)^2}$
$\Rightarrow \dfrac{l}{BC} = \dfrac{z + x}{x + y + z}$
$\Rightarrow \dfrac{l}{65} = \dfrac{z + x}{x + y + z}$
Similarly, considering $\triangle s\ BMN$ and $CST$ we get:
$\dfrac{l}{78} = \dfrac{x + y}{x + y + z}$
and
$\dfrac{l}{39} = \dfrac{y + z}{x + y + z}$
Adding the three equations, we get:
$\dfrac{l}{65} + \dfrac{l}{78} + \dfrac{l}{39} = \dfrac{z + x}{x + y + z} + \dfrac{x + y}{x + y + z} + \dfrac{y + z}{x + y + z}$
$\Rightarrow l\left(\dfrac{1}{65} + \dfrac{1}{78} + \dfrac{1}{39}\right) = \dfrac{2(x + y + z)}{(x + y + z)}$
$\Rightarrow l \left(\dfrac{6 + 5 + 10}{13 \times 5 \times 6}\right) = 2$
$\Rightarrow l = 2 \times \dfrac{13 \times 5 \times 6}{21} = \dfrac{260}{7} = 37\dfrac{1}{7}$
The integer nearest to the value of $l$ is $37$
Problem 25 of 30
Let $ABCD$ be a rectangle and let $E$ and $F$ be points on $CD$ and $BC$ respectively such that $area(ADE) = 16$, $area(CEF)=9$ and $area(ABF) = 25$. What is the area of triangle $AEF$?Solution:
Let us draw the diagram for the problem as shown below:
Let $AB = DC = x$, $AD = BC = y$, $CE = a$ and $CF = b$
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Based on the given conditions,
$ar[\triangle ADE] = \dfrac{1}{2} \times ya = 16$
$\Rightarrow ya = 32$
$\Rightarrow a = \dfrac{32}{y}$
$ar[\triangle ABF] = \dfrac{1}{2} \times xb = 25$
$\Rightarrow xb = 50$
$\Rightarrow b = \dfrac{50}{x}$
$ar[\triangle CEF] = \dfrac{1}{2} \times (x - a)(y - b) = 9$
$\Rightarrow (x - a)(y - b) = 18$
$\Rightarrow \left(x - \dfrac{32}{y}\right)\left(y - \dfrac{50}{x}\right) = 18$
$\Rightarrow (xy - 32)(xy - 50) = 18xy$
$\Rightarrow x^2y^2 -82xy + 1600 = 18xy$
$\Rightarrow (xy)^2 - 100xy + 1600 = 0$
$\Rightarrow (xy)^2 -80xy -20xy + 1600 = 0$
$\Rightarrow (xy - 80)(xy - 20) = 0$
Therefore, $xy = 80$ or $xy = 20$
$xy$ is the area of rectangle $ABCD$, and cannot be less than $16 + 25 + 9 = 50$.
Therefore, we ignore the solution $xy = 20$
Therefore, $ar[ABCD] = xy = 80$
$ar[\triangle AEF] = ar[ABCD] - (ar[\triangle ADE] + ar[\triangle ABF] + ar[\triangle CEF])$
$= 80 - 50 = 30$
Problem 26 of 30
Let $AB$ and $CD$ be two parallel chords in a circle with radius $5$ such that the centre $O$ lies between these chords. Suppose $AB = 6$, $CD = 8$. Suppose further that the area of the part of the circle lying between the chords $AB$ and $CD$ is $(m\pi + n)/k$, where $m,\ n,\ k$ are positive integers with $gcd(m, n, k)=1$. What is the value of $m + n + k$?Solution:
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Let us draw the diagram for the problem as shown below:
Let $O$ be the centre of the circle, and $PQ$ be a diameter drawn parallel to $AB$ and $CD$.
Let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively. We have drawn the lines $OA$, $OB$, $OC$, $OD$, $OM$ and $ON$.
$\triangle AOB$ is isosceles, with $M$ as the midpoint of $AB$, therefore $OM \perp AB$.
Similarly $ON \perp CD$
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$MB = \dfrac{AB}{2} = \dfrac{6}{2} = 3$
$\therefore OM = \sqrt{OB^2 - MB^2} = \sqrt{5^2 - 3^2} = 4$
Similarly,
$ND = 4$, and $ON = \sqrt{5^2 - 4^2} = 3$
Considering $\triangle s OMB$ and $OND$
$OD = OB = 5$
$MD = ON = 3$
$OM = ND = 4$
$\therefore \triangle OMB \cong \triangle OND$
$\therefore \angle MOB = \angle ODN$ and $\angle MBO = \angle NOD$
Since $PQ \parallel AB$
$\angle BOQ = \angle MBO$
Since $PQ \parallel CD$
$\angle DOQ = \angle ODN = \angle MOB$
$\therefore \angle BOQ + \angle DOQ = \angle MBO + \angle MOB$
$\angle MBO + \angle MOB = 180 - 90 = 90^\circ$
Therefore $\angle BOD = \angle BOQ + \angle DOQ = 90^\circ$
Similarly we can show that $\angle AOC = 90^\circ$
Area of the circle enclosed by $AB$ and $CD$
$= ar[\triangle OAB] + ar[\triangle OCD] + ar[sector\ BOD] + ar[sector\ AOC]$
$= \dfrac{1}{2} \times 4 \times 6 + \dfrac{1}{2} \times 3 \times 8 + \dfrac{1}{4}\times \pi 5^2 + \dfrac{1}{4}\times \pi 5^2$
$= 12 + 12 + \dfrac{1}{2} \times 25\pi$
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$= \dfrac{25\pi + 48}{2}$
Therefore, $m = 25,\ n = 48,\ k = 2$ and
$m + n + k = 75$
Problem 27 of 30
Let $\Omega_1$ be a circle with centre $O$ and let $AB$ be a diameter of $\Omega_1$. Let $P$ be a point on the segment $OB$ different from $O$. Suppose another circle $\Omega_2$ with centre $P$ lies in the interior of $\Omega_1$. Tangents are drawn from $A$ and $B$ to the circle $\Omega_2$ intersecting $\Omega_1$ again at $A_1$ and $B_1$ respectively such that $A_1$ and $B_1$ are on the opposite sides of $AB$. Given that $A_1B = 5$, $AB_1 = 15$ and $OP = 10$, find the radius of $\Omega_1$.Solution:
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We can draw the diagram for this problem as shown below.
$AA_1$ and $BB_1$ are tangents to the circle $\Omega_2$ at $M$ and $N$ respectively. We have drawn the radii $PM$ and $PN$ of circle $\Omega_2$
Let the radii of $\Omega_1$ and $\Omega_2$ be $r_1$ and $r_2$ respectively.
Considering $\triangle s AMP$ and $AA_1B$, $\angle M = \angle A_1= 90^\circ$.
Therefore, $MP \parallel A_1B$
Therefore, $\triangle AMP \sim AA_1B$
Therefore, $\dfrac{MP}{A_1B} = \dfrac{AP}{AB}$
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$\Rightarrow \dfrac{r_2}{5} = \dfrac{r_1 + 10}{2r_1}$ $...\ eqn\ (i)$
Similarly, $\triangle BPN \sim \triangle BAB_1$
Therefore, $\dfrac{PN}{AB_1} = \dfrac{BP}{AB}$
$\Rightarrow \dfrac{r_2}{15} = \dfrac{r_1-10}{2r_1}$ $...\ eqn\ (i)$
Dividing equation $(i)$ by equation $(ii)$, we get:
$\dfrac{15}{5} = \dfrac{r_1 + 10}{r_1 - 10}$
$\Rightarrow 3 = \dfrac{r_1 + 10}{r_1 - 10}$
$\Rightarrow 2r_1 = 40$
$\Rightarrow r_1 = 20$
Problem 28 of 30
Let $p,\ q$ be prime numbers such that $n^{3pq}-n$ is a multiple of $3pq$ for $all$ positive integers $n$. Find the least possible value of $p + q$.Solution:
We know from Fermat's Little Theorem that if $m$ is a prime number then, $n^m - n$ is divisible by all integers $m$ for all integers $n$.
But the given exponent $3pq$ is obviously a composite number. We also know from , that Carmichael Numbers satisfy Fermat's Theorem despite being composite.
Therefore, $3pq$ is a Carmichael Number which is a multiple of $3$. The smallest Carmichael number that is a multiple of $3$, is $561$.
Therefore $3pq = 561 = 3 \times 11 \times 17$
Since $p$ and $q$ are primes, $p$ and $q$ must be $11$ and $17$ in any order.
Therefore $p + q = 28$
Problem 29 of 30
For each positive integer $n$, consider the highest common factor $h_n$ of the two numbers $n!+1$ and $(n + 1)!$. For $n \lt 100$, find the largest value of $h_n$.Solution:
For the given problem, the number $n!$ is the product of all numbers $2 \longrightarrow n$, $n! + 1$ will leave a remainder of $1$ when divided by these numbers.
Therefore, $n! + 1$ and $(n+1)!$ will be coprimes, unless $n!+1$ is divisible by $n+1$.
From Wilson's Theory explained , we know that, $(p - 1)! \equiv -1 (mod\ p)$, if and only if $p$ is a prime. Therefore we can say that , $(p-1)! + 1$ is divisible by $p$ if and only if $p$ is a prime.
Therefore, $n! + 1$ is divisible by $(n+1)$ only if $n - 1$ is a prime number.
The largest prime number is $97$, therefore $96! + 1$ is divisible by $97$ and so is $97!$. Since there will be no other factors, the hcf will be $97$.
Hence the correct answer is $97$.
Problem 30 of 30
Consider the areas of the four triangles obtained by drawing the diagonals $AC$ and $BD$ of a trapezium $ABCD$. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are $1296$ and $576$, determine the square root of the maximum possible area of the trapezium to the nearest integer.Solution:
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Let us draw a trapezium $ABCD$ with diagonals $AC$ and $BD$ as shown below:
We know that the opposite triangles formed by the two unequal sides of a trapezium are equal in area.
Let us say $area[\triangle AOD] = area[\triangle BOC] = y$
$area[\triangle COD] = x$, and $area[\triangle AOB] = z$
Also, that the product of two areas of each pair of opposite triangles are equal.
Therefore, $y^2 = xz$
We can say that $x,\ y,\ z$ are in $G.P$ series, with $a$ as the first term and $r$ as the common ratio.
$x = a$
$y = ar$
$z = ar^2$
The area of the trapezium is given by $a + ar + ar + ar^2$
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Since we have two triangles with equal areas, out of the $6$ possible products only $3$ will be distinct values, which are:
$a^2r,\ a^2r^2,\ a^2r^3$
$\underline{Case\ 1}:$
$a^2r = 576$ $a^2r^2 = 1296$
$r = \dfrac{1296}{576} = \dfrac{9}{4}$
$\therefore a^2 = \dfrac{576 \times 4}{9}$
$\Rightarrow a = 16$
Therefore, the area of the four triangles are:
$16 + 36 + 36 + 81 = 169$
$\underline{Case\ 2}:$
$a^2r = 576$ $a^2r^3 = 1296$
$r^2 = \dfrac{1296}{576} = \dfrac{9}{4}$
$\Rightarrow r = \dfrac{3}{2}$
$\therefore a^2 = \dfrac{576 \times 2}{3}$
$\Rightarrow a = 8\sqrt{6}$
Therefore, the area of the four triangles are:
$8\sqrt{6} + 12\sqrt{6} + 12\sqrt{6} + 18\sqrt{6} = 50\sqrt{6}$
This is smaller than the area we got for $Case\ 1$, so we can ignore this.
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$\underline{Case\ 3}:$
Here we can simply argue that for an increasing $G.P$ series, if we are given the first two terms, we will get higher values for $a$ and $r$ as compared to what we will get if we treat the values as the last two terms.
We can ignore this case as well.
Therefore, the square root of the largest possible area is $\sqrt{169} = 13$