Problem 1 of 30
A book is published in three volumes, the pages being numbered from $1$ onwards. The page numbers are continued from first volume to the second volume to the third. The number of pages in the second volume is $50$ more than that in the first volume, and the number of pages in the third volume is one and a half times that in the second. The sum of the page numbers on the first pages of the three volumes is $1709$. If $n$ is the last page number, what is the largest prime factor of $n$?Solution:
In this case, we will take the number of pages in the second volume as $p$.The number of pages in the third volume, is $\dfrac{3}{2}p$ and the number of pages in the first volume is $p - 50$.
The first page number of the first volume is $1$.
The first page number of the second volume is $(p - 50) + 1 = p - 49$.
The first page number of the third volume is $(p - 50) + p + 1 = 2p - 49$
Adding all the first page numbers, we get:
$1 + (p - 49) + (2p - 49) = 1709$
$\Rightarrow 3p - 97 = 1709$
$\Rightarrow 3p = 1806$
$\Rightarrow p = 602$
The last page number is the total number of pages, which is:
$(p - 50) + p + \dfrac{3}{2}p$
$= \dfrac{7}{2}p - 50$
$= \dfrac{7}{2}602 - 50$
$= 7 \times 301 - 50$
$= 2107 - 50$
$= 2057$
We can see that the number $2057$ is divisible by $11$
$2057 = 11 \times 187 = 11 \times 11 \times 17$
Therefore the largest prime factor of the last page number is $17$
Problem 2 of 30
In a quadrilateral $ABCD$, it is given that $AB = AD = 13$, $BC = CD = 20$, $BD = 24$. If $r$ is the radius of the circle inscribable in the quadrilateral, then what is the integer closest to $r$?Solution:
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Let us draw the diagram as shown below:
$ABCD$ is a kite, therefore, diagonal $BD \perp AC$ and $AC$ bisects $BD$.
$BO = \dfrac{24}{2} = 12$
$AO = \sqrt{13^2 - 12^2} = \sqrt{(1)(25)} = 5$
$OC = \sqrt{20^2 - 12^2} = \sqrt{32 \times 8} = \sqrt{2^8} = 2^4 = 16$
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$ar[\triangle ABD] = \dfrac{1}{2}AO \times BD = \dfrac{1}{2} \times 5\times 24 = 60$
$ar[\triangle ACD] = \dfrac{1}{2}OC \times BD = \dfrac{1}{2} \times 16\times 24 = 192$
$\therefore ar[ABCD] = 252$
We know from , that the inradius of a tangential quadrilateral is given by:
$r = \dfrac{2 \times area}{perimeter}$
In this case the perimeter of $ABCD = 66$
$\therefore r = \dfrac{2 \times 252}{66} = \dfrac{84}{11} = 7\dfrac{7}{11}$
The closest integer to the inradius is $8$.
Problem 3 of 30
Consider all $6$-digit numbers of the form $abccba$, where $b$ is odd. Determine the number of all such $6$-digit numbers that are divisible by $7$.Solution:
Since $a,\ b$ and $c$ are digits $0 \le a,b,c \le 9$ and $a \ne 0$ since this is a $6$-digit number, and it is given that $b$ is odd.Therefore the value of the number is:
$10^5a + 10^4b + 10^3c + 10^2c + 10b + a$
$= 100001a + 10010b + 1100c$
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We can see that $10010 \equiv 0\ (mod\ 7)$, therefore no matter what the value of $b$ is, $10010b$ is always divisible by $7$.
$100001 \equiv 6\ (mod\ 7)$
$\therefore 100001a \equiv 6a\ (mod\ 7)$
Similarly,
$1100 \equiv 1\ (mod\ 7)$
$1100c \equiv c\ (mod\ 7)$
We need to find values of $a$ and $c$ such that:
$6a + c \equiv 0\ (mod\ 7)$
For each digit $a$, $c$ can be either equal to $a$ or equal to $a \pm 7$, if $a \pm 7$ is a valid decimal digit.
If $a = 1$, $c = 1$ or $8$ $(2\ cases)$
If $a = 2$, $c = 2$ or $9$ $(2\ cases)$
If $a = 3$, $c = 3$ $(1\ case)$
If $a = 4$, $c = 4$ $(1\ case)$
If $a = 5$, $c = 5$ $(1\ case)$
If $a = 6$, $c = 6$ $(1\ case)$
If $a = 7$, $c = 7$ or $c = 0$ $(2\ cases)$
If $a = 8$, $c = 8$ or $c = 1$ $(2\ cases)$
If $a = 9$, $c = 9$ or $c = 2$ $(2\ cases)$
We get a total of $14$ cases.
For each of these cases, $b$ can be any odd digit, $1,\ 3,\ 5,\ 7,\ or\ 9$
So, there are $14 \times 5 = 70$ possible combinations satisfying the given conditions.
Problem 4 of 30
The equation $166 \times 56 = 8590$ is valid in some base $b \ge 10$ (that is $1,\ 6,\ 5,\ 8,\ 9,\ 0$ are digits in base $b$ in the above equation).Find the sum of all possible values of $b \ge 10$ satisfying the equation.
Solution:
If we take modulo base $b$ of the given equation, we get:$6 \times 6 \equiv 0\ (mod\ b)$
$36 \equiv 0\ (mod\ b)$
The above condition will be satisfied by any divisor of $36$
Therefore, the divisors can be:
$1,\ 2,\ 3,\ 4,\ 6,\ 9,\ 12,\ 18,\ 36$
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From this we eliminate all values $\lt 10$ based on the given condition.
We are left with $12,\ 18,\ 36$
We can convert the given numbers to polynomials in $b$ and convert this to an equation:
$(b^2 + 6b + 6)(5b + 6) = 8b^3 + 5b^2 + 9b + 0$
$\Rightarrow 5b^3 + 30b^2 + 30b + 6b^2 + 36b + 36 = 8b^3 + 5b^2 + 9b$
$\Rightarrow 3b^3 - 31b^2 - 57b - 36 = 0$
Since, we know that any one of the possible values could be a factor, we can try and factor the polynomial using $b = 12$
$3b^3 - 36b^2 + 5b^2 - 60b + 3b - 36 = 0$
$\Rightarrow 3b^2(b - 12) + 5b(b - 12) + 3(b - 12) = 0$
$\Rightarrow (b-12)(3b^2 + 5b + 3) = 0$
The discriminant for the equation $3b^2 + 5b + 3 = 0$ is $\sqrt{25 - 36} = \sqrt{-11}$ which makes the other two roots complex numbers.
Therefore, the only possible value of $b$ is $12$
Problem 5 of 30
Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AD \perp AB$. Suppose $ABCD$ has an incircle which touches $AB$ at $Q$ and $CD$ at $P$. Given that $PC = 36$ and $QB = 49$, find $PQ$.Solution:
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We can draw the diagram as shown below:
Let the tangent points of $BC$ and $AD$ with the incircle be $S$ and $R$ respectively.
Let $CN$ be the perpendicular drawn from vertex $C$ to the side $AB$.
$CR = CP = 36$ (tangents from the same point to a circle)
$BR = BQ = 49$ (tangents from the same point)
$\therefore CB = 36 + 49 = 85$
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$BN = 49 - 36 = 13$
$\therefore CN = \sqrt{85^2 - 13^2}$
$= \sqrt{(85 - 13)(85 + 13)}$
$= \sqrt{72 \times 98}$
$= \sqrt{36 \times 2 \times 2 \times 49}$
$= 6 \times 2 \times 7$
$= 84$
$PQ = CN = 84$
Problem 6 of 30
Integers $a,\ b,\ c$ satisfy the $a + b - c = 1$ and $a^2 + b^2 -c^2 = 1$What is the sum of all possible values of $a^2 + b^2 + c^2$?
Solution:
Given$a + b - c = 1$
$\Rightarrow a + b = 1 + c$
$\Rightarrow (a + b)^2 = (1 + c)^2$
$\Rightarrow a^2 + 2ab + b^2 = 1 + c^2 + 2c$
$\Rightarrow a^2 + b^2 - c^2 + 2ab = 1 + 2c$
Substituting $a^2 + b^2 - c^2 = -1$
$-1 + 2ab = 1 + 2c$
$\Rightarrow 2ab = 2(1 + c)$
$1 + c = a + b$
$ab = a + b$
$ab - a - b + 1 = 1$
$(a - 1)(b - 1) = 1$
$\because a$ and $b$ are integers, $a$ and $b$ can each be equal to $2$, or $0$
If $a = b = 2$, then $2 + 2 - c = 1 \Rightarrow c = 3$
If $a = b = 0$, then $0 + 0 - c = 1 \Rightarrow c = -1$
The possible values of $a,\ b,\ c$ are $(2,\ 2,\ 3)$ or $(0, 0,\ -1)$
Sum of $a^2 + b^2 + c^2$ for all possible values is:
$\{2^2 + 2^2 + 3^2\} + \{0^2 + 0^2 + (-1)^2\} = 18$
Problem 7 of 30
A point $P$ in the interior of a regular hexagon is at a distance $8,\ 8,\ 16$ units from three consecutive vertices of the hexagon, respectively. If $r$ is the radius of the circumscribed circle of the hexagon, what is the integer closest to $r$?Solution:
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Let diagram for this problem is given below:
We have marked the three adjacent vertices as $A,\ B$ and $C$ and joined these vertices with the circumcentre $O$.
We have joined $OP$ and produced it to meet the side $AB$ at $M$.$OA = OB = OC = r$
$\triangle PAB$ is isosceles, since $PA = PB = 8$
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$\triangle OAB$ is equilateral, since $O$ is the circumcentre of a regular hexagon. $OM$ is the median, as well as the altitude of $\triangle OAB$. $OM$ also bisect $\angle AOB$
$\therefore \angle MOB = \dfrac{1}{2} 60^\circ = 30^\circ$
$\therefore \angle MOC = \angle MOB + \angle BOC = 30 + 60 = 90^\circ$
$OP = \sqrt{PC^2 - r^2} = \sqrt{16^2 - r^2}$
$OM = \dfrac{\sqrt{3}}{2} r$
$AB = OB = r$
$\therefore MB = \dfrac{r}{2}$
$\therefore PM = \sqrt{PB^2 - MB^2}$
$\therefore PM = \sqrt{8^2 - \dfrac{r^2}{4}}$
$PM + OP = OM$
$\Rightarrow \sqrt{8^2 - \dfrac{r^2}{4}} + \sqrt{16^2 - r^2} = \dfrac{\sqrt{3}}{2} r$
$\Rightarrow \sqrt{\dfrac{16^2 - r^2}{4}} + \sqrt{16^2 - r^2} = \dfrac{\sqrt{3}}{2} r$
$\Rightarrow \dfrac{1}{2}\sqrt{16^2 - r^2} + \sqrt{16^2 - r^2} = \dfrac{\sqrt{3}}{2} r$
$\Rightarrow \dfrac{3}{2}\sqrt{16^2 - r^2} = \dfrac{\sqrt{3}}{2} r$
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$\Rightarrow \sqrt{3}\sqrt{16^2 - r^2} = r$
$\Rightarrow 3 (16^2 - r^2) = r^2$
$\Rightarrow 3 \times 16^2 = 4r^2$
$\Rightarrow 3 \times 8^2 = r^2$
$\Rightarrow r = 8\sqrt{3} \approx 13.85 \approx 14$
Problem 8 of 30
Let $AB$ be a chord of circle with centre $O$. Let $C$ be a point on the circle such that $\angle ABC = 30^\circ$ and $O$ lies inside the $\triangle ABC$. Let $D$ be a point on $AB$ such that $\angle DCO = \angle OCB = 20^\circ$. Find the measure of $\angle CDO$ in degrees.Solution:
Let us draw the diagram for the given problem as follows:
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Given that $\angle ABC = 30^\circ$
$\therefore AOC = 60^\circ$
$\therefore \triangle AOC$ is equilateral.
$\angle ACD = 60 - 20 = 40^\circ$
$\angle BAC = 180 - (80 + 30) = 70^\circ$
$\angle ADC = 180 - (40 + 70) = 70^\circ$
$\therefore \triangle ACD$ is isosceles, with $AC = CD$
$\because AC = CO = OA = r$, $CD = r$
$\therefore \triangle CDO$ is isosceles with $CD = CO$
$\therefore \angle CDO = \angle COD = \dfrac{180 - 20}{2} = 80^\circ$
Problem 9 of 30
Suppose $a,\ b$ are integers $a + b$ is a root of the equation $x^2 + ax + b = 0$. What is the maximum possible value of $b^2$?Solution:
Since $a + b$ is a root of the given equation, substituting $x = a + b$ we get:$(a + b)^2 + a(a + b) + b = 0$
$\Rightarrow a^2 + 2ab + b^2 + a^2 + ab + b = 0$
$\Rightarrow 2a^2 + 3ab + b^2 + b = 0$
Treating the above as a quadratic equation in $a$ we get the discriminant as:
$\sqrt{(3b)^2 - 4\times2(b^2 + b)}$
$= \sqrt{9b^2 - 8b^2 - 8b}$
$= \sqrt{b^2 - 8b}$
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For $a$ to be an integer, the discriminant needs to be a perfect square, therefore, we can write:
$b^2 - 8b = n^2$ for some integer value $n$
$\Rightarrow b^2 - 2.b.4 + 16 - 16 = n^2$
$\Rightarrow (b - 4)^2 - 16 = n^2$
$\Rightarrow (b - 4)^2 - n^2 = 16$
$\Rightarrow (b - 4 - n)(b - 4 + n) = 16$
Since $b$ and $n$ are integers, both the factors on the left hand side must be integers.
We can factor $16$ in the following ways:
$16 = 1 \times 16$
$16 = 4 \times 4$
$16 = 2 \times 8$
Taking $b - 4 - n = 1$ and $b - 4 + n = 16$ we get:
$b = \dfrac{25}{2}$ which is not an integer value.
Taking $b - 4 - n = 4$ and $b - 4 + n = 4$ we get:
$b = \dfrac{16}{2} = 8$ which is an integer value.
Taking $b - 4 - n = 2$ and $b - 4 + n = 8$ we get:
$b = \dfrac{18}{2} = 9$ which is an integer value.
For, each of the above cases, we can change the order of the factors, but will always get the same result for $b$, only $n$ will change its sign.
Therefore, the maximum value of $b^2$ is $9^2 = 81$
Problem 10 of 30
In a triangle $ABC$, the median from $B$ to $CA$ is perpendicular to the median from $C$ to $AB$. If the median from $A$ to $BC$ is $30$, determine $(BC^2 + CA^2 + AB^2)/100$Solution:
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Let us draw the diagram for the given problem as shown below.
Let $P$, $Q$ and $R$ be the midpoints of $AB$, $BC$ and $CA$ respectively. We join the $3$ medians.
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$P$ and $R$ are the midpoints of $AB$ and $AC$, $PR \parallel BC$ and $PR = \dfrac{1}{2} BC$
$\triangle GBC \sim \triangle GRP$
Therefore,
$\dfrac{GB}{GR} = \dfrac{GC}{GP} = \dfrac{GQ}{GM} = \dfrac{RP}{BC} = \dfrac{2}{1}$
If we consider $\triangle APM$ and $\triangle ABQ$,
$PM \parallel BQ$,
$\therefore \triangle APM \sim ABQ$
$\therefore M$ is the midpoint of $AQ$
$\Rightarrow MQ = 15$
$GM:GQ = 1:2 \Rightarrow GQ = 10$
$GQ$ is the median of $\triangle GBC$, and $BC$ is the hypotenuse,
$\therefore BC = 20$ and $PR = 10$
$BG^2 + CG^2 = BC^2 = 20^2$
Since $G$ is the centroid of $\triangle ABC$, $BG = \dfrac{2}{3}BR$ and $CG = \dfrac{2}{3}CP$
$\left(\dfrac{2}{3}BR\right)^2 + \left(\dfrac{2}{3}PC\right)^2 = 20^2$
$\Rightarrow BR^2 + CP^2 = \dfrac{9}{4}20^2$
$\therefore AQ^2 + BR^2 + CP^2 = 30^2 + \dfrac{9}{4}20^2$
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Using the corollary to Apollonius' Theorem explained , we know that thrice the sum of the squares of the sides equals four times the sum of the squares of the median.
$\therefore 3(AB^2 + BC^2 + CA^2) = 4(AQ^2 + BR^2 + CP^2)$
$\Rightarrow 3(AB^2 + BC^2 + CA^2) = 4\left(30^2 + \dfrac{9}{4} 20^2\right)$
$\Rightarrow 3(AB^2 + BC^2 + CA^2) = 4 \times 3\left(10 \times 30 + \dfrac{3}{4} 20^2\right)$
$\Rightarrow AB^2 + BC^2 + CA^2 = 4 \left(10 \times 30 + \dfrac{3}{4} 20^2\right)$
$\Rightarrow AB^2 + BC^2 + CA^2 = 40 \times 30 + 3 \times 20^2 = 2400$
$\therefore \dfrac{AB^2 + BC^2 + CA^2}{100} = 24$
Problem 11 of 30
There are several tea cups in the kitchen, some with handles and the others without handles. The number of ways of selecting two cups without a handle and three with a handle is exactly $1200$. What is the maximum possible number of cups in the kitchen?Solution:
Let the number of cups with handles be $m$ and the number of cups without handles be $n$.Therefore, the number of ways three cups with handle and two cups without handle can be selected is:
$\xacomb{m}{3} \times \xacomb{n}{2}$
Therefore,
$\dfrac{m(m-1)(m-2)}{2\times 3} \times \dfrac{n(n-1)}{2} = 1200$
$\Rightarrow m(m-1)(m-2)n(n-1) = 12 \times 1200$
We can write $12 \times 1200$ in the following ways:
$12 \times 1200 = 5 \times 4 \times 3 \times 16 \times 15$
$12 \times 1200 = 4 \times 3 \times 2 \times 25 \times 24$
$12 \times 1200 = 10 \times 9 \times 8 \times 5 \times 4$
Therefore, the maximum possible value of $m + n = 25 + 4 = 29$
Problem 12 of 30
Determine the numbers of $8$-tuples $(\epsilon_1,\ \epsilon_2,\ ...\ \epsilon_8)$ such that $\epsilon_1,\ \epsilon_2,\ ...\ \epsilon_8 \ \Large{\epsilon} \normalsize \ [1,\ -1]$and
$\epsilon_1 + 2\epsilon_2 + 3\epsilon_3 + ... + 8\epsilon_8$ is divisible by $3$.
Solution:
Observe, that the terms $3\epsilon_3$ and $6\epsilon_6$ are divisible by $3$, for $\epsilon = \pm 1$, therefore will not change the divisibility by $3$ for the whole expression, therefore both these values can either be $+1$ or $-1$.We can now analyse the remaining expression:
$\epsilon_1\ + 2\epsilon_2 + 4\epsilon_4 + 5\epsilon_5 + 7\epsilon_7 + 8\epsilon_8$
$= \epsilon_1\ + (3\epsilon_2 - \epsilon_2) + (3\epsilon_4 + \epsilon_4) + (6\epsilon_5 - \epsilon_5) + (6\epsilon_7 + \epsilon_7) + (9\epsilon_8 - \epsilon_9)$
$= \epsilon_1 + 3\epsilon_2 - \epsilon_2 + 3\epsilon_4 + \epsilon_4 + 6\epsilon_5 - \epsilon_5 + 6\epsilon_7 + \epsilon_7 + 9\epsilon_8 - \epsilon_8$
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Now we can again eliminate the terms with coefficients that are multiples of $3$, since they will not change the divisibility property of the rest of the expression:
$= \epsilon_1 - \epsilon_2 + \epsilon_4 - \epsilon_5 + \epsilon_7 - \epsilon_8$
$= (\epsilon_1 - \epsilon_2) + (\epsilon_4 - \epsilon_5) + (\epsilon_7 - \epsilon_8)$
For $\epsilon_i = \pm 1$, each of the above group can either be $0$, $2$ or $-2$. No other value is possible.
Each group can take the value of $0$ in two ways $\epsilon_i = \epsilon_{i+1}$
Whereas, they can take the value of $2$ or $-2$, each in $1$ way, $\epsilon_i = +1$ and $\epsilon_{i+1} = -1$ and $\epsilon_i = -1$ and $\epsilon_{i+1} = +1$.
For the entire expression to be divisible by $3$, each group can have the same value $0$, $2$ or $-2$ or any one of them can be $0$, remaining two can be $-2$ and $2$
All groups $0:$ $2 \times 2 \times 2 = 8$ ways
All groups $2:$ $1 \times 1 \times 1 = 1$ way
All groups $-2:$ $1$ way
Groups $0$, $2$ and $-2$ can occur in any order. We can choose the order in $3! = 6$ ways, and for each order, there are $2 \times 1 \times 1$ ways of getting the values of the groups. Therefore, we get $6 \times 2 = 12$ ways for this selection.
Totally, there are $8 + 1 + 1 + 12 = 22$ ways for getting this expression divisible by $3$.
Initially we had eliminated $3\epsilon_3$ and $6\epsilon_6$, and each of them can take two values, for any combination of all other $\epsilon_i$.
Therefore, there are totally $22 \times 2 \times 2 = 88$ $8$-tuples of $\epsilon_i$ for which the given expression will be divisible by $3$.
Problem 13 of 30
In a $\triangle ABC$, right-angled at $A$, the altitude through $A$ and the internal bisector of $\angle A$ have lengths $3$ and $4$ respectively. Find the length of the median through $A$.Solution:
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We can draw the diagram for the given problem as shown below, where $AD$, $AE$ and $AF$ are the altitude, angle bisector and median respectively, of $\triangle ABC$ right angled at $A$.
We know that
$ar[\triangle ABC] = \dfrac{1}{2} AC \times AB = \dfrac{1}{2}AD \times BC$
$\Rightarrow \dfrac{1}{2} AC \times AB = \dfrac{1}{2} \times 3 \times BC$
$\Rightarrow AC \times AB = 3 \times BC$
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Also,
$ar[\triangle ABC] = ar[\triangle AEC] + ar[\triangle AEB]$
Using the area rule from ,
$ar[\triangle ABC] = \dfrac{1}{2}AE \times AC \times sin(45) + \dfrac{1}{2}AE \times AB \times sin(45)$
$\Rightarrow ar[\triangle ABC] = \dfrac{1}{2\sqrt{2}} AE (AC + AB)$
$\Rightarrow ar[\triangle ABC] = \sqrt{2} (AC + AB)$
$\therefore \dfrac{1}{2}AC \times AB = \sqrt{2} (AC + AB)$
$\Rightarrow AC \times AB = 2\sqrt{2}(AC + AB)$
$\Rightarrow AC \times AB = 2\sqrt{2}(AC + AB)$
$\Rightarrow AC^2AB^2 = 8(AC^2 + AB^2 + 2AC.AB)$
$\Rightarrow 9BC^2 = 8(BC^2 + 6BC)$
$\Rightarrow 9BC^2 = 8BC^2 + 48BC$
$\Rightarrow 9BC = 8BC + 48$
$\Rightarrow BC = 48$
Since $F$ is the midpoint of the hypotenuse of a right angled triangle $AF = \dfrac{1}{2} BC = \dfrac{1}{2} 48 = 24$
Problem 14 of 30
If $x = cos1^\circ cos2^\circ cos3^\circ\ ...\ cos89^\circ$ and$y = cos2^\circ cos6^\circ cos10^\circ\ ...\ cos86^\circ$,
then what is the integer nearest to
$\dfrac{2}{7} \log_{2}\left(\dfrac{y}{x}\right)$?
Solution:
$x$ has $89$ terms, with the middle term being $cos45^\circ$
We can rewrite $x$ as:
$x = cos1^\circ cos2^\circ cos3^\circ...\ cos44^\circ cos45^\circ cos(90-44)^\circ ... $$\ \ \ cos(90 - 3)^\circ cos(90-2)^\circ cos(90 - 1)^\circ$
Since $cos(90 - x) = sin(x)$
$x = cos1^\circ cos2^\circ cos3^\circ...\ cos44^\circ cos45^\circ sin44^\circ ... sin3^\circ sin2^\circ sin1^\circ$
Now we can pair up each $cos$ term with the corresponding $sin$, except for $cos45^\circ$.
Therefore, we get:
$x = (cos1^\circ sin1^\circ) (cos2^\circ sin2^\circ) (cos3^\circ sin3^\circ) ...\ (cos44^\circ sin44^\circ) cos45^\circ$
$cos(A)sin(A)$ can be written as:
$cos(A)sin(A) = \dfrac{sin2A}{2}$.
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Therefore, we get:
$x = \dfrac{sin2^\circ}{2} \dfrac{sin4^\circ}{2} \dfrac{sin6^\circ}{2}... \dfrac{sin88^\circ}{2} \times cos45^\circ $
$\Rightarrow x = \dfrac{1}{2^{44}\sqrt{2}}(sin2^\circ sin4^\circ sin6^\circ ...sin88^\circ)$
$\Rightarrow x = \dfrac{1}{2^{44}\sqrt{2}}(sin2^\circ sin4^\circ sin6^\circ... sin44^\circ sin46^\circ...sin88^\circ)$
$\Rightarrow x = \dfrac{1}{2^{44}\sqrt{2}}(sin2^\circ sin4^\circ sin6^\circ... sin44^\circ cos44^\circ...cos2^\circ)$
$\Rightarrow x = \dfrac{1}{2^{44}\sqrt{2}}(sin2^\circ cos2^\circ sin4^\circ cos4^\circ sin6^\circ cos6^\circ... sin44^\circ cos44^\circ)$
$\Rightarrow x = \dfrac{1}{2^{44}\sqrt{2}}\dfrac{1}{2^{22}}(sin4^\circ sin8^\circ sin12^\circ ... sin88^\circ)$
$\Rightarrow x = \dfrac{1}{2^{66}\sqrt{2}}(sin(90-86)^\circ sin(90 - 82)^\circ sin(90 - 78)^\circ ... sin(90 - 2)^\circ)$
$\Rightarrow x = \dfrac{1}{2^{66}\sqrt{2}}(cos86^\circ cos82^\circ cos78^\circ ... cos6^\circ cos2^\circ)$
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$\therefore \dfrac{y}{x} = \dfrac{cos2^\circ cos6^\circ cos10^\circ\ ...\ cos86^\circ}{\frac{1}{2^{66}\sqrt{2}}(cos86^\circ cos82^\circ cos78^\circ ... cos6^\circ cos2^\circ)} = 2^{66}\sqrt{2}$
$\therefore log_2{\dfrac{y}{x}} = log_2{\left(2^{66.5}\right)} = 66.5$
$\therefore \dfrac{2}{7}log_2{\dfrac{y}{x}} = \dfrac{2}{7}66.5 = 2 \times 9.5 = 19$
Problem 15 of 30
Let $a$ and $b$ are natural numbers, such that $2a - b$, $a - 2b$ and $a + b$ are distinct squares.What is the smallest possible value of $b$?
Solution:
Let,$2a - b = p^2$ $eqn\ (i)$
$a - 2b = q^2$ $eqn\ (ii)$
$a + b = r^2$ $eqn\ (iii)$
where $p,\ q,\ r$ are integers.
Subtracting equation $(ii)$ from $(i)$, we get:
$(2a - b) - (a - 2b) = p^2 - q^2$
$\Rightarrow a + b = p^2 - q^2$
$\Rightarrow r^2 = p^2 - q^2$
$\Rightarrow r^2 + q^2 = p^2$
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Therefore, $r$, $q$ and $p$ is a Pythagorean triplet, with $p$ as the hypotenuse.
Again, subtracting equation $(ii)$ from $(iii)$ we get:
$3b = r^2 - q^2$
$b = \dfrac{r^2 - q^2}{3}$
A perfect square, when divided by $3$, can leave a remainder of either $0$ or $1$. It cannot leave a remainder of $2$.
(You can attempt to prove this separately. Any integer can be expressed as one of $3N$, $3N + 1$ or $3N + 2$. Take their squares and check the congruence modulo $3$)
Therefore, difference of two perfect squares can be divisible by $3$, if and only if they are each divisible by $3$. Therefore, $p$, $q$ and $r$, each is divisible by $3$
The smallest Pythagorean triplet where all three sides are divisible by $3$ are:
$9,\ 12$ and $15$
Therefore, $b = \dfrac{12^2 - 9^2}{3} = \dfrac{21 \times 3}{3} = 21$
Problem 16 of 30
What is the value of$\sum\limits_{\begin{matrix} 1 \le i \lt j \le 10 \\ i+j = odd\end{matrix}}(i + j) - \sum\limits_{\begin{matrix} 1 \le i \lt j \le 10 \\ i+j = even\end{matrix}}(i + j)$
Solution:
We can break up the first part which is$\sum\limits_{\begin{matrix} 1 \le i \lt j \le 10 \\ i+j = odd\end{matrix}}(i + j)$ into multiple series as follows:
For $i = 1,\ j = 2 \longrightarrow 10$, we get the series:
$(1+2),\ (1 + 4),\ ...,\ (1+ 10)$ $(Series\ a1,\ 5\ terms)$
For $i = 2,\ j = 3 \longrightarrow 9$, we get:
$(2 + 3),\ (2 + 5),\ ...,\ (2 + 9)$ $(Series\ a2,\ 4\ terms)$
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For $i = 3,\ j = 4 \longrightarrow 10$, we get:
$(3 + 4),\ (3 + 6),\ ...,\ (3 + 10)$ $(Series\ a3,\ 4\ terms)$
$.$
$.$
$.$
$i = 9,\ j = 10 \longrightarrow 10$
$(9+10)$ $(Series\ a-9,\ 1\ term)$
Similarly for the second part $\sum\limits_{\begin{matrix} 1 \le i \lt j \le 10 \\ i+j = even\end{matrix}}(i + j)$ can be broken down as multiple series as:
For $i = 1,\ j = 3 \longrightarrow 9$, we get:
$(1 + 3),\ (1 + 5),\ ...,\ (1 + 9)$ $(Series\ b1,\ 4\ terms)$
For $i = 2,\ j = 4 \longrightarrow 10$, we get:
$(2+4),\ (2+6),\ ...,\ (2 + 10)$ $(Series\ b2,\ 4\ terms)$
For $i = 3,\ j = 5 \longrightarrow 7$, we get:
$(3+5),\ (3+7),\ ...,\ (3+9)$ $(Series\ b3,\ 3\ terms)$
$.$
$.$
$.$
$i = 8$
$(8+10)$ $(Series\ b-8,\ 1\ term)$
The $j$ values $3,\ 5,\ ...,\ 9$ of $Series\ b1$ will cancel out with corresponding $j$ values in $Series\ a2$
All the four $1$s ($i$ values) in $Series\ b1$ four will cancel out with the $1$ in $Series\ a1$ leaving only one instance of $1$.
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Similarly, the $j$ values $4,\ 6,\ ...,\ 10$ of $Series\ b2$ will cancel out with the corresponding values of $Series\ a3$
Out of the four $2$s in $Series\ b2$ all of them will cancel out with the $2$s of $Series\ a2$
Likewise, all the even numbers of the $i$ values in $Series\ b$ will cancel out with $Series\ a$, while one instance of the odd numbers of the $i$ values will remain in $Series a$ after cancellation.
Also, none of the $j$ values of $Series\ a1$ will cancel out and all of them will remain.
Therefore, we will be left with:
$1,\ 3\ 5\ ...,\ 9$ from the $i$ values, and $2,\ 4,\ 8\ ...,\ 10$ of the $j$ values.
The sum of these numbers are:
$\sum\limits_{k=1}^{10} k = \dfrac{10\times 11}{2} = 55$
Problem 17 of 30
Triangle $ABC$ and $DEF$ are such that $\angle A = \angle D$, $AB = DE = 17$, $BC = EF = 10$ and $AC - DF = 12$. What is $AC + DF$?Solution:
We can draw the diagram for $\triangle ABC$ and $\triangle DEF$ separately as follows:
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We have marked the equal sides and the equal angles as shown above.
Now we can do the following construction for $\triangle ABC$.
Since $AC - DF = 12$, we will extend $DF$ by $12$ units, to $G$, join $EG$ and draw a line $EP$ from $E$ perpendicular to $BG$. We will get the following figure:
$\because AC = DF + 12 = DG$, $AC = DG$, and given that $AB = DE$ and $\angle A = \angle D$
Therefore, $\triangle DEG \cong \triangle ABC$, and $EG = BC$
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$\because BC = EF$ (given),
$EG = EF$
$\therefore \triangle EGF$ is isosceles.
Since $EP \perp FG$, $P$ is the midpoint of $FG$ and
$FP = PG$.
$FG = DG - DF = AC - DF = 12$
$FP = PG = \dfrac{FG}{2} = 6$
$\because \triangle EPF$ is right angled at $P$,
$EP^2 = EF^2 - PF^2 = 10^2 - 6^2 = 64$
$\Rightarrow EP = 8$
$\because \triangle DPE$ is right angled at $P$,
$DP^2 = DE^2 - EP^2 = 17^2 - 8^2$
$\Rightarrow DP = \sqrt{(17 - 8)(17 + 8)} = \sqrt{9 \times 25} = 15$
Therefore,
$AC + DF = DG + DF = (DP - 6) + (DP + 6) = 2DP = 30$
Problem 18 of 30
If $a,\ b,\ c \ge 4$ are integers, not all equal, and $4abc = (a + 3)(b+3)(c+3)$, then what is the value of $a + b + c$?Solution:
From the given relation we get:$4abc = (a + 3)(b+3)(c+3)$
$\Rightarrow 4abc = 27 + 3(ab + bc + ca) + 9(a + b + c) + abc$
$\Rightarrow 3abc = 27 + 3(ab + bc + ca) + 9(a + b + c)$
$\Rightarrow abc = 9 + (ab + bc + ca) + 3(a + b + c)$
We know that $(x - 1)(y - 1)(z - 1) = x + y + z - xy - yz - zx + xyz - 1$. We can rearrange our relation in the following way to utilise this identity.
$abc - (ab + bc + ca) + (a + b + c) - 1 = 8 + 4(a + b + c)$
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$\Rightarrow abc - ab - bc - ca + a + b + c - 1 = 8 + 4(a + b + c)$
$\Rightarrow (a - 1)(b - 1)(c - 1) = 8 + 4(a + b + c)$
Let $a - 1 = p$, $b - 1 = q$ and $c - 1 = r$
Therefore we get:
$pqr = 8 + 4(p + q + r + 3)$
$\Rightarrow pqr = 20 + 4(p + q + r)$
$\Rightarrow pqr = 4(5 + p + q + r)$
$\Rightarrow pqr - 4p = 20 + 4(q + r)$
$\Rightarrow p = \dfrac{4(5 + q + r)}{qr - 4}$
For $p$ to be an integer, $qr$ must be divisible by $4$.
It is given that $a,\ b,\ c \ge 4$ therefore, $p,\ q,\ r \ge 3$
We can try the smallest value of $q$ which is divisible by $4$ by taking $q = 4$
We get $p = \dfrac{4(5 + 4 + r)}{4r - 4} = \dfrac{9 + r}{r - 1}$
The smallest possible value of $r$ for which the right hand side is an integer, satisfying $r \ge 3$ is $r = 3$
Therefore $p = 6$
Therefore, $a = p + 1 = 7$, $b = q + 1 = 4 + 1 = 5$, and $c = r + 1 = 4$
Therefore, $a + b + c = 7 + 5 + 4 = 16$
Observe, that the values of $a$, $b$ and $c$ obtained here are interchangeable, since the given equation is symmetric for $a,\ b,\ c$
Problem 19 of 30
Let $N = 6 + 66 + 666 +\ ...\ + 666...66$, where there are hundred $6$'s in the last term in the sum. How many times does the digit $7$ occur in the number $N$?Solution:
$6 + 66 + 666 + ... + \underbrace{666...66}_{100\ times}$$= 6(1 + 11 + 111 + ... + 111...11)$
$= \dfrac{6}{9}(9 + 99 + 999 + ... + 999...99)$
$= \dfrac{6}{9}\left\{(10 - 1) + (100 - 1) + (1000 - 1) + ... + (100...000 - 1)\right\}$
$= \dfrac{6}{9}\underbrace{\left\{(10^1 - 1) + (10^2 - 1) + (10^3 - 1) + ... + (10^{100} - 1)\right\}}_{100\ terms}$
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$= \dfrac{6}{9} \left\{(\underbrace{10^1 + 10^2 + 10^3 + ... + 10^{100}}_{100\ terms}) - (\underbrace{1 + 1 + 1 + ... + 1}_{100\ terms})\right\}$
$= \dfrac{6}{9} \left\{(10^1 + 10^2 + 10^3 + ... + 10^{100}) - 100\right\}$
$= \dfrac{6}{9} \left\{10 + (\underbrace{10^2 + 10^3 + ... + 10^{100}}_{99\ terms}) - 100\right\}$
$= \dfrac{6}{9} \left\{(10^2 + 10^3 + ... + 10^{100}) - 90\right\}$
$= \dfrac{6}{9} (10^2 + 10^3 + ... + 10^{100}) - 60$
$= \dfrac{2}{3} \left\{\dfrac{10^2(10^{99} - 1)}{10 - 1}\right\} - 60$
$= \dfrac{200}{3} \left\{\dfrac{\overbrace{999...999}^{99\ digits}}{9}\right\} - 60$
$= \dfrac{200}{3} \left\{111...111\right\} - 60$
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$= \dfrac{\overbrace{222...222}^{99\ digits} \times 100}{3} - 60$
Since the number $222$ is divisible by $3$ we will create groups of $3$ digits in the following way:
$= \dfrac{(\overbrace{222 \times 10^{96} + 222 \times 10^{93} + ... + 222 \times 10^3 + 222}^{33\ terms}) \times 100}{3} - 60$
$= (074 \times 10^{96} + 074 \times 10^{93} + ... + 074 \times 10^3 + 074)\times 100 - 60$
$= 074 \times 10^{98} + 074 \times 10^{95} + ... + 074 \times 10^5 + 07400 - 60$
$= 074 \times 10^{98} + 074 \times 10^{95} + ... + 074 \times 10^5 + 07340$
There are $33$ groups each containing one $7$, hence the digit $7$ occurs $33$ times.
Problem 20 of 30
Determine the sum of all possible positive integers $n$, the product of whose digits equals $n^2 - 15n - 27$.Solution:
The expression $n^2 - 15n - 27$ is odd for either odd or even $n$, therefore all digits of the number $n$ has to be odd.Let $P(n) = n^2 - 15n - 27$ be a polynomial in $n$.
Let us find the minimum value of the polynomial.
$n^2 - 15n - 27$
$= n^2 - 2.\dfrac{15}{2}n + \left(\dfrac{15}{2}\right)^2 - \left(\dfrac{15}{2}\right)^2 - 27$
$= \left(n - \dfrac{15}{2}\right)^2 - \dfrac{225}{4} - 27$
$= \left(n - \dfrac{15}{2}\right)^2 - \dfrac{69}{2}$
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By putting $n = \dfrac{15}{2}$, we can make the square term vanish, and be left with $-\dfrac{69}{2}$
Therefore, for $n \ge \dfrac{15}{2}$ it is an increasing function.
The zeros of the polynomial are at $n = \dfrac{15 \pm \sqrt{225 + 4\times 27}}{2}$
Therefore, from $n = \dfrac{15}{2}$ to $n = \dfrac{15 + \sqrt{333}}{2} = \dfrac{15 + 3\sqrt{37}}{2} \approx \dfrac{15 + 3\times 6.1}{2} = 16.1$ the values of $P(n)$ will be negative.
Therefore, the minimum possible value of $n$ is $17$.
Putting $n = 17$, in the given expression we get:
$P(17) = 17^2 - 15\times 17 - 27 = 7$, which is same as the the product of the digits $1 \times 7 = 7$
The next possible candidate is $19$ (all odd digits)
$P(19) = 19^2 - 15\times 19 - 27 = 49$ This does not satisfy the condition, since the product of the digits is $1 \times 9 = 9$
The next candidate will be $21$
$P(21) = 21^2 - 15 \times 21 - 27 = 99$
The maximum possible value of the product of the digits of a two-digit number is $9 \times 9 = 81$.
Therefore, $21$ or any two-digit number greater than $21$ is not a possible valid value.
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Let us look at the $3$-digit numbers. The minimum $3$-digit number with all odd digits is $111$.
$P(111) = 10629$, which is much larger than the product of the digits of the maximum $3$-digit number, which is $9 \times 9 \times 9 = 729$
Therefore, the only value satisfying the given conditions is $17$
Problem 21 of 30
Let $ABC$ be an acute-angled triangle and let $H$ be its orthocentre. Let $G_1$, $G_2$ and $G_3$ be the centroids of the triangles $HBC$, $HCA$ and $HAB$ respectively. If the area of triangle $G_1G_2G_3$ is $7$ units, what is the area of triangle $ABC$?Solution:
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Let us draw the diagram for the given problem as shown below:
Let us join $HG_1$ and extend it to meet $BC$ at $P$, similarly let extended $HG_2$ and $HG_3$ meet sides $CA$ and $AB$ at $Q$ and $R$ respectively.
Since $G_1$ is the centroid of $\triangle HBC$, $HP$ is the median, and $P$ is the midpoint of $BC$. Similarly, $Q$ and $R$ are the midpoints of $CA$ and $AB$.
Since $P$ and $Q$ are midpoints of $BC$ and $CA$ respectively, $PQ \parallel BC$ and $PQ = \dfrac{1}{2} AB$.
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Similarly, $RQ = \dfrac{1}{2} BC$ and $RP = \dfrac{1}{2} CA$.
$ar[\triangle PQR] = \dfrac{1}{4} ar[\triangle ABC]$
We know from , that the centroid divides the median in the ratio $2:1$, therefore,
$HG_1:G_1P = 2:1$ $\because HP$ is the median and $G_1$ is the centroid of $\triangle HBC$.
Similarly $HG_2:G_2Q = 2:1$
Considering $\triangle HPQ$,
$HG_1:G_1P = HG_2:G_2P$
$\angle PHQ$ is common. Therefore:
$\triangle HPQ \sim \triangle HG_1G_2$
And $G_1G_2 \parallel PQ$ also, $G_1G_2 = \dfrac{2}{3} PQ$
Similarly $G_2G_3 = \dfrac{2}{3} QR$ and
$G_3G_1 = \dfrac{2}{3} AB$
Therefore, $\triangle G_1G_2G_3 \sim \triangle PQR$ with
$ar[\triangle G_1G_2G_3] = \left(\dfrac{2}{3}\right)^2 ar[\triangle PQR] = \dfrac{4}{9} ar[\triangle PQR]$
Therefore,
$ar[\triangle G_1G_2G_3] = \dfrac{4}{9} \times \dfrac{1}{4} ar[\triangle ABC]$
Given that $ar[\triangle G_1G_2G_3] = 7$, $ar[\triangle ABC] = 63$
Problem 22 of 30
A positive integer $k$ is said to be good if there exists a partition of $\{1,\ 2,\ 3,\ ...\ 20\}$ into disjoint proper subsets such that the sum of the numbers in each subset of the partition is $k$. How many good numbers are there?Solution:
The sum of the numbers $1 + 2 + ... + 20 = \dfrac{20 \times 21}{2} = 210$We need to figure out the possible factors of $210$
$210$ can be written as $2 \times 3 \times 5 \times 7$
Counting the divisors of $210$ using the method described , we know that it has $(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 16$ factors, which are:
$1,\ 2,\ 3,\ 5,\ 6,\ 7,\ 10,\ 14,\ 15,\ 21,\ 30,\ 35,\ 42,\ 70,\ 105,\ 210$
Since, the largest number in the given set is $20$, $k$ cannot be less than $20$
If $k = 210$, then there can be only one subset which will contain all the elements of the given set, hence will not be a proper subset.
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Therefore, the possible values of $k$ are $105,\ 70,\ 42,\ 35,\ 30,\ 21$
For $k = 105$, there will be $210 \div 105 = 2$ subsets, and the subsets can be:
$\{1,\ 3,\ 5,\ 7,\ 9,\ 12,\ 14,\ 16,\ 18,\ 20\}$ and $\{2,\ 4,\ 6,\ 8,\ 10,\ 11,\ 13,\ 15,\ 17,\ 19\}$
For $k = 70$, there will be $210 \div 70 = 3$ subsets, which can be:
$\{1,\ 7,\ 11,\ 14,\ 17,\ 20\}$, $\{2,\ 3,\ 4,\ 5,\ 8,\ 13,\ 16,\ 19\}$ and $\{6,\ 9,\ 10,\ 12,\ 15,\ 18\}$
For $k = 42$, there will be $210 \div 42 = 5$ subsets, which can be:
$\{1,\ 6,\ 15,\ 20\}$, $\{2,\ 7,\ 14,\ 19\}$, $\{3,\ 8,\ 13,\ 18\}$, $\{4,\ 9,\ 12,\ 17\}$ and $\{5,\ 10,\ 11,\ 16\}$
For $k = 35$, there will be $6$ subsets, which can be:
$\{1,\ 14,\ 20\}$, $\{16,\ 19\}$, $\{3,\ 5,\ 9,\ 18\}$, $\{2,\ 6,\ 10,\ 17\}$, $\{4,\ 8,\ 11,\ 12\}$ and $\{7,\ 13,\ 17\}$
For $k = 30$, there will be $7$ subsets, and the following subsets are possible:
$\{10,\ 20\}$, $\{11,\ 19\}$, $\{12,\ 18\}$, $\{13,\ 17\}$, $\{5,\ 9,\ 16\}$, $\{6,\ 8,\ 15\}$ and $\{2,\ 3,\ 4,\ 7,\ 14\}$
If $k = 21$, there will be $10$ subsets, which can be:
$\{1,\ 20\}$, $\{2,\ 19\}$, $\{3,\ 18\}$, $\{4,\ 17\}$, $\{5,\ 16\}$, $\{6,\ 15\}$, $\{7,\ 14\}$, $\{8,\ 13\}$, $\{9,\ 12\}$ and $\{10,\ 11\}$
Since all the $6$ values of $k$ are possible, the correct answer is $6$.
Observe that each of the above cases except $k = 21$, there can be more than one possible choice of subsets. But our aim is to find if there is at least one possible subset for each value of $k$ satisfying the given condition.
Problem 23 of 30
What is the largest positive integer $n$ such that$\dfrac{a^2}{\frac{b}{29} + \frac{c}{31}} + \dfrac{b^2}{\frac{c}{29} + \frac{a}{31}} + \dfrac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge n(a + b + c)$
holds for all positive real numbers $a,\ b,\ c$?
Solution:
From Titu's Lemma explained ,we know that:
$\dfrac{a_1^2}{b_1} + \dfrac{a_2^2}{b_2} + ... + \dfrac{a_n^2}{b_n} \ge \dfrac{(a_1 + a_2 + ... + a_3)^2}{b_1 + b_2 + ... + b_n^2}$
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To be able to use this theory, we can use the following mappings:
$a = a_1$
$b = a_2$
$c = a_3$
$\dfrac{b}{29} + \dfrac{c}{31} = b_1$
$\dfrac{c}{29} + \dfrac{a}{31} = b_2$
$\dfrac{a}{29} + \dfrac{b}{31} = b_3$
Therefore, we get:
$\dfrac{a^2}{\frac{b}{29} + \frac{c}{31}} + \dfrac{b^2}{\frac{c}{29} + \frac{a}{31}} + \dfrac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge \dfrac{(a + b + c)^2}{\frac{b}{29} + \frac{c}{31} + \frac{c}{29} + \frac{a}{31} + \frac{a}{29} + \frac{b}{31}}$
$\Rightarrow \dfrac{a^2}{\frac{b}{29} + \frac{c}{31}} + \dfrac{b^2}{\frac{c}{29} + \frac{a}{31}} + \dfrac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge \dfrac{(a + b + c)^2}{\frac{a + b + c}{29} + \frac{a + b + c}{31}}$
$\Rightarrow \dfrac{a^2}{\frac{b}{29} + \frac{c}{31}} + \dfrac{b^2}{\frac{c}{29} + \frac{a}{31}} + \dfrac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge \dfrac{(a + b + c)^2}{(a + b + c)\left(\frac{1}{29} + \frac{1}{31}\right)}$
$\Rightarrow \dfrac{a^2}{\frac{b}{29} + \frac{c}{31}} + \dfrac{b^2}{\frac{c}{29} + \frac{a}{31}} + \dfrac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge \dfrac{(a + b + c)}{\frac{60}{29 \times 31}}$
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$\Rightarrow \dfrac{a^2}{\frac{b}{29} + \frac{c}{31}} + \dfrac{b^2}{\frac{c}{29} + \frac{a}{31}} + \dfrac{c^2}{\frac{a}{29} + \frac{b}{31}} \ge \dfrac{899}{60}(a + b + c)$
$\Rightarrow \dfrac{a^2}{\frac{b}{29} + \frac{c}{31}} + \dfrac{b^2}{\frac{c}{29} + \frac{a}{31}} + \dfrac{c^2}{\frac{a}{29} + \frac{b}{31}}$
Since, $a,\ b,\ c$ are positive, this will hold for all $n \le \dfrac{899}{60} = 14\dfrac{59}{60}$.
The largest integer not exceeding $14\dfrac{59}{60}$ is $14$
Problem 24 of 30
If $N$ be the number of triangles of different shapes (i.e. not similar) whose angles are all integers (in degrees), what is $\dfrac{N}{100}$.Solution:
We can consider each degree as an identical object, and $180$ of these objects can be distributed into $3$ distinct bins in, such that each bin contains at least $1$ degree. To do that we can first assign $1^\circ$ to each of the three angles, and then distribute the remaining $177^\circ$ amongst the three angles using the method described .
The total number of ways of distributing these degrees is $\xacomb{177 + 2}{2} = \xacomb{179}{2}$
This distribution will include $3! = 6$ repeats of the scalene triangles, $3$ repeats of each isosceles triangle and no repeat of the equilateral triangle.
There is one equilateral triangle in the distribution.
An isosceles triangle can be formed for all base angles less than $90^\circ$, that is $1^\circ$ to $89^\circ$, except the base angle value of $60^\circ$ therefore, $88$ different triangles.
If every triangle repeated $6$ times, we would have,
$\xacomb{179}{2} + 88 \times 3 + 5$
(In the above step we added $3$ more repeats for all the isosceles triangles and $5$ more repeats for the equilateral triangle)
Since the above figure contains all triangles repeated exactly $6$ times, the total number of non-similar triangles is:
$= \dfrac{179 \times 89 + 88 \times 3 + 5}{6}$
$= 2700$
Therefore, the answer is $\dfrac{2700}{100} = 27$.
Problem 25 of 30
Let $T$ be the smallest positive integer which, when divided by $11$, $13$, $15$ leaves remainders in the sets $\{7,\ 8,\ 9\}$, $\{1,\ 2,\ 3\}$, $\{4,\ 5,\ 6\}$ respectively. What is the sum of the squares of the digits of $T$?Solution:
Let $T$ be a number of the form $15N + 4$ or $15N + 5$ or $15N + 6$ where $N$ is an integer.$\underline{Case\ 1\ (T = 15N + 4)}$:
$T = 15N + 4 \equiv 4N + 4\ (mod\ 11)$
$\therefore 4N + 4 \equiv\ 7\ or\ 8\ or\ 9\ (mod\ 11)$
$\Rightarrow 4N \equiv 3\ or\ 4\ or\ 5\ (mod\ 11)$
$\therefore 4N\ \epsilon\ \{14,\ 15,\ 16,\ 25,\ 26,\ 27,\ 36,\ 37,\ 38,\ 47,\ 48,\ 49,\ 60,\ 61,\ 62,\ ...\}$
$\therefore N\ \epsilon\ \{4,\ 9,\ 12,\ 15,\ ...\}$
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Similarly,
$T = 15N + 4 \equiv 2N + 4\ (mod\ 13)$
$\therefore 2N + 4 \equiv 1\ or\ 2\ or\ 3\ (mod\ 13)$
$\therefore 2N \equiv -3\ or\ -2\ or\ -1\ (mod\ 13)$
$\therefore 2N\ \epsilon\ \{10,\ 11,\ 12,\ 23,\ 24,\ 25,\ 36,\ 37,\ 38,\ ...\}$
$\therefore N\ \epsilon\ \{5,\ 6,\ 12,\ 18,\ 19,\ ...\}$
We can see that the smallest value of $N$ satisfying both conditions above is $N = 12$
Therefore, $T = 15 \times 12 + 4 = 184$
$\underline{Case\ 2\ (T = 15N + 5)}$:
$T = 15N + 5 \equiv 4N + 5\ (mod\ 11)$
$\therefore 4N + 5 \equiv\ 7\ or\ 8\ or\ 9\ (mod\ 11)$
$\Rightarrow 4N \equiv 2\ or\ 3\ or\ 4\ (mod\ 11)$
$\therefore 4N\ \epsilon\ \{13,\ 14,\ 15,\ 24,\ 25,\ 26,\ 35,\ 36,\ 37,\ 46,\ 47,\ 48,\ 57,\ 58,\ 59,\ ...\}$
$\therefore N\ \epsilon\ \{6,\ 9,\ 12,\ ...\}$
Similarly,
$T = 15N + 5 \equiv 2N + 5\ (mod\ 13)$
$\therefore 2N + 5 \equiv 1\ or\ 2\ or\ 3\ (mod\ 13)$
$\Rightarrow 2N \equiv -4\ or\ -3\ or\ -2\ (mod\ 13)$
$\therefore 2N\ \epsilon\ \{9,\ 10,\ 11,\ 22,\ 23,\ 24,\ 35,\ 36,\ 37,\ ...\}$
$\therefore N\ \epsilon\ \{5,\ 11,\ 12,\ 18,\ ...\}$
Again, the smallest value of $N$ satisfying both cases is $N = 12$.
Since we already have considered $N = 12$ as the minimum value, we can ignore this.
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$\underline{Case\ 3\ (T = 15N + 6)}$:
$T = 15N + 6 \equiv 4N + 6\ (mod\ 11)$
$\therefore 4N + 6 \equiv\ 7\ or\ 8\ or\ 9\ (mod\ 11)$
$\Rightarrow 4N \equiv 1\ or\ 2\ or\ 3\ (mod\ 11)$
$\therefore 4N\ \epsilon\ \{1,\ 2,\ 3,\ 12,\ 13,\ 14, 23,\ 24,\ 25,\ 34,\ 35,\ 36,\ 45,\ 46,\ 47,\ 56,\ 57,\ 58,\ 67,\ 68,\ 69...\}$
$\therefore N\ \epsilon\ \{3,\ 8,\ 9,\ 14,\ 17...\}$
Since, we have obtained the minimum value of $N$ as $12$, we can check for each $N = 3,\ 8,\ 9$, if that satisfies the condition
$T = 15N + 6 \equiv\ 1\ or\ 2\ or\ 3\ (mod\ 13)$
For $N = 3$, $15N + 6= 51$
$51 \equiv 12\ (mod\ 13)$
For $N = 8$, $15N + 6 = 126$
$126 \equiv 9\ (mod\ 13)$
For $N = 9$, $15N + 6 = 141$
$141 \equiv 11\ (mod\ 13)$
None of these values satisfy the above condition.
Therefore, the minimum value of $N$ is $12$, and the minimum value of $T$ satisfying the given condition is $184$
Sum of the squares of the digits of $T$ is $1^2 + 8^2 + 4^2 = 81$
Problem 26 of 30
What is the number of ways in which one can choose $60$ unit squares from a $11 \times 11$ chessboard, such that no two chosen squares have a side in common?Solution:
There are a total number of $11 \times 11 = 121$ squares.Let's assume these are coloured, like a normal chessboard, alternate black and white. If the first square is black, then there will be $61$ black squares and $60$ white squares.
To be able to choose $60$ squares which do not share any side we should select either all black squares or all white squares.
Therefore, we can choose $60$ black squares in $\xacomb{61}{60} = \xacomb{61}{1} = 61$ ways, or,
$60$ white squares in $\xacomb{60}{60} = 1$ way.
The total number of selecting the squares is $62$
Problem 27 of 30
What is the number of ways in which one can colour the squares of a $4 \times 4$ chessboard with colours red and blue such that each row as well as each column has exactly two red squares and two blue squares?Solution:
We can solve this problem easily if we assign digits to the individual colours.
We can assign the digit $1$ for red and $2$ for blue.
For each row to have two reds and two blues, we can form four digit numbers using digits $1$ and $2$ only.
For example the number $1221$ represents a valid row with colour arrangement $RBBR$
If each column has exactly two $1$s and two $2$s, then the sum of each column will be $6$ and the four numbers will always add up to $6666$
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Now, let us see how many four digit numbers we can form with two $1$s and two $2$s.
$1122$
$1212$
$2112$
$1221$
$2121$
$2211$
Now we can select four of the above numbers, repetitions allowed, such that the sum is $6666$.
Before that we can observe the restrictions of the selection method.
- We can select a number at most twice. A number repeated more than twice will cause the digits (colours) to repeat in the row more than twice.
- If we select a number twice, we need to repeat the other number twice as well, to maintain $2 + 2$ arrangement in a row.
Writing the above numbers in ascending order:
$1122,\ 1212,\ 1221,\ 2112,\ 2121,\ 2211$
We will see the possible selection where each number repeats twice:
$1122 + 1122 + 2211 + 2211 = 6666$
$1212 + 1212 + 2121 + 2121 = 6666$
$2112 + 2112 + 1221 + 1221 = 6666$
Next we will find the selections where no number repeats:
$1122 + 1212 + 2121 + 2211 = 6666$
$1122 + 1221 + 2112 + 2211 = 6666$
$1212 + 1221 + 2112 + 2121 = 6666$
Each number in the above selection corresponds to a row. Now we need to find how many ways we can arrange these numbers.
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For the first $3$ selections, each of the selected set of numbers can be arranged in $\dfrac{4!}{2! \times 2!} = 6$
For the $3$ selections without repetition, each selected set can be arranged in $4! = 24$ ways.
Therefore, the total number of ways to fill the digits (colours) is: $6 + 6 + 6 + 24 + 24 + 24 = 90$
$\underline{Alternate\ Approach}$
This approach is similar but requires knowledge of number bases, esp. binary numbers.
An alternate approach can be devised using binary digits $0$ and $1$ each representing one of the two colours.
Each number will contain two $0$s and two $1$s, leading zeros are counted.
Since each column has two $0$s and two $1$s, the sum of all the rows of a valid combination of numbers will always be:
$11110_2 = 30$
A valid row can be any of the following binary numbers:
$0011_2 = 3$
$0101_2 = 5$
$1001_2 = 9$
$0110_2 = 6$
$1010_2 = 10$
$1100_2 = 12$
So our valid rows can be represented using the numbers $3,\ 5,\ 6,\ 9,\ 10,\ 12$
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Now, if we select any four numbers from this set that add up to $30$, where repetitions are allowed, we will have a valid arrangement for the board.
We can take an example to understand this better. Let us say we select the numbers $3,\ 3,\ 12,\ 12$, which add up to $30$ and represent them as binary numbers, we get:
$0011$
$0011$
$1100$
$1100$
or if we choose $3,\ 5,\ 10,\ 12$ which also add up to $30$, we will get:
$0011$
$0101$
$1010$
$1100$
As we can see that both the above arrangements have exactly two $0$s and two $1$s in each row and column.
We can select the numbers in the following ways, with repetitions:
$3 + 3 + 12 + 12 = 30$
$5 + 5 + 10 + 10 = 30$
$6 + 6 + 12 + 12 = 30$
And without repetitions in the following ways:
$3 + 5 + 10 + 12 = 30$
$3 + 6 + 9 + 12 = 30$
$5 + 6 + 9 + 10 = 30$
As before, we can arrange each of the sets with repetitions in $\dfrac{4!}{2! \times 2!}$ ways and each of the sets without repetition in $4! = 24$ ways.
Total number of ways $= 6 + 6 + 6 + 24 + 24 + 24 = 90$
Problem 28 of 30
Let $N$ be the number of ways of distributing $8$ chocolates of different brands among $3$ children such that each child gets at least one chocolate, and no two children get the same number of chocolates. Find the sum of the digits of $N$.Solution:
We can break up the problem into two parts.Dividing $8$ distinct objects into $3$ groups, such that no group contains the same number of objects and each group contains at least $1$ object.
Distributing each of these distinct groups amongst $3$ children.
Based on the given conditions, $8$ objects can be divided into $3$ groups in $2$ ways, that is:
$(1,\ 3,\ 4)$ and $(1,\ 2,\ 5)$
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For the first case, there are:
$\xacomb{8}{1} \times \xacomb{7}{3} \times \xacomb{4}{4}$ ways of dividing the chocolates.
and for the second case there are:
$\xacomb{8}{1} \times \xacomb{7}{2} \times \xacomb{5}{5}$ ways of dividing the chocolates.
Therefore, there are a total of:
$\xacomb{8}{1} \times \xacomb{7}{3} \times \xacomb{4}{4} + \xacomb{8}{1} \times \xacomb{7}{2} \times \xacomb{5}{5}$
$= 8 \times 35 \times 1 + 8 \times 21 \times 1$
$= 8 \times 56 = 448$ ways of dividing the chocolates into $3$ groups.
For each of the above ways, there are $3!$ ways of distributing the $3$ groups amongst $3$ children.
Therefore, there are a total of $6 \times 448 = 2688$ ways of distributing these chocolates.
Therefore $N = 2688$
The sum of the digits of $N$ is $2 + 6 + 8 + 8 = 24$
Problem 29 of 30
Let $D$ be an interior point of the side $BC$ of a triangle $ABC$. Let $I_1$ and $I_2$ be the incentres of triangles $ABD$ and $ACD$ respectively. Let $AI_1$ and $AI_2$ meet $BC$ in $E$ and $F$ respectively.If $\angle BI_1E = 60^\circ$, what is the measure of $\angle CI_2F$ in degrees?
Solution:
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Let's draw the diagram for this problem as shown below:
$\angle ABC + \angle BAD + \angle DAC + \angle ACB = 180^\circ$
$\Rightarrow \dfrac{\angle ABC}{2} + \dfrac{\angle BAD}{2} + \dfrac{\angle DAC}{2} + \dfrac{\angle ACB}{2} = 90^\circ$ $...eqn\ (i)$
$\because I_1$ is the incentre of $\triangle BAD$, $BI_1$ and $AI_1$ are the bisectors of $\angle ABD$ and $\angle BAD$ respectively. Similarly, $AI_2$ and $CI_2$ are bisectors of $\angle DAC$ and $\angle DCA$ respectively.
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$\therefore \angle BI_1E = \angle ABI_1 + \angle BAI_1 = \dfrac{\angle ABD}{2} + \dfrac{\angle BAD}{2} = 60^\circ$
Substituting this in $eqn\ (i)$ we get:
$60 + \dfrac{\angle DAC}{2} + \dfrac{\angle ACB}{2} = 90^\circ$
$\dfrac{\angle DAC}{2} + \dfrac{\angle ACB}{2} = 30^\circ$
$\angle CI_2F = \dfrac{\angle DAC}{2} + \dfrac{\angle ACB}{2} = 30^\circ$
Problem 30 of 30
Let $P(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ be a polynomial in which $a_i$ is non-negative integer for each $i\ \epsilon\ \{0,\ 1,\ 2,\ 3,\ ...,\ n\}$.If $P(1) = 4$ and $P(5) = 136$, what is the value of $P(3)$
Solution:
Since $P(1) = 4$$a_0 + a_1 + a_2 + ... + a_n = 4$
Since it is given that $a_i$ is non-negative integer for all $i$, and their sum is $4$, we can conclude that:
$0 \le a_i \le 4$ for all $i$
$P(5) = 136$
$\Rightarrow a_0 + a_15 + a_25^2 + ... + a_n5^n = 136$
$\Rightarrow a_0 + 5(a_1 + a_25 + ... + a_n5^{n-1}) = 136$
$\Rightarrow 5(a_1 + a_25 + ... + a_n5^{n-1}) = 136 - a_0$
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Therefore, $136 - a_0$ is divisible by $5$
Therefore, $a_0$ can be $1$ or $6$ or $11$ or any number of the form $5m + 1$ where $m$ is an integer.
Since $a_0 \le 4$, the only possible value of $a_0$ is $1$.
Therefore,
$1 + 5(a_1 + a_25 + ... + a_n5^{n-1}) = 136$
$\Rightarrow a_1 + a_25 + ... + a_n5^{n - 1} = \dfrac{136 - 1}{5} = 27$
Using similar logic as before:
$a_1 + 5(a_2 + a_35 + ... + a_n5^{n-2}) = 27$
Therefore $27 - a_1$ must be divisible by $5$.
The possible values of $a_1$ are $2$, $7$, $12...$
Since $a_1 \le 4$, the only possible value of $a_1$ is $2$.
Therefore,
$a_2 + a_35 + ... + a_n5^{n-2} = \dfrac{27 - 2}{5} = 5$
$\Rightarrow a_2 + 5(a_3 + ... + a_n5^{n-3}) = 5$
Therfore, the possible values of $a_2$ are $0$, $5...$ and the only value $\le 4$ is $0$
Therefore $a_2 = 0$, and
$a_3 + ... + a_n5^{n-3} = \dfrac{5 - 0}{5} = 1$
The only possible value of $a_3$ is $1$ and all remaining coefficients $a_i$ are $0$
Therefore we get:
$a_0 = 1$, $a_1 = 2$ and $a_3 = 1$
and the given polynomial is:
$1 + 2x + x^3$
Therefore, $P(3) = 1 + 2\times 3 + 3^3 = 34$