PRMO-2015

Problem 1 of 20

A man walks a certain distance and rides back in $3\dfrac{3}{4}$ hours; he could ride both ways in $2 \dfrac{1}{2}$ hours. How many hours would it take him to walk both ways?

Solution:

The person's riding time for riding one way is $2 \dfrac {1}{2} \div 2 = 1\dfrac {1}{4}$.

Therefore, his walking time for one way is $3\dfrac{3}{4} - 1 \dfrac{1}{4} = 2 \dfrac{1}{2}$

Therefore, he will take $2 \times 2\dfrac{1}{2} = 5$ hours to walk both ways.

Problem 2 of 20

Positive integers $a$ and $b$ are such that $a + b = \dfrac{a}{b} + \dfrac{b}{a}$. What is the value of $a^2 + b^2$?

Solution:

Given that:

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

$\Rightarrow a + b = \dfrac{a^2 + b^2}{ab}$

$\Rightarrow a^2b + b^2a = a^2 + b^2$

$\Rightarrow a^2(b - 1) + b^2(a - 1) = 0$

$a$ and $b$ are positive integers, therefore, $a^2$ and $b^2$ are also positive integers, and

$a \ge 1$

$\therefore a-1 \ge 0$

and similarly,

$b - 1 \ge 0$

Therefore, $a^2(b - 1)$ and $b^2(a - 1)$ are non-negative integers, and the only way their sum can be $0$ is that both of them are $0$.

Therefore, $b - 1 = 0 \Rightarrow b = 1$

and $a - 1 = 0 \Rightarrow a = 1$

Therefore, $a^2 + b^2 = 1^2 + 1^2 = 2$

Problem 3 of 20

The equations $x^2 - 4x + k = 0$ and $x^2 + kx - 4 = 0$, where $k$ is a real number, have exactly one common root. What is the value of $k$?

Solution:

Let the common root be $\alpha$, therefore,

$\alpha^2 - 4\alpha + k = 0$

and

$\alpha^2 + k\alpha - 4 = 0$

$k\alpha + 4\alpha - 4 - k = 0$

$\alpha(k + 4) - (k + 4) = 0$

$(\alpha - 1)(k+4) = 0$

Therefore, $\alpha = 1$ and $k = -4$

Replacing $k = -4$ in the given equations, both equations become identical, that is $x^2 - 4x - 4 = 0$, and both roots will be same.

Since it is given that they have exactly one root common, therefore $\alpha = 1$ is the common root of the given equations.

Substituting $x = 1$ in any of the given equations we get $k = 3$

Problem 4 of 20

Let $P(x)$ be a non-zero polynomial with integer coefficients. If $P(n)$ is divisible by $n$ for each positive integer $n$, what is the value of $P(0)$?

Solution:

Let the polynomial $P(x)$ be defined as:

$a_mx^m + a_{m-1}x^{m-1} + ... + a_1x^1 + a_0$

Therefore,

$P(n) = a_mn^m + a_{m-1}n^{m-1} + ... + a_1n^1 + a_0$

$= n(a_mn^{m-1} + a_{m-1}n^{m-2} + ... + a_1) + a_0$

The first part is divisible by $n$, therefore, for the whole polynomial to be divisible by $n$, $a_0$ has to be divisible by $n$.

This has to be true for $all$ positive integers $n$.

The only finite number which is divisible by $n$ for all integers $n$ is $0$.

Therefore $a_0 = 0$

Therefore,

$P(0) = a_m0^m + a_{m-1}0^{m-1} + ... + a_10^1 + a_0$

$= 0$

Problem 5 of 20

How many line segments have both their endpoints located at the vertices of a given cube?

Solution:

A cube has $8$ vertices. Choosing any two of them will give us a unique line segment.

Therefore, there are $\xacomb{8}{2}$ such line segments.

$\xacomb{8}{2} = \dfrac{8 \times 7}{2 \times 1} = 28$ line segments.

Problem 6 of 20

Let $E(n)$ denote the sum of the even digits of $n$. For example, $E(1243) = 2 + 4 = 6$. What is the value of $E(1) + E(2) + E(3) + ... + E(100)$?

Solution:

The number $100$ does not have a non-zero even number, therefore the sum of $E(1) + ... + E(100)$ will be same as $E(1) + ... + E(99)$

If we consider $0$ to $99$ all as two digit numbers, that is, we consider the single digit numbers as two digit numbers with a leading $0$,

we get:

$00,\ 01,\ 02,\ ...,\ 99$ $...$ a total of $100$ numbers.

In this series each of the digits $0\ ...,\ 9$ will occur an equal number of times in the unit's place and in the ten's place.

That is, for example the digit $2$ will occur $\dfrac{100}{10} = 10$ times in the unit's place and $10$ times in the ten's place.

Similarly, each of the even digits $2, 4, 6, 8$ will occur $10$ times in the ten's place and $10$ times in the unit's place.

Therefore the sum of the even digits is in the ten's place is:

$(2 + 4 + 6 + 8) \times 10$ and the the sum of the even digits in the unit's place is:

$(2 + 4 + 6 + 8) \times 10$

Therefore, the total sum of the even digits is:

$(2 + 4 + 6 + 8) \times 10 + (2 + 4 + 6 + 8) \times 10$

$= 20 \times 10 + 20 \times 10$

$= 400$

Problem 7 of 20

How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of $N$ is a perfect square?

Solution:

Let $N$ be a number of the form $10d_1 + d_0$ where $d_1$ and $d_0$ are digits, be such a number that satisfies the given condition.

The number obtained by reversing the digits is $10d_0 + d_1$

Therefore,

$10d_1 + d_0 + 10d_0 + d_1$ is a perfect square.

Therfore,

$10d_1 + d_0 + 10d_0 + d_1$

$= 10(d_1 + d_0) + (d_1 + d_0)$

$= 11(d_1 + d_0)$ is a perfect square.

$11(d_1 + d_0)$ can be a perfect square only if $d_1 + d_0$ is an integer of the form $11p$ where $p$ is a perfect square.

Therefore, $d_1 + d_0$ can be $11,\ 44,\ 99,\ ...$

Since $d_1$ and $d_0$ are decimal digits, their sum can never be greater than $18$.

Therefore, the only possible case is where $d_1 + d_0 = 11$

The only possible values of $d_1$ and $d_0$, where $d_1 + d_0 = 11$ are:

$(9,\ 2)$, $(2,\ 9)$

$(8,\ 3)$, $(3,\ 8)$

$(7,\ 4)$, $(4,\ 7)$

$(6,\ 5)$, $(5,\ 6)$

Therefore, there are $8$ two-digit numbers whose sum with its reverse, is a perfect square.

Problem 8 of 20

The figure below shows a broken piece of a circular plate made of glass.

$C$ is the midpoint of $AB$, and $D$ is the midpoint of arc $AB$. Given that $AB = 24$ cm and

$CD = 6$ cm, what is the radius of the plate in centimetres? (The figure is not drawn to scale.)

Solution:

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We can complete the circle as shown in the following figure, and the extend $DC$ to meet the circle at $E$.

Since, $C$ is the the midpoint of chord $AB$ and $D$ is the midpoint of arc $AB$, $CD$ is part of the diameter of the circle, therefore, $DE$ is the diameter.

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From the secant rule of triangles explained , we have:

$AC \times CB = CD \times CE$

$\therefore 12 \times 12 = 6 \times (2r - 6)$ where $r$ is the radius of the circle

$\Rightarrow 2r - 6 = 24$

$\Rightarrow 2r = 30$

$\Rightarrow r = 15$

Problem 9 of 20

A $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any interior point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?

Solution:

Let us observe that whatever arrangement we choose, the side of the containing square will be an integer.

The total number of squares in the given rectangles is $2 \times 3 + 3 \times 4 = 18$

Therefore, it will not be possible to fit the rectangles in a square of side less than $5$, because:

$4^2 = 16 \lt 18$

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If we arrange the two rectangles as shown below, each side of the containing square will have a length of $5$.

Therefore, the minimum area of the containing square is $25$

Problem 10 of 20

What is the greatest possible perimeter of a right-angled triangle with integer side lengths if one of the sides has length $12$?

Solution:

To maximise the perimeter of the triangle with one side as $12$ units, this side should be the smallest side of the triangle.

We know that $3,\ 4,\ 5$ is a Pythagorean triplet.

We can multiply each side by $4$, and get the triplet as $12,\ 16,\ 20$ which will also be a Pythagorean triplet.

The perimeter of this triangle is $12 + 16 + 20 = 48$

Problem 11 of 20

In rectangle $ABCD$, $AB = 8$ and $BC = 20$. Let $P$ be a point on $AD$ such that $\angle BPC = 90^\circ$. If $r_1$, $r_2$, $r_3$ are the radii of the incircles of triangles $APB$, $BPC$ and $CPD$, what is the value of $r_1 + r_2 + r_3$?

Solution:

Let us draw the diagram as shown below. We have drawn the altitude $PQ$ from the point $P$, to the side $BC$.

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Triangles $BAP$, $BPC$ and $PDC$ are right-angled triangles.

We know from $Corollary\ 5$ of this , that the inradius of a right-angled triangle is given by:

$r = \dfrac{1}{2} (\text{sum of the right angle arms} - \text{hypotenuse})$

Therefore,

$r_1 = \dfrac{1}{2} (AB + AP - BP)$

$r_2 = \dfrac{1}{2} (BP + PC - BC)$

$r_3 = \dfrac{1}{2} (PD + DC - PC)$

Therefore,

$r_1 + r_2 + r_3$

$= \dfrac{1}{2}(AB + AP - \cancel{BP} + \cancel{BP} + \cancel{PC} - BC + PD + DC - \cancel{PC})$

$= \dfrac{1}{2}(AB + AP - BC + PD + DC)$

$= \dfrac{1}{2}(AB + AP + PD + DC - BC)$

$= \dfrac{1}{2}(AB + AD + AB - AD)$

$= \dfrac{1}{2}(2AB) = AB = 8$

Problem 12 of 20

Let $a$, $b$, and $c$ be real numbers such that $a - 7b + 8c = 4$ and $8a + 4b - c = 7$. What is the value of $a^2 - b^2 + c^2$?

Solution:

Let us observe that there are three variables and two equations are given.

Therefore, it is not possible to find the values of the three variables individually, but we can try to find the value of the required expression by manipulating the two equations.

The required expression is $a^2 - b^2 + c^2$.

If we directly square the given equations $b^2$ will have a positive sign, but if we take $b$ to the other side and bring it back after squaring, $b^2$ will have a negative sign while $a^2$ and $c^2$ will have a positive sign.

Therefore, let us try that approach.

Using the first equation we get:

$a - 7b + 8c = 4$

$\Rightarrow a + 8c = 4 + 7b$

$\Rightarrow (a + 8c)^2 = (4 + 7b)^2$

$\Rightarrow a^2 + 16ac + 64c^2 = 16 + 56b + 49b^2$ $...eqn\ (i)$

Similarly, using the second equation, we get:

$8a + 4b - c = 7$

$\Rightarrow 8a - c = 7 - 4b$

$\Rightarrow (8a - c)^2 = (7 - 4b)^2$

$\Rightarrow 64a^2 - 16ac + c^2 = 49 - 56b + 16b^2$ $...eqn\ (ii)$

Adding equations $(i)$ and $(ii)$ we get:

$65a^2 + 65c^2 = 65 + 65b^2$

$\Rightarrow a^2 + c^2 = 1 + b^2$

$\Rightarrow a^2 + c^2 - b^2 = 1$

Problem 13 of 20

Let $n$ be the largest integer that is the product of exactly $3$ distinct prime numbers, $x$, $y$ and $10x + y$, where $x$ and $y$ are digits. What is the sum of the digits of $n$?

Solution:

As per the given conditions $x$ and $y$ are single digit prime numbers. The single digit prime number are $2,\ 3,\ 5,\ 7$

Therefore, $x,\ y \in \{2,\ 3,\ 5,\ 7\}$

The number $10x + y$ is a two-digit prime number formed using the digits $x$ and $y$. To maximise the value of $10x + y$ we can start by putting the largest digit in the ten's place and try with the other digits.

$75$ is not a prime number.

$73$ is a prime number.

Therefore the largest prime number that can be formed using $x$ and $y$ is $73$, with $x = 7$ and $y = 3$.

Therefore the product of these three prime numbers is:

$7 \times 3 \times 73 = 1533$

The sum of the digits of $1533$ is $12$

Problem 14 of 20

At a party, each man danced with exactly four women and each woman danced with exactly three men. Nine men attended the party. How many women attended the party?

Solution:

Since each men danced with $4$ women, there where $9 \times 4 = 36$ dances.

Let the number of women be $w$, and since each had $3$ dances, the total number of dances the women had was $3w$.

Therefore, $3w = 36$

$\Rightarrow w = 12$

Problem 15 of 20

If $3^x + 2^y = 985$ and $3^x - 2^y = 473$, what is the value of $xy$?

Solution:

Adding the two given equations we get:

$2.3^x = 985 + 473 = 1458$

$\Rightarrow 3^x = 729 = 3^6$

$\Rightarrow x = 6$

Substituting $3^x = 729$ in the first equation, we get:

$729 + 2^y = 985$

$\Rightarrow 2^y = 256 = 2^8$

$\Rightarrow y = 8$

Therefore, $xy = 6 \times 8 = 48$

Problem 16 of 20

In triangle $ABC$, let $D$ be the foot of the altitude from $A$, and $E$ be the midpoint of $BC$. Let $F$ be the midpoint of $AC$. Suppose $\angle BAE = 40^\circ$. If $\angle DAE = \angle DFE$, what is the magnitude of $\angle ADF$ in degrees?

Solution:

We can draw the diagram for the given problem as shown below.

Since $E$ and $F$ are the midpoints of $BC$ and $CA$ respectively, $EF \parallel AB$

$\therefore \angle AEF = \angle BAE = 40^\circ$

Since $\angle DAE = \angle DFE$, and they are subtended by the same line $DE$, $ADEF$ is a cyclic quadrilateral.

Therefore, $\angle ADF = \angle AEF = 40^\circ$

Note: This problem, as published originally for PRMO, mentions $\triangle ABC$ as acute-angled. It can be shown that $\triangle ABC$ is right-angled at $A$.
$ADEF$ is cyclic quadrilateral with $\angle ADE = 90^\circ$, therefore $AFE = 180 - 90 = 90^\circ$ and $\angle BAC = \angle EFC = 90^\circ$. We have corrected this to omit the word acute-angled triangle.

Problem 17 of 20

A subset $B$ of the set of first $100$ positive integers has the property that no two elements of $B$ sum to 125. What is the maximum possible number of elements in $B$?

Solution:

The first $100$ positive integers will include integers from $1$ to $100$.

If we consider integers $1$ to $24$, no other number in this set, when added to any of these numbers will add up to $125$

Starting with $25$, if we add one number $n$ to the set, we have to exclude $125 - n$ from the set.

For $n = 25$, we have to exclude $125 - 25 = 100$

For $n = 26$, we have to exclude $125 - 26 = 99$

Likewise, as we keep on increasing $n$ we will reach a number that has already been excluded, and we should stop counting there.

We can find that number using the equation:

$n = 125 - n$

$\Rightarrow 2n = 125$

$\Rightarrow n = 62.5$

Therefore, for $n = 62$ we have to exclude $125 - 62 = 63$.

Since all numbers greater than $62$ will already have been excluded, $62$ is the last number that we can include.

Since we have included all numbers from $1$ to $62$, the maximum size of our set is $62$.

Problem 18 of 20

Let $a$, $b$ and $c$ be such that $a + b + c = 0$ and

$P = \dfrac{a^2}{2a^2 + bc} + \dfrac{b^2}{2b^2 + ca} + \dfrac{c^2}{2c^2 + ab}$ is defined.

What is the value of $P$?

Solution:

$a + b + c = 0$

$\Rightarrow a = -(b + c)$

$\Rightarrow a^2 = (b + c)^2$

Using the above substitution, we can simplify the first term $\dfrac{a^2}{2a^2 + bc}$ of $P$.

$\dfrac{a^2}{2a^2 + bc}$

$= \dfrac{a^2}{2(b+ c)^2 + bc}$

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$= \dfrac{a^2}{2b^2 + 4bc + 2c^2 + bc}$

$= \dfrac{a^2}{2b^2 + 4bc + bc + 2c^2}$

$= \dfrac{a^2}{2b(b + 2c) + c(b + 2c)}$

$= \dfrac{a^2}{(2b + c)(b + 2c)}$

$= \dfrac{a^2}{(b + b + c)(c + b + c)}$

We can again substitute $b + c = -a$ and get this term as:

$\dfrac{a^2}{(b - a)(c - a)}$

Since the other two terms are symmetric, using similar method we will get:

$\dfrac{b^2}{2b^2 + ca} = \dfrac{b^2}{(a - b)(c - b)}$

and

$\dfrac{c^2}{2c^2 + ab} = \dfrac{c^2}{(a - c)(b - c)}$

Therefore,

$P = \dfrac{a^2}{2a^2 + bc} + \dfrac{b^2}{2b^2 + ca} + \dfrac{c^2}{2c^2 + ab}$

$= \dfrac{a^2}{(b - a)(c - a)} + \dfrac{b^2}{(a - b)(c - b)} + \dfrac{c^2}{(a - c)(b - c)}$

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Maintaining cyclic order we can rewrite this as:

$P = -\dfrac{a^2}{(a - b)(c - a)} - \dfrac{b^2}{(a - b)(b - c)} - \dfrac{c^2}{(c - a)(b - c)}$

$= -\dfrac{a^2(b-c) + b^2(c-a) + c^2(a - b)}{(a - b)(b - c)(c - a)}$

$= -\dfrac{a^2(b-c) + b^2c- b^2a + c^2a - c^2b}{(a - b)(b - c)(c - a)}$

$= -\dfrac{a^2(b-c) + b^2c - c^2b - b^2a + c^2a}{(a - b)(b - c)(c - a)}$

$= -\dfrac{a^2(b-c) + bc(b - c) - a(b^2 - c^2)}{(a - b)(b - c)(c - a)}$

$= -\dfrac{(b - c)(a^2 + bc - ab - ac)}{(a - b)(b - c)(c - a)}$

$= -\dfrac{(b - c)(a^2 - ac + bc - ab)}{(a - b)(b - c)(c - a)}$

$= -\dfrac{(b - c)\{a(a - c) - b(a - c)\}}{(a - b)(b - c)(c - a)}$

$= -\dfrac{(b - c)(a-b)(a - c)}{(a - b)(b - c)(c - a)}$

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$= \dfrac{(a - b)(b - c)(c - a)}{(a - b)(b - c)(c - a)}$

Since it is given that $P$ is defined, we can conclude that $P \ne \dfrac{0}{0}$, $\therefore a \ne b \ne c$

Therefore, $P = 1$

Problem 19 of 20

The circle $\omega$ touches the circle $\Omega$ internally at $P$. The centre $O$ of $\Omega$ is outside $\omega$. Let $XY$ be a diameter of which is also tangent to $\omega$. Assume $PY > PX$. Let $PY$ intersect $\omega$ at $Z$. If $YZ = 2PZ$, what is the magnitude of $\angle PYX$ in degrees?

Solution:

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Let $O_1$ and $O_2$ be the centres of circles $\Omega$ and $\omega$ respectively. Let us draw the diagram for this problem as shown below.

Line $PO_2$ has been joined and extended to meet $XY$ at $O_1$. Let $Q$ be the point of contact of $XY$ with $\omega$. $PS$ is the perpendicular drawn from point $P$ to $XY$. $PS$ has been extended to $P'$ such that $P'S$ is equal to $PS$. Points $O_2$ and $Z$ have been joined.

$O_2P$ is the radius of $\omega$, therefore $O_2P$ is perpendicular to the tangent to $\omega$ at $P$. This tangent is same as the tangent to $\Omega$ at $P$. Therefore, $O_1P$ is the radius of $\Omega$

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Given that $\dfrac{PZ}{ZY} = \dfrac{1}{2}$

$\therefore \dfrac{PZ}{PY} = \dfrac{1}{3}$

Considering isosceles $\triangle$s $O_2PZ$ and $O_1PY$,

$\angle O_2PZ$ is common

$\angle O_1YP = \angle O_2ZP$ (both are equal to $\angle O_2PZ$)

$\therefore \triangle O_2PZ \sim \triangle O_1PY$

$\therefore \dfrac{PO_2}{PO_1} = \dfrac{PZ}{PY} = \dfrac{1}{3}$

Let $r$ be the radius of $\omega$, therefore $PO_2 = r$ and $PO_1 = 3r$

$O_1O_2 = PO_1 - PO_2 = 3r - r = 2r$

Since $O_2Q \parallel PS$, $\triangle O_1O_2Q \sim \triangle O_1PS$.

$\therefore \dfrac{O_1O_2}{O_1P} = \dfrac{O_2Q}{PS}$

$\Rightarrow \dfrac{2r}{3r} = \dfrac{r}{PS}$

$\Rightarrow PS = \dfrac{3}{2}r$

$\therefore \dfrac{PS}{O_1P} = \dfrac{\frac{3}{2}r}{3r} = \dfrac{1}{2}$

Now, if we reflect point $P$ using the line $XY$ to point $P'$, then we can prove that $\triangle OPP'$ is equilateral.

Therefore, $\angle PO_1P' = 60^\circ$

Therefore, $\angle PO_1X = 30^\circ$

Therefore, $\angle PYX = \dfrac{1}{2} \angle PO_1X = 15^\circ$

Problem 20 of 20

The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$?

Solution:

Let the smallest digit be $d$.

Therefore, the number can be written as:

$N = 1000(d + 3) + 100(d + 2) + 10(d+1) + d$

$= 1000d + 3000 + 100d + 200 + 10d + 10 + d$

$= 1111d + 3210$

Using the concept of number congruence from ,

$1111 \equiv 1\ (mod\ 37)$

$\Rightarrow 1111d \equiv d\ (mod\ 37)$

Similarly,

$3210 \equiv 28\ (mod\ 37)$

$\therefore N \equiv d + 28\ (mod\ 37)$

The value of $d$ can range from $0$ to $6$

Sum of remainders

$\sum\limits_{d = 0}^6 (d + 28)$

$= \dfrac{6(6+1)}{2} + 28 \times 7$

$= 21 + 196$

$= 217$