PRMO (IOQM) 2020

Problem 1 of 30

Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AB = 3CD$. Let $E$ be the midpoint of the diagonal $BD$. If $[ABCD] = n \times [CDE]$, what is the value of $n$?

(Here $[\Gamma]$ denotes the area of the geometrical figure $\Gamma$.)

Solution:

Let $h$ be the height of the given trapezium, as shown in the following diagram:

Therefore, height of $\triangle CDE = \dfrac{h}{2}$

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$[ABCD] = h \times \dfrac{CD + AB}{2}$

$= h \times \dfrac{CD + 3CD}{2}$

$= h \times \dfrac{4CD}{2}$

$= 2h \times CD$

$[CDE] = \dfrac{1}{2} \times \dfrac{h}{2} \times CD$

$= \dfrac{1}{4} h \times CD$

Therefore,

$n = \dfrac{[ABCD]}{[CDE]}$

$= \dfrac{2h \times CD \times 4}{h \times CD}$

$= 8$

Problem 2 of 30

A number $N$ in base $10$, is $503$ in base $b$ and $305$ in base $b+2$. What is the product of the digits of $N$?

Solution:

$503_b = (5 \times b^2 + 0 \times b + 3)_{10}$

$305_{b+2} = (3 \times (b+2)^2 + 0 \times (b+2) + 5)_{10}$

$503_b = 305_{b+2}$

$\Rightarrow (5 \times b^2 + 0 \times b + 3)_{10} = (3 \times (b+2)^2 + 0 \times (b+2) + 5)_{10}$

$\Rightarrow 5b^2 + 3 = 3 \times (b^2 + 4b + 4) + 5$

$\Rightarrow 5b^2 + 3 = (3b^2 + 12b + 12) + 5$

$\Rightarrow 5b^2 + 3 = 3b^2 + 12b + 17$

$\Rightarrow 2b^2 -12b -14 =0$

$\Rightarrow b^2 - 6b -7 =0$

$\Rightarrow (b -7)(b + 1) =0$

$\Rightarrow b = 7 \ \{\because b \gt 0\}$

$N_{10} = 503_7 = (5 \times 7^2 + 3)_{10} = 248_{10}$

Product of digits of $N = 2 \times 4 \times 8 = 64$

Problem 3 of 30

If $\sum\limits_{k=1}^{N}\dfrac{2k + 1}{(k^2 + k)^2} = 0.9999$ then determine the value of $N$

Solution:

We can solve this problem using the method of telescoping explained .

We can take the $n\xasuper{th}$ term $T_n$, for any $n$ and split it up as follows:

$T_n = \dfrac{2n + 1}{(n^2 + n)^2}$

$\Rightarrow T_n = \dfrac{(n+1)^2 - n^2}{n^2(n+1)^2}$

$\Rightarrow T_n = \dfrac{1}{n^2} - \dfrac{1}{(n+1)^2}$

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Therefore, the given equation becomes:

$\sum\limits_{k=1}^{N}\dfrac{2k + 1}{(k^2 + k)^2} = 0.9999$

$\Rightarrow \sum\limits_{k=1}^{N}\dfrac{1}{k^2} - \dfrac{1}{(k+1)^2} = 1 - \dfrac{1}{10^4}$

$\Rightarrow \left(\dfrac{1}{1^2} - \dfrac{1}{2^2}\right) + \left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right) + ... + \left(\dfrac{1}{N^2} - \dfrac{1}{(N+1)^2}\right) = 1 - \dfrac{1}{10^4}$

$\Rightarrow \dfrac{1}{1^2} - \cancel{\dfrac{1}{2^2}} + \cancel{\dfrac{1}{2^2}} - \cancel{\dfrac{1}{3^2}} + ... + \cancel{\dfrac{1}{N^2}} - \dfrac{1}{(N+1)^2} = 1 - \dfrac{1}{10^4}$

$\Rightarrow 1 - \dfrac{1}{(N+1)^2} = 1 - \dfrac{1}{10^4}$

$\Rightarrow \dfrac{1}{(N+1)^2} = \dfrac{1}{10^4}$

$\Rightarrow (N+1)^2 = 10^4$

$\Rightarrow N + 1= 100$

$\Rightarrow N = 99$

Problem 4 of 30

Let $ABCD$ be a rectangle in which $AB + BC + CD = 20$ and $AE = 9$ where $E$ is the mid-point of the side $BC$. Find the area of the rectangle.

Solution:

In the above diagram, let,

$AB = CD = x$ and $BC = AD = y$

From the given data,

$2x + y = 20$ $...eqn(i)$

and

$x^2 + \left(\dfrac{y}{2}\right)^2 = 9^2 \Rightarrow x^2 + \dfrac{y^2}{4} = 81$ $...eqn(ii)$

and we need to find out $xy$ without having to solve for $x$ and $y$ individually.

$\{eqn(i)\}^2 - 4 \times eqn(ii)$ gives us:

$4x^2 + y^2 + 4xy - 4x^2 - y^2 = 400 - 4 \times 81$

$\Rightarrow 4xy = 4 \times 19$

$\Rightarrow xy = 19$

Problem 5 of 30

Find the number of integer solutions to ${\Large{|}}|x| - 2020{\Large{|}} \lt 5$.

Solution:

We can divide the given inequality into two cases:

$|x| - 2020 < 5 \Rightarrow |x| \lt 2025$ $...(i)$

$-(|x| - 2020) < 5 \Rightarrow 2020 - |x| < 5 \Rightarrow |x| \gt 2015$ $...(ii)$

Combining the above two,

$2015 \lt |x| \lt 2025$

$\Rightarrow |x| \in \{2016, 2017, ..., 2024\}$

$\therefore x \in \{2016, -2016, 2017, -2017, ..., 2024, -2024\}$

There are $9 \times 2 = 18$ integer solutions for $x$

Problem 6 of 30

What is the least positive integer by which $2^5 \cdot 3^6 \cdot 4^3 \cdot 5^3 \cdot 6^7$ should be multiplied so that, the product is a perfect square.

Solution:

We can first convert the given expression as powers of its prime factors as follows:

$2^5 \cdot 3^6 \cdot 4^3 \cdot 5^3 \cdot 6^7$

$= 2^5 \cdot 3^6 \cdot (2 \cdot 2)^3 \cdot 5^3 \cdot (2 \cdot 3)^7$

$= 2^5 \cdot 3^6 \cdot 2^3 \cdot 2^3 \cdot 5^3 \cdot 2^7 \cdot 3^7$

$= 2^{18} \cdot 3^{13} \cdot 5^3$

Since, $2^{18}$ is already a perfect power.

$3^{13}$ needs to be multiplied by $3$ to give it an even power, similarly $5^3$ needs to be multiplied by $5$.

Therefore, we need to multiply the number by at least $15$ to make it a perfect square.

Problem 7 of 30

Let $ABC$ be a triangle with $AB = AC$. Let $D$ be a point on the segment $BC$ such that $BD = 48\dfrac{1}{61}$ and $DC = 61$. Let $E$ be a point on $AD$ such that $CE$ is perpendicular to $AD$ and $DE = 11$. Find $AE$.

Solution:

Let us draw the diagram for the problem as below, without any additional construction:

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Let $AB = AC = x$

$BD = 48\dfrac{1}{61}$

$DC = 61$

$ED = 11$

Let $AE = y$

$CE^2 = DC^2 - ED^2$

$\Rightarrow CE = \sqrt{61^2 - 11^2} = \sqrt{50 \times 72} = 60$

Applying Stewart's theorem, explained , to $\triangle ABC$ and cevian $AD$ we can write:

$AC^2(48\dfrac{1}{61}) + AB^2 (61) = (BC)\left(AD^2 + 48\dfrac{1}{61} \times 61\right)$

$\Rightarrow x^2 \cancel{\left(48\dfrac{1}{61} + 61\right)} = \cancel{\left(48\dfrac{1}{61} + 61\right)}\left(AD^2 + \dfrac{2929}{61} \times 61\right)$

$\Rightarrow x^2 = (11 + y)^2 + 2929$

Substituting $x^2 = y^2 + 60^2$ from $\triangle AEC$

$y^2 + 60^2= y^2 + 22y + 121 + 2929$

$\Rightarrow 60^2= 22y + 121 + 3050$

$\Rightarrow 22y = 3600 - 3050 = 550$

$\Rightarrow y = 25$

Problem 8 of 30

A $5$-digit number (in base $10$) has digits $k$, $k + 1$, $k + 2$, $3k$, $k + 3$ in that order, from left to right. If this number if $m^2$ for some natural number $m$, find the sum of the digits of $m$.

Solution:

We can significantly reduce the amount of calculations in this problem by using elimination.

Since, $3k$ is a digit, $0 \le k \le 3$

If $k = 0$ then the number is a $4$-digit number,

$\therefore k \in \{1, 2, 3\}$

Now, we can take a look at the two least significant digits of the number:

For $k = 1$, these two digits are:

$34$

Any number ending with $34$ is divisible by $2$ but not by $4$, therefore, cannot be a perfect square.

For $k = 2$, these two digits are

$65$

Similarly as before, a number ending with $65$ is divisible by $5$ and not by $25$, therefore, cannot be a perfect square.

For $k = 3$, these two digits are:

$96$ which can be the last two digits of a perfect square:

Therefore, $3$ is the only possible value of $k$, with the given $5$-digit number being:

$34596$

Now, finding the square root of $34596$ using the long division method explained , we get

$m = 186$

Therefore, the sum of the digits of $m$ is $1 + 8 + 6 = 15$

Problem 9 of 30

Let $ABC$ be a triangle with $AB = 5$, $AC = 4$, $BC = 6$. The internal angle bisector of $C$ intersects the side $AB$ at $D$. Points $M$ and $N$ are taken on sides $BC$ and $AC$, respectively, such that $DM \parallel AC$ and $DN \parallel BC$. If $(MN)^2 = \dfrac{p}{q}$ where $p$ and $q$ are relatively prime positive integers then what is the sum of the digits of $|p - q|$?

Solution:

We draw the diagram as follows:

In the above diagram,

$AD:DB = AC:CB = 4:6 \Rightarrow AD = 2$ and $DB = 3$

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We can use Stewart's theorem from , to find the length of the cevian $CD$.

$CB^2\cdot AD + AC^2 \cdot DB = AB(CD^2 + AD \cdot DB)$

$\Rightarrow 6^2 \cdot 2 + 4^2 \cdot 3 = 5(CD^2 + 2 \cdot 3)$

$\Rightarrow 120 = 5(CD^2 + 6)$

$\Rightarrow CD = \sqrt{18}$

Also, $AN:NC = AD:DB$

$\Rightarrow AN = 4 \times \dfrac{2}{5} = \dfrac{8}{5}$ and

$NC = 4 \times \dfrac{3}{5} = \dfrac{12}{5}$

$CMDN$ is a parallelogram, and diagonal $CD$ bisects $\angle C$

Therefore, $CMDN$ is a rhombus, with $MN \perp CD$

Therefore, $\left(\dfrac{MN}{2}\right)^2 + \left(\dfrac{CD}{2}\right)^2 = NC^2$

$\Rightarrow \dfrac{MN^2}{4} = \left(\dfrac{12}{5}\right)^2 - \dfrac{18}{4}$

$\Rightarrow \dfrac{MN^2}{4} = \dfrac{12^2\cdot 4 - 18 \cdot 5^2}{5^2 \cdot 4}$

$\Rightarrow MN^2 = \dfrac{144 \times 4 - 18 \times 25}{25} = \dfrac{126}{25}$

Therefore,

$p = 126$, $q = 25$ and $|p - q| = 101$

Sum of digits $= 2$

Problem 10 of 30

Five students take a test on which any integer score from $0$ to $100$ inclusive is possible. What is the largest possible difference between the median and the mean of the scores? (The median of a set of scores is the middlemost score when the data is arranged in increasing order. It is exactly the middle score when there are an odd number of scores and it is the average of the two middle scores when there are and even number of scores.)

Solution:

The difference between the median and mean can be maximised by:

$I)$ Maximising the median and minimising the mean, or

$II)$ Maximising the mean and minimising the median

$\underline{Case\ I}:$

The maximum value for the median $(3^{rd} term)$ is $100.$ This implies that the $4^{th}$ and $5^{th}$ term also have to be $100$.

For the mean to be minimum the sum has to be minimum and this is possible when the $1^{st}$ and the $2^{nd}$ term have to be minimum, that is $0.$

Therefore, the terms are $0, 0, 100, 100$ and $100.$

The median of these $5$ terms is $100$ and the mean is $60.$ The maximum difference of median and mean is $40$.

$\underline{Case\ II}:$

The median can be set to $0$, implying that the first and the second term will also be $0$.

To maximise the mean we can set the fourth and the fifth term to $100$.

Therefore, the distribution becomes:

$0, 0, 0, 100$ and $100$

The mean of the above distribution is $200 \div 5 = 40$ and the median is $0$.

Therefore, the difference between the mean and the median is $40$.

Combining the above two cases we get the maximum possible difference between the mean and the median as $40$.

Problem 11 of 30

Let $X = \{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$ and

$S = \{(a,b) \in X \times X : x^2 + ax + b$ and $x^3 + bx + a$ have at least a common real zero$\}$.

How many elements are there in $S$?

Solution:

Let us find the GCD of the two given polynomials using the Euclid's method applied to polynomials as follows:

$\begin{array}{rl}\phantom{000000000}x - a \\ \phantom{00}x^2+ax+b \enclose{longdiv}{x^3 + 0x^2 + bx + a} \\ \underline{-(x^3+ax^2+bx)} \phantom{000} \\ -ax^2+a \\ \underline{-(-ax^2 -a^2x - ab)}\\ a^2x + ab + a \end{array}$

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The polynomial $x^2 + ax + b$ must be divisible by $a^2x + ab + a$

$a^2x + ab + a = 0$ must be a zero for the polynomial $x^2 + ax + b$

Therefore,

$a^2x = -a(b + 1)$

For $a = 0$

the two polynomials will reduce to:

$x^2 + b$

$x^3 + bx = x(x^2 + b)$

Therefore, $x^2 + b$ is the common root.

If $x^2 + b = 0$

$x^2 = -b$, for $x$ to be real, $b$ has to be negative or zero.

Therefore, the possible values of $b$ are $-5, -4, -3, -2, -1, 0$

For $a \ne 0$

$x = \dfrac{-(b + 1)}{a}$

Therefore,

$\left(-\dfrac{b+1}{a}\right)^2 - a\dfrac{b+1}{a} + b = 0$

$\left(\dfrac{b+1}{a}\right)^2 - b - 1 + b = 0$

$\dfrac{b + 1}{a} = \pm 1$

$b = a - 1$ or $b = -a - 1$

If $b = a-1$, then

for $a \in \{-4, -3,..., 4, 5\}$ $b$ will be in $\{-5, -4, ..., 3, 4\}$ $...9$ cases (not counting $a = 0$)

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if $b = -a - 1$, then

for $a \in \{-5, -4, ... 3, 4\}$ $b$ will be in $\{4, 3,...-4, -5\}$ $...9$ cases, excluding $a = 0$

Therefore, there are $9 + 9 + 6 = 24$ ordered pair $(a, b)$ satisfying the given condition.

Problem 12 of 30

Given a pair of concentric circles, chords $AB$, $BC$, $CD,\ ...$ of the outer circle are drawn such that they all touch the inner circle.

If $\angle ABC = 75^\circ$, how many chords can be drawn before returning to the starting point?

Solution:

As shown in the figure above, every tangent makes an angle of $105^\circ$ with the center, in other words, the point rotates by an angle of $105^\circ$ along the outer circle with each tangent.

If we assume that the tangent reaches the starting point $A$, after $m$ tangents are drawn, that is after $m$ rotations of $105^\circ$ each, then,

$m \times 105 = n \times 360$ for some integer $m$ and $n$.

$\Rightarrow n = \dfrac{105 \times m}{360} = \dfrac{7 \times m}{24}$

For $n$ to be an integer, $m$ must be a multiple of $24$ therefore, the minimum value of $m$ is $24$

Problem 13 of 30

Find the sum of all positive integers $n$ for which $|2^n + 5^n - 65|$ is a perfect square.

Solution:

$5^n - 65 \equiv 0\ (mod\ 10)$

And,

$2^n \equiv \{2, 4, 8, 6, 2, 4, 8, 6,...\}\ (mod\ 10)$

Therefore,

$2^n + 5^n - 65 \equiv \{2, 4, 8, 6\}\ (mod\ 10)$

For $n = 1, 3, 5...$ the given expression will end with $2$ or $8$ which cannot be perfect squares, and for $n = 2, 4, 6...$ it will end with $4$ or $6$ which can be perfect squares.

Therefore, there is no odd integer solution for $n$.

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For even solutions, let $n = 2k$

$5^{2k} + 2^{2k} - 65$

For, $k > 3$,

$2k > 6$

$2^{2k} > 65$

$\Rightarrow 2^{2k} - 65 > 0$

$\Rightarrow 5^{2k} + 2^{2k} - 65 > 5^{2k}$

$\Rightarrow 5^{2k} + 2^{2k} - 65 > (5^k)^{2}$

Now

$(5^k + 1)^2$

$= 5^{2k} + 2\cdot 5^k + 1$

$= 5^{2k} + 2(2^2 + 1)^k + 1$

$= 5^{2k} + 2(2^{2k} + \xacomb{k}{1} 2^{2k - 2} + ...+ 1) + 1$

$= 5^{2k} + 2\cdot 2^{2k} + 2\cdot \xacomb{k}{1} 2^{2k - 2} + ...+ 2 + 1$

Now,

$2\cdot 2^{2k} \gt 2^{2k}$

$\Rightarrow 2\cdot 2^{2k} + 2\cdot \xacomb{k}{1} 2^{2k - 2} + ...+ 2 + 1 \gt 2^{2k} - 65$

(In the above step we added a number of positive terms to the LHS and subtracted a positive quantity from the RHS. Therefore, the inequality relation $LHS \gt RHS$ still holds)

$\Rightarrow 5^{2k} + 2\cdot 2^{2k} + 2\cdot \xacomb{k}{1} 2^{2k - 2} + ...+ 2 + 1 \gt 5^{2k} + 2^{2k} - 65$

$\Rightarrow (5^k + 1)^2 \gt 5^{2k} + 2^{2k} - 65$

Therefore, for all $k \gt 3$,

$(5^k)^2 \lt 5^{2k} + 2^{2k} - 65 \lt (5^k + 1)^2$

Since $5^{2k} + 2^{2k} - 65$ lies between two consecutive perfect squares, it cannot be a perfect square itself.

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Therefore, there is no integer solution for $k > 3$

For, $1 \le k \le 3$,

For $k = 1$, $n = 2$,

$|5^2 + 2^2 - 65| = |-36| = 36$ which is a perfect square.

For $k = 2$, $n = 4$,

$|5^4 + 2^4 - 65| = |576| = 24^2$ , which is also a perfect square.

For $k = 3$, $n = 6$

$5^6 + 2^6 - 65$

$= 5^6 + 2^6 - 2^6 - 1$

$= 5^6 - 1$

$= (5^3 - 1)(5^3 + 1)$

$= 124 \times 126$

$= 2^2 \times 31 \times 2 \times 63$ which is not a perfect square.

Therefore, there are only two integer solution for $n$,

$n = 2$ and $n = 4$

The answer is $2 + 4 = 6$

Problem 14 of 30

The product $55 \times 60 \times 65$ is written as the product of five distinct positive integers. What is the least possible value of the largest of these integers?

Solution:

We can start by representing the given expression as a product of its prime factors as follows:

$55 \times 60 \times 65$

$= 5 \times 11 \times 2 \times 2 \times 3 \times 5 \times 5 \times 13$

Now we can for five groups, starting with the maximum prime numbers:

$13 \times 11 \times 5 \times (5) \times (5 \times 2 \times 2 \times 3)$

Now we start by moving the smallest number from the fifth group to the smallest of the remaining groups:

$= 13 \times 11 \times 5 \times (5 \times 2) \times (5 \times 2 \times 3)$

$= 13 \times 11 \times (5 \times 3) \times (5 \times 2) \times (5 \times 2)$

Since all the integers are not distinct, we need to move one more $2$ from the last group to the previous group, giving us:

$= 13 \times 11 \times (5 \times 3) \times (5 \times 2 \times 2) \times (5)$

$= 13 \times 11 \times 15 \times 20 \times 5$

Therefore, the minimum possible value of largest of the factors is $20$

Problem 15 of 30

Three couples sit for a photograph in $2$ rows of three people each such that no couple is sitting in the same row next to each other or in the same column one behind the other. How many arrangements are possible?

Solution:

We can start by looking at the different ways we can fill up the front row.

There are two mutually exclusive ways of filling up the front row. These are:

$\underline{Case\ 1}$:

- All three individuals in this row are from different couples

Let us denote the individuals as $X_1$, $X_2$, $X_3$, where $X_i$ can either be the husband or wife from the $i\xasuper{th}$ couple.

There are $3 \times 2 = 6$ arrangements of the $X$s

Now, each of these $X$s can be either the husband or wife.

Therefore, there are $6 \times 2 \times 2 \times 2 = 48$ possible arrangements.

For each of these arrangements, there is exactly one arrangement for the second row, where each person occupies the same column as his/her spouse in the first row, that is all the persons in the second row are in the wrong positions.

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Any derangement of this second row placement will give us a valid re-arrangement where nobody is occupying the same column as his/her spouse in the first row. The number of derangements of any $3$ objects can be obtained as explained .

$!3 = 3!\left(\dfrac{1}{0!} - \dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!}\right)$

$= 3!\left(\dfrac{6 - 6 + 3 - 1}{6}\right)$

$= 3! \times \dfrac{2}{6} = 2$

Therefore, there are $48 \times 2 = 96$ different arrangements in this case.

$\underline{Case\ 2}:$

- One couple (not sitting next to each other) and one individual.

If the $i\xasuper{th}$ couple and an individual from the $j\xasuper{th}$ couple is occupying the front row, then they can only be arranged in the following ways:

$H_iX_jW_i$ or $W_iX_jH_i$ (where $X_j$ could be either the husband or the wife of the $j\xasuper{th}$ couple).

We can select one couple in $3$ ways, and the remaining individual in $4$ ways, and for each of these selections there are $2$ ways of arranging them.

Therefore, there are $3 \times 4 \times 2 = 12$ possible arrangements for the first row.

In this case, the spouse of $X_j$ cannot occupy the center seat of the second row and has to occupy one of the side seats, which means the remaining couple, has to occupy two adjacent seats.

Therefore, there is no valid arrangement for the second row in this case.

Therefore, the total number of arrangements satisfying the given conditions is $96$.

Problem 16 of 30

The sides $x$ and $y$ of a scalene triangle satisfy $x + \dfrac{2\Delta}{x} = y + \dfrac{2\Delta}{y}$, where $\Delta$ is the area of the triangle. If $x = 60$, $y = 63$, what is the length of the largest side of the triangle?

Solution:

Given that:

$x + \dfrac{2\Delta}{x} = y + \dfrac{2\Delta}{y}$

$\Rightarrow x - y = 2\Delta \left(\dfrac{1}{y} - \dfrac{1}{x}\right)$

$\Rightarrow \cancel{x - y} = 2\Delta \dfrac{\cancel{(x - y)}}{xy}$ $\because$ The triangle is scalene $x \ne y$

$\Rightarrow \dfrac{1}{2}xy = \Delta$

Since the area of the triangle is half the product of two sides, the angle between them is $90^\circ$

Therefore, the third side $(z)$ is the hypotenuse, therefore, is the largest side, and can be found using:

$z = \sqrt{63^2 + 60^2}$

$= 3 \sqrt(21^2 + 20^2)$

$= 3 \sqrt{841}$

$= 3 \times 29$

$= 87$

Problem 17 of 30

How many two digit numbers have exactly $4$ positive factors? (Here $1$ and the number $n$ itself are also considered as factors of $n$.)

Solution:

As we have seen , when a number $N$ can be expressed in terms of its prime factors as:

$I)$ $N = {p_1}^1 \times {p_2} ^ 1$ where $p_1$ and $p_2$ are distinct prime numbers, or

$II)$ $N = p^3$ where $p$ is any prime number,

then $N$ has exactly $4$ divisors, including $1$ and the number itself, where $p_1$, $p_2$ and $p$ are prime numbers.

$\underline{Case\ 1}:$

As $p_1, p_2 > 1$ and $p_1 \times p_2 < 100$,

$1 \lt p_1, p_2 \le 50$

$p_1, p_2 \in \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$

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Without loss of generality, we assume $p_1 \lt p_2$

For $p_1 = 2$

$p_2 \in \{5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$

Total $13$ numbers

For $p_1 = 3$

$p_2 \in \{5, 7, 11, 13, 17, 19, 23, 29, 31\}$

Total $9$ numbers

For $p_1 = 5$

$p_2 \in \{7, 11, 13, 17, 19\}$

Total $5$ numbers

For $p_1 = 7$

$p_2 \in \{11, 13\}$

Total $2$ numbers

$\underline{Case\ II}:$

Since $2^3 \lt 10$ and $5^3 > 99$,

$p \in \{3\}$

Total $1$ number

Therefore, there is a total of $13 + 9 + 5 + 2 + 1 = 30$ two digit numbers having exactly $4$ positive divisors.

Problem 18 of 30

If,

$\sum\limits_{k = 1}^{40}\left(\sqrt{1 + \dfrac{1}{k^2} + \dfrac{1}{(k+1)^2}}\right) = a + \dfrac{b}{c}$

where $a, b, c \in \mathbb{N}, b \lt c, gcd(b, c) = 1$, then what is the value of $a + b$

Solution:

We will use the method of telescoping explained , to solve this problem.

We can split up the $k\xasuper{th}$ term as follows:

$\sqrt{1 + \dfrac{1}{k^2} + \dfrac{1}{(k+1)^2}}$

$= \sqrt{\dfrac{k^2(k+1)^2 + (k+1)^2 + k^2}{k^2(k+1)^2}}$

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$= \sqrt{\dfrac{k^2(k^2 + 2k+1) + k^2 + 2k +1 + k^2}{k^2(k+1)^2}}$

$= \sqrt{\dfrac{k^4 + 2k^3+k^2 + k^2 + 2k +1 + k^2}{k^2(k+1)^2}}$

$= \sqrt{\dfrac{k^4 + 2k^3+3k^2 + 2k +1}{k^2(k+1)^2}}$

$= \sqrt{\dfrac{(k^2+k+1)^2}{k^2(k+1)^2}}$

$= \sqrt{\dfrac{(k^2+2k+1-k)^2}{k^2(k+1)^2}}$

$= \sqrt{\dfrac{((k+1)^2-k)^2}{k^2(k+1)^2}}$

$= \dfrac{(k+1)^2-k}{k(k+1)}$

$= \dfrac{(k+1)^2}{(k)(k+1)} - \dfrac{k}{k(k+1)}$

$= \dfrac{k+1}{k} - \dfrac{1}{k+1}$

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$= 1 + \dfrac{1}{k} - \dfrac{1}{k+1}$

Therefore,

$\sum\limits_{k = 1}^{40}\left(\sqrt{1 + \dfrac{1}{k^2} + \dfrac{1}{(k+1)^2}}\right)$

$= \sum\limits_{k = 1}^{40}\left(1 + \dfrac{1}{k} - \dfrac{1}{k+1}\right)$

$\sum\limits_{k = 1}^{40} 1 + \sum\limits_{k = 1}^{40}\left(\dfrac{1}{k} - \dfrac{1}{k+1}\right)$

$= 40 + \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{40} - \dfrac{1}{41}$

$= 40 + 1 - \dfrac{1}{41}$

$= 40 +\dfrac{40}{41}$

Therefore, $a=40,$ $b=40$ and $c=41.$

$\Rightarrow a+b = 40 + 40 = 80$

Problem 19 of 30

Let $ABCD$ be a parallelogram. Let $E$ and $F$ be midpoints of $AB$ and $BC$ respectively. The lines $EC$ and $FD$ intersect in $P$ and form four triangles $APB$, $BPC$ and $CPD$ and $DPA$. If the area of the parallelogram is $100$ sq. units, what is the maximum area in sq. units of a triangle among this four triangles?

Solution:

Let the areas be denoted as follows:

$[\triangle PFB] = a$, $[\triangle PEB] = b$, $[\triangle PAD] = c$, $[\triangle PDC] = d$

And, we get the diagram as follows:

-----------book page break-----------

$[\triangle PFC] = [\triangle PFB] = a$ (equal bases and height)

$[\triangle PEA] = [\triangle PEB] = b$ (equal bases and height)

$2a + c = \dfrac{100}{2} = 50$ $...eqn(i)$

$2b + d = \dfrac{100}{2} = 50$ $...eqn(ii)$

$a + d = \dfrac{100}{4} = 25$ $...eqn(iii)$

$2a + b = \dfrac{100}{4} = 25$ $...eqn(iv)$

$2 \times (iii) - (iv)$ gives us

$2d - b = 25$ $...eqn(v)$

$(ii) + 2 \times (v)$ gives us:

$5d = 100$

$\Rightarrow d = 20$

$\therefore a = 5$, $c = 40$ and $b = 15$

Therefore,

$[\triangle APB] = 2b = 30$

$[\triangle BPC] = 2a = 10$

$[\triangle CPD] = d = 20$

$[\triangle DPA] = c = 40$

Therefore, the maximum area out of the three triangles is $40$

Problem 20 of 30

A group of women working together at the same rate can build a wall in $45$ hours. When the work started, all the women did not start working together. They joined the work over a period of time, one by one, at equal intervals. Once at work, each one stayed till the work was complete. If the first woman worked $5$ times as many hours as the last woman, for how many hours did the first woman work?

Solution:

Let the time worked by the last woman be $a$ hours, and the interval between any two of them joining the work be $d$ hours and $n$ be the total number of woman in the team.

Therefore, the first woman worked for

$a + (n - 1)d$ hours

Therefore,

$a + (n - 1)d = 5a$

$\Rightarrow 4a = (n - 1)d$

Also,

$\dfrac{n}{2}\left\{2a + (n-1)d\right\} = 45n$

$\Rightarrow 2a + (n-1)d = 45$

$2a + 4a = 90$

$a = 15$

Therefore, the first woman worked for $15 \times 5 = 75\ hours$

Problem 21 of 30

A total fixed amount of $N$ thousand rupees is given to three persons $A$, $B$, $C$, every year, each being given an amount proportional to her age. In the first year, $A$ got half the total amount. When the sixth payment was made, $A$ got six-seventh of the amount that she had in the first year; $B$ got $Rs\ 1000$ less than that she had in the first year; and $C$ got twice of that she had in the first year. Find $N$.

Solution:

Let the total amount of money, that is $N \times 1000 = n$

Let the ages of $A$, $B$ and $C$ during the first year be $a$, $b$ and $c$ respectively.

In the first year $A$ received half the money, that is $\dfrac{n}{2}$, therefore,

$\dfrac{a}{a + b + c} = \dfrac{1}{2}$

$\Rightarrow a = b + c$

During the $6\xasuper{th}$ payment, $A$ got $\dfrac{6}{7}$ of her first year payment, that is $\dfrac{6n}{14}$, therefore,

$\dfrac{a + 5}{a + b + c + 15} = \dfrac{6}{14}$

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$\Rightarrow \dfrac{a + 5}{a + a + 15} = \dfrac{6}{14}$

$\Rightarrow 14a + 70 = 12a + 90$

$\Rightarrow a = 10$ and $a + b + c = a + a = 20$

Since $C$ got twice the money during the $6\xasuper{th}$ year,

$\dfrac{c + 5}{a + b + c + 15} = 2 \times \dfrac{c}{a + b + c}$

$\Rightarrow \dfrac{c + 5}{35} = 2 \times \dfrac{c}{20} = \dfrac{c}{10}$

$\Rightarrow 10c + 50 = 35c$

$\Rightarrow c = 2$

Therefore, $b = b + c - c = a - c = 10 - 2 = 8$

Since, $B$ got $1000$ less in the sixth year,

$n \times \dfrac{b}{a + b + c} - n \times \dfrac{b + 5}{a + b + c + 15} = 1000$

$\Rightarrow n \times \left(\dfrac{8}{20} - \dfrac{13}{35}\right) = 1000$

$\Rightarrow n = \dfrac{1000 \times 70}{2} = 35000$

$\therefore N = \dfrac{n}{1000} = 35$

Problem 22 of 30

In triangle $ABC$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$, respectively. If $PQ = 7$, $QR = 6$ and $RS = 8$ what is the area of triangle $ABC$?

Solution:

We can draw the diagram for the problem as shown below:

We can denote the intersection point of the two angle bisectors as $I$, and join $AI$.

$AI$ is the angle bisector of $\angle A$

-----------book page break-----------

$\angle PBR = 90^\circ$ (angle between bisectors of internal and external angle)

Therefore, $APBR$ is a rectangle.

If we draw the diagonal $PR$ and name its intersection point with diagonal $AB$ as $M$, then:

$\angle MPB = \angle MBP$

Therefore,

$\angle MPB + \angle PBC$

$= \angle MBP + \angle MBP + \angle ABC = 2\angle MBP + 2\angle ABR = 2 \angle PBR$

$= 180^\circ$

Therefore,

$PR \parallel BC$

Similarly, if we join the diagonal $QS$ of rectangle $AQCS$ we can show that $QS \parallel BC$.

(We omitted this in our diagram for clarity's sake)

We can join $QR$ for the next part.

$\angle RBA = \dfrac{\angle B}{2}$

$\therefore \angle BAR = 90 - \dfrac{\angle B}{2}$

$\angle BAI = \dfrac{\angle A}{2} = \dfrac{180 - \angle B - \angle C}{2} = 90 - \dfrac{\angle B}{2} - \dfrac{\angle C}{2}$

$\angle IAR = \angle BAR - \angle BAI$

$= \left(90 - \dfrac{\angle B}{2}\right) - \left(90 - \dfrac{\angle B}{2} - \dfrac{\angle C}{2}\right)$

$= \dfrac{\angle C}{2}$

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$AQIR$ is a cyclic quadrilateral, because opposite angles $\angle AQI + \angle ARI = 180^\circ$

$\therefore \angle IQR = \angle IAR = \dfrac{\angle C}{2}$

Since alternate angles, $\angle IQR = \dfrac{\angle C}{2} = \angle ICB$

$QR \parallel BC$

Because, $PR || BC$, $QS \parallel BC$ and $QR || BC$, points $P$, $Q$, $R$ and $S$ are collinear.

Now that we have shown that these points are collinear, we can redraw the diagram for more clarity as follows:

Let the intersection points of $PS$ with $AB$ and $AC$ be named $M$ and $N$ respectively.

$PR = PQ + QR = 7 + 6 = 13$

$\Rightarrow AB = PR = 13$ (diagonal of rectangle $APBR$)

-----------book page break-----------

Similarly, in rectangle $AQCS$,

$AC = QS = QR + RS = 6 + 8 = 14$

$PM = MB = \dfrac{AB}{2} = \dfrac{13}{2}$

$NS = NC = \dfrac{AC}{2} = 7$

$\therefore MN = PS - \dfrac{13}{2} - 7 = 6 + 7 + 8 - \dfrac{13}{2} - 7 = \dfrac{15}{2}$

Since $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively,

$MN = \dfrac{BC}{2}$

$\therefore BC = MN \times 2 = 15$

Semiperimeter of $\triangle ABC = \dfrac{13 + 14 + 15}{2} = 21$

$\therefore ar[\triangle ABC] = \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$

$= \sqrt{21 \times 8 \times 7 \times 6}$

$= 3 \times 4 \times 7 = 84$

Problem 23 of 30

The incircle $\Gamma$ of a scalene triangle $ABC$ touches $BC$ at $D$, $CA$ at $E$ and $AB$ at $F$. Let $r_A$ be the radius of the circle inside $ABC$ which is tangent to $\Gamma$ and the sides $AB$ and $AC$. Define $r_B$ and $r_C$ similarly. If $r_A = 16$, $r_B = 25$ and $r_C = 36$, determine the radius of $\Gamma$.

Solution:

Using the corner circle property of incircle explained , we have:

$r = \sqrt{16 \cdot 25} + \sqrt{25 \cdot 36} + \sqrt{36 \cdot 16}$

$= 4 \times 5 + 5 \times 6 + 6 \times 4 = 74$

Problem 24 of 30

A light source at the point $(0, 16)$ in the coordinate plane casts light in all directions. A disc (a circle along with its interior) of radius $2$ with center at $(6, 10)$ casts a shadow on the $X$ axis. The length of the shadow can be written in the form $m\sqrt{n}$ where $m$, $n$ are positive integers and $n$ is square-free. Find $m + n$.

Solution:

To simplify this problem we will give it an axes translation such that the new origin becomes the center of the circle.

To bring the origin to the center we need to translate the axes by $(+6, +10)$

Therefore, the new coordinates for the source of the light becomes $(0-6, 16 - 10) = (-6, 6) $

The equation of the old $X$-axis $y = 0$ becomes $y = -10$

The following diagram shows the coordinates after the above axes translation:

-----------book page break-----------

The equations of the tangents, from the given point to the circle as explained , are given by:

$y - 6 = \left(\dfrac{-(-6)(6) \pm 2\sqrt{(-6)^2 + 6^2 - 2^2}}{2^2 - 36}\right)(x + 6)$

$\Rightarrow y - 6 = \left(\dfrac{36 \pm 2\sqrt{36 + 36 - 4}}{2^2 - 36}\right)(x + 6)$

$\Rightarrow y - 6 = \left(\dfrac{36 \pm 2\sqrt{68}}{-32}\right)(x + 6)$

-----------book page break-----------

Substituting $y = -10$ in the above equation, we get:

$-16 = \left(\dfrac{36 \pm 2\sqrt{68}}{-32}\right)(x + 6)$

$\Rightarrow x + 6 = -16 \times \left(\dfrac{-32}{36 \pm 2\sqrt{68}}\right)$

$\Rightarrow x = \dfrac{32 \times 16}{36 \pm 2\sqrt{68}} - 6$

Let,

$x_1 = \dfrac{32 \times 16}{36 + 2\sqrt{68}} - 6$

$x_2 = \dfrac{32 \times 16}{36 - 2\sqrt{68}} - 6$

Therefore,

$x_2 - x_1 = \dfrac{32 \times 16}{36 - 2\sqrt{68}} - \dfrac{32 \times 16}{36 + 2\sqrt{68}}$

$= 32 \times 16 \left(\dfrac{1}{36 - 2\sqrt{68}} - \dfrac{1}{36 + 2\sqrt{68}}\right)$

$= 32 \times 16 \left(\dfrac{4\sqrt{68}}{36^2 - 4\times 68}\right)$

$= 32 \times 16 \left(\dfrac{\sqrt{68}}{18^2 - 68}\right)$

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$= 32 \times 4 \left(\dfrac{\sqrt{68}}{9^2 - 17}\right)$

$= 32 \times 4 \left(\dfrac{\sqrt{68}}{64}\right)$

$= 2\sqrt{68}$

$= 4 \sqrt{17}$

Therefore, $m = 4$ and $n = 17$ and

$m + n = 21$

Problem 25 of 30

For a positive integer $n$, let $\langle n \rangle$ denote the perfect square integer closest to $n$. For example, $\langle 74 \rangle = 81$, $\langle 18 \rangle = 16$. If $N$ is the smallest positive integer such that

$\langle 91 \rangle \cdot \langle 120 \rangle \cdot \langle 143 \rangle \cdot \langle 180 \rangle \cdot \langle N \rangle = 91 \cdot 120 \cdot 143 \cdot 180 \cdot N$

find the sum of the squares of the digits of $N$.

Solution:

$\langle n \rangle$ is the perfect square closest to $n.$ Therefore,

$\langle 91 \rangle = 100$

$\langle 120 \rangle =121$

$\langle 143 \rangle = 144$

$\langle 180 \rangle = 169$

$\langle 91 \rangle \cdot \langle 120 \rangle \cdot \langle 143 \rangle \cdot \langle 180 \rangle \cdot \langle N \rangle = 91 \cdot 120 \cdot 143 \cdot 180 \cdot N$

$\Rightarrow 100 \cdot 121 \cdot 144 \cdot 169 \cdot \langle N \rangle = 91 \cdot 120 \cdot 143 \cdot 180 \cdot N$

$\Rightarrow \langle N \rangle = \dfrac{91 \cdot 120 \cdot 143 \cdot 180 \cdot N}{100 \cdot 121 \cdot 144 \cdot 169}$

$\Rightarrow \langle N \rangle = \dfrac{21}{22} N$

For $\langle N \rangle$ to be a perfect square $N$ has to be $22 \times 21 = 462$

Sum of squares of digits of $N = 4^2 + 6^2 + 2^2 = 16 + 36 + 4 = 56$

Problem 26 of 30

In the figure below, $4$ of the $6$ disks are to be colored black and $2$ are to be colored white. Two colorings that can be obtained by a rotation or a reflection of the entire figure are considered the same.

There are only four such colorings for the given two colors, as shown in Figure 1.

In how many ways can we color the $6$ disks such that $2$ are colored black, 2 are colored white, $2$ are colored blue with the given identification condition?

Solution:

-----------book page break-----------

In this question part of the answer is given in the question itself. The given example uses $4$ black and $2$ white, and has illustrated all the possible arrangements.

Now, all we need to do is select any two black and color them blue, after which if we eliminate the reflections we should have the necessary number of arrangements. Also, note that since none of the $4$ given arrangements is a rotation of each other, coloring any two of the given blacks will also not lead to a rotation of any other arrangement.

For each of the given $4$ arrangements we can choose two blue circles in $\xacomb{4}{2}$ ways, therefore, there are

$4 \times \xacomb{4}{2} = 4 \times 6 = 24$ ways to paint before eliminating the reflections.

Now let us observe the given illustrations. Other than $(ii)$ all are symmetric about their respective vertical axis.

If we select any one of the arrangements $(i)$, $(iii)$ or $(iv)$ and draw all the $6$ cases, we will see that the arrangements that are still symmetric about the vertical do not have a distinct reflection, while the arrangements that are non-symmetric about the median have a distinct reflection.

We will draw all the $6$ arrangements of illustration $(i)$ for this purpose.

and

does not have any separate reflection since they are symmetric about the vertical axis and they are their own reflection.

-----------book page break-----------

While the following arrangements are not symmetric about the vertical axis and have a distinct reflection included in the set.

$\stackrel{reflection}{\Longleftrightarrow}$

and

$\stackrel{reflection}{\Longleftrightarrow}$

Similarly, each of illustrations $(iii)$ and $(iv)$ will also have $2$ arrangements that are reflections of another arrangement.

Therefore, the total number of reflections in the new coloring arrangements is $2 \times 3 = 6$

After eliminating the reflections there will be:

$24 - 6 = 18$ arrangements.

Problem 27 of 30

A bug travels in the coordinate plane moving only along the lines that are parallel to the $x$ axis or $y$ axis. Let $A = (-3, 2)$ and $B = (3, -2)$. Consider all possible paths of the bug from $A$ to $B$ of length atmost $14$. How many points with integer coordinates lie on at least one of these paths.

Solution:

We can see that the minimum path length for the ant to reach point $B$ from point $A$ is length $\{3 - (-3)\} + \{2 - (-2)\} = 10$, and the bug can cover this distance by moving $6$ units to the right and $4$ units downwards without any particular order.

Therefore all points on and inside the rectangle with $A$ and $B$ as two opposite corners can be included in one or more paths.

Since the path length of $14$ is allowed, the bug can take a path in the upward or the left direction, whose maximum length will be $2$, that is because it needs another $2$ extra lengths to come back to the original path.

Therefore, we can draw a rectangle with $A' = -5, 4$ and $B' = 5, -4$ as the opposite corners and see what happens.

-----------book page break-----------

We can draw the diagram of the top left corner of this grid as shown below:

We can see that from point $A$ to $A'$ the minimum distance is $2 + 2 = 4$, and the minimum distance from point $A'$ to $B$ is $\{3 - (-5)\} + \{4 - (-2)\} = 14$

Therefore, the minimum path from $A$ to $B$ via $A'$ is $14 + 4 = 18$ which is not within the given limit.

Similarly the minimum distance from $A$ to $B$ via points $P$ or $Q$ is $3 + 13 = 16$.

Therefore, the point $A'$, $P$ and $Q$ are the three points in the top left corner of the grid which cannot be included in a path of length $14$.

Similarly, $3$ points on each corner of the grid cannot be included in the given path limit, all other points on or inside the rectangle $A'B'$ can be covered within the given path limits.

The grid has a total of $\{5 - (-5) + 1\} \times \{4 - (-4) + 1\} = 11 \times 9 = 99$ points.

The total number of excluded points is $4 \times 3 = 12$

Therefore, the number of points that can be included is $99 - 12 = 87$

Problem 28 of 30

A natural number $n$ is said to be good if $n$ is the sum of $r$ consecutive positive integers, for some $r \ge 2$. Find the number of good numbers in the set $\{1, 2, ... , 100\}$.

Solution:

We can show that a number can expressed as a sum of at least two consecutive positive integers $if\ and\ only\ if$ it has at least one odd factor.

$\underline{Step\ 1}:$

We will first show that it is impossible to express a number which does not have at least one odd factor as a sum of consecutive positive integers.

A number greater than $1$ which does not have any odd factor can only be a power of two.

Any sum of consecutive integer, from $m$ to $n$ can be written as

$\sum\limits_{i = m}^{n} i$

$= \sum\limits_{i = 1}^{n} i - \sum\limits_{i = 1}^{m-1} i$

-----------book page break-----------

$= \dfrac{n(n+1)}{2} - \dfrac{(m - 1)m}{2}$

$= \dfrac{n^2 + n - m^2 + m}{2}$

$= \dfrac{(n - m)(n+m) + (n + m)}{2}$

$= \dfrac{(n + m)(n - m + 1)}{2}$

If $m$ and $n$ are either both odd or both even, that is, they are of the same parity, then $(n - m + 1)$ is odd and $(n + m)$ is even.

If they are of different parity, that is, one of them is odd and the other one is even, then $(n + m)$ is odd and $(n - m + 1)$ is even.

Therefore, in all cases there has to be an odd factor of the sum.

Therefore, any number of the the form $2^a$ where $a$ is a positive integer, cannot be expressed as a sum of at least two consecutive integers.

$\underline{Step\ 2:}$

Here we will prove that all numbers that have at least one odd factor can be expressed as a sum of two or more consecutive positive integers.

Let us consider any odd $n > 1$ first.

We can write $n$ as $2p + 1$ for some positive integer $p$.

Therefore, $n = 2p + 1 = p + (p + 1)$

$p$ and $p+1$ are two consecutive positive integers. Therefore, our assertion is true for any odd $n$.

-----------book page break-----------

However, we will prove this for a more generic case where the number $n$ can be odd or even, but has at least one non trivial odd factor.

Let the number be $a \times b$ where $a$ is odd, and greater than $1$, and $b$ can either be odd or even.

We can write $a = 2p + 1$

Therefore, we can write the number as $n = b \times a = \overbrace{b + b + ... + b}^{a\ times} = \overbrace{b + b + ... + b}^{2p + 1\ times}$

$= (\overbrace{b + b + ... + b}^ {p\ times}) + \overbrace{b}^{1\ term} + (\overbrace{b + b + ... + b}^ {p\ times})$

In the above step we have grouped the $(2p + 1)$ count of $b$s into three groups, the first and the third containing $p$ occurrences of $b$ and the second (middle) group containing only one $b$. Now for the third group, we will add $1, 2, 3..., p$ respectively to each of the $b$s starting with the first.

For the first group we subtract $p, p-1, ..., 2, 1$ from each of the $b$s. Therefore, the sum of all these terms still remain the same, that is $n$

$\therefore n = \{(b - p) + (b - (p-1)) + ... + (b - 1)\} + b + $

$\{(b + 1) + ... + (b + (p-1)) + (b + p)\}$

Now, we can see that the above expression is a sum of integers starting with $(b - p)$ and ending with $b + p$, and each term exceeding the previous term by $1$

Therefore, this is a sequence of consecutive integers, all of them may or may not be positive. However, we can see that the middle term, $b$, is positive. Therefore, there are more positive numbers in the series than negative numbers.

Each negative number will cancel out with the corresponding positive number, leaving us with at least two consecutive positive integers in the sequence.

-----------book page break-----------

We can try this with a small example, let's say $14$, which has only one odd factor $7$.

$14 = 2 \times 7 = \overbrace{2 + 2 + ... + 2}^{7\ terms}$

$= \{\overbrace{2 + 2 + 2}^{3\ terms}\} + \overbrace{2}^{1\ term} + \{\overbrace{2 + 2 + 2}^{3\ terms}\}$

$= \{(2 - 3) + (2 - 2) + (2 - 1)\} + 2 + \{(2 + 1) + (2 + 2) + (2 + 3)\}$

$= (-1) + 0 + 1 + 2 + 3 + 4 + 5$

$= \cancel{(-1)} + \cancel{0} + \cancel{1} + 2 + 3 + 4 + 5$

Coming back to our problem, the largest value of $n$, such that $2^n < 100$ is $n = 6$

Therefore, there are $7$ numbers $(2^0, 2^1, ..., 2^6)$ that are $not$ good in the given set and the remaining $93$ numbers are good.

Problem 29 of 30

Positive integers $a, b, c$ satisfy $\dfrac{ab}{a - b} = c$. What is the largest possible value of $a + b + c$ not exceeding $99$?

Solution:

Given that:

$\dfrac{ab}{a - b} = c$

$\Rightarrow \dfrac{a - b}{ab} = \dfrac{1}{c}$

$\Rightarrow \dfrac{1}{b} - \dfrac{1}{a} = \dfrac{1}{c}$

Therefore, $2, 3, 6$ can be one possible set of values of $a, b, c$

Any set of numbers of the form $2k, 3k$ and $6k$ will satisfy the above equation.

Therefore,

$2k + 3k + 6k \le 99$

$\Rightarrow 11k \le 99$

$\Rightarrow k \le 9$

For $k = 9$, we get the three numbers as $18, 27, 54$ which gives us a sum of $99$.

Therefore, the maximum sum of the three integers, not exceeding $99$, is $99$.

Problem 30 of 30

Find the number of pairs $(a, b)$ of natural numbers such that $b$ is a $3$-digit number, $a+1$ divides $b - 1$ and $b$ divides $a^2 + a + 2$.

Solution:

Given that:

$a + 1 \bigm\lvert b - 1$

$\therefore b - 1 = m(a + 1)$ where $m \in \mathbb{N}$

$\Rightarrow$ $b = ma + m + 1$ $...eqn(i)$

and

$b \bigm\lvert a^2 + a + 2$

$\therefore a^2 + a + 2 = nb$ where $n \in \mathbb{N}$

$\Rightarrow a^2 + a + 2 = n(ma + m + 1)$

$\Rightarrow ma^2 + ma + 2m = mn(ma + m + 1)$

$\Rightarrow ma^2 + ma + a + 2m - a = mn(ma + m + 1)$

$\Rightarrow a(ma + m + 1) + 2m - a = mn(ma + m + 1)$

$\Rightarrow$ $a + \dfrac{2m - a}{ma + m + 1} = mn$ $...eqn(ii)$

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$\because a, m \in \mathbb{N}$

$m + m \le ma + m$

$\Rightarrow m + m - a \lt ma + m + 1$

$\Rightarrow 2m - a \lt ma + m + 1$

Going back to $eqn\ (ii)$,

The R.H.S $mn$ is a natural number, and $a$ on the L.H.S is also a natural number, and

$\dfrac{2m-a}{ma + m + 1} < 1$.

This is possible only if $2m - a = 0$

$\therefore 2m = a$

Substituting $2m = a$ in $eqn(i)$,

$b = 2m^2 + m + 1$

Since $b$ is a $3$-digit number,

$100 \le 2m^2 + m + 1 \le 999$

Therefore,

$0 \le 2m^2 + m - 99$

$\Rightarrow 0 \le \left(m - \dfrac{-1 + \sqrt{793}}{4} \right) \left(m - \dfrac{-1 - \sqrt{793}}{4} \right)$

Since $m$ is positive integer,

$m \ge \dfrac{-1 + \sqrt{793}}{4}$

$\Rightarrow m \ge 6.7$

$\Rightarrow m \ge 7$

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Likewise,

$2m^2 + m + 1 \le 999$

$\Rightarrow 2m^2 + m - 998 \le 0$

$\Rightarrow \left(m - \dfrac{-1 + \sqrt{7985}}{4}\right) \left(m - \dfrac{-1 - \sqrt{7985}}{4}\right) \le 0$

$\Rightarrow \left(m - \dfrac{-1 + 89.3}{4}\right) \left(m - \dfrac{-1 - 89.3}{4}\right) \le 0$

$\Rightarrow \left(m - 22.07\right) \left(m + 22.5\right) \le 0$

Since $m$ is positive integer,

$m \le 22$

Therefore,

$7 \le m \le 22$

Therefore, the number of values of $m$ satisfying the given condition is $22 - 7 + 1 = 16$