NSEJS-2016 (Maths Section)

Problem 1 of 20

The number of natural numbers $n \le 30$ for which $\sqrt{n + \sqrt{n + \sqrt{n + ...}}}$ is a natural number is

Solution:

Let $m = \sqrt{n + \sqrt{n + \sqrt{n + ...}}}$

Therefore,

$m^2 = n + \sqrt{n + \sqrt{n + \sqrt{n + ...}}}$

$\Rightarrow m^2 = n + m$

$\Rightarrow m^2 - m - n = 0$

Using the generic form of quadratic equation from , we get:

$m = \dfrac{1 \pm \sqrt{(1 + 4n)}}{2}$

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For, $m$ to be a natural number, $1 + 4n$ has to be a perfect square. Therefore, $1 + 4n$ can only be square of odd numbers, and the value of $m$ will be an integer.

Here we have the choice of starting with $n = 1$ and find all values of $n$ which gives us a perfect square, however, we can try to reduce the cases by doing some more observations.

Let $1 + 4n = r^2$

$n = \dfrac{r^2 - 1}{4}$

Since $n \le 30$

$\therefore \dfrac{r^2 - 1}{4} \le 30$

$\Rightarrow r^2 \le 121$

$\Rightarrow r \le 11$

$r^2 - 1 = (r - 1)(r+1)$.

If $r$ is odd, and $r \gt 1$, $r-1$ and $r+1$ are even, therefore, $r^2 - 1$ will be divisible by $4$.

Therefore,

$r$ can be $11,\ 9,\ 7,\ 5,\ 3$

and the corresponding values of $n$ are:

$30,\ 20,\ 12,\ 6,\ 2$

Therefore, there are $5$ possible values of $n \le 30$ where the given expression is a natural number.

Problem 2 of 20

The number of natural numbers $n \le 30$ for which $\sqrt{n + \sqrt{n + \sqrt{n + ...}}}$ is a prime number is

Solution:

Let $m = \sqrt{n + \sqrt{n + \sqrt{n + ...}}}$

Therefore,

$m^2 = n + \sqrt{n + \sqrt{n + \sqrt{n + ...}}}$

$\Rightarrow m^2 = n + m$

$\Rightarrow m^2 - m - n = 0$

Using the generic form of quadratic equations from , we get:

$m = \dfrac{1 \pm \sqrt{(1 + 4n)}}{2}$

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For, $m$ to be a natural number, $1 + 4n$ has to be a perfect square. Therefore, $1 + 4n$ can only be square of odd numbers, and the value of $m$ will be an integer.

Here we have the choice of starting with $n = 1$ and find all values of $n$ which gives us a perfect square, however, we can try to reduce the cases by doing some more observations.

Let $1 + 4n = r^2$

$n = \dfrac{r^2 - 1}{4}$

Since $n \le 30$

$\therefore \dfrac{r^2 - 1}{4} \le 30$

$\Rightarrow r^2 \le 121$

$\Rightarrow r \le 11$

$r^2 - 1 = (r - 1)(r+1)$.

If $r$ is odd, and $r \gt 1$, $r-1$ and $r+1$ are even, therefore, $r^2 - 1$ will be divisible by $4$.

Therefore,

$r$ can be $11,\ 9,\ 7,\ 5,\ 3$

and the corresponding values of $n$ are:

$30,\ 20,\ 12,\ 6,\ 2$

For each of these values of $n$, the corresponding value of $m$ is:

For $n = 30$, $m = \dfrac{1 + \sqrt{1 + 120}}{2} = 6$

For $n = 20$, $m = \dfrac{1 + \sqrt{1 + 80}}{2} = 5$

For $n = 12$, $m = \dfrac{1 + \sqrt{1 + 48}}{2} = 4$

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For $n = 6$, $m = \dfrac{1 + \sqrt{1 + 24}}{2} = 3$

For $n = 2$, $m = \dfrac{1 + \sqrt{1 + 8}}{2} = 2$

Out of the above values of $m$, three are prime.

Problem 3 of 20

The number of squares formed by $5$ vertical and $4$ horizontal lines (all are equispaced) is

Solution:

The number of squares can be found by counting row-wise or column-wise. But this method may be error prone, specially for larger values of horizontal and vertical lines.

So we will learn a better approach to solve this problem.

Let us first assume that there are $m$ vertical lines and $n$ horizontal lines.

Let us draw the $45^\circ$ diagonals for passing through all points except for the two extreme points on both sides, as shown in the following diagram:

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Since each diagonal is $45^\circ$, choosing any two points on any one of these diagonals will give us the diagonal of a square.

The first diagonal passes through $2$ points, the second one passes through $3$ points, the third one passes through $4$ points.

It is easy to see that the number of points the longest diagonal will pass through the smaller of $m$ and $n$.

And when $m \ne n$ there will be two such diagonals, and if $m = n$ there will be only one such diagonal.

In this case the diagonals will pass through, $2,\ 3,\ 4,\ 4,\ 3,\ 2$ points respectively.

Therefore, using the concept of choosing from , the number of possible squares is:

$\xacomb{2}{2} + \xacomb{3}{2} + \xacomb{4}{2} + \xacomb{4}{2} + \xacomb{3}{2} + \xacomb{2}{2}$

$= 1 + 3 + 6 + 6 + 3 + 1 = 20$

Problem 4 of 20

The number of integers $a,\ b,\ c$ for which $a^2 + b^2 - 8c = 3$ is

Solution:

The given expression can be written as:

$a^2 + b^2 = 3 + 8c$

$3 + 8c$ is an odd number.

Therefore, one of $a$ and $b$ must be odd and the other one must be even.

Without loss of generality, let us take $a$ as even and $b$ as odd.

Therefore we can write $a = 2p$ and $b = 2q + 1$ where $p$ and $q$ are integers.

Substituting these values, we get:

$(2p)^2 + (2q + 1)^2 = 3 + 8c$

$\Rightarrow 4p^2 + 4q^2 + 4q + 1 = 3 + 8c$

$\Rightarrow 4p^2 + 4q^2 + 4q - 2 = 8c$

$\Rightarrow 2p^2 + 2q^2 + 2q - 1 = 4c$

$\Rightarrow 2(p^2 + q^2 + q) - 1 = 4c$

The expression on the left hand side is odd, and the expression on the right hand side is even.

Therefore, no integer value set for $p$ and $q$ will satisfy this equation.

Hence there are $zero$ solutions to this problem for integer $a,\ b,\ c$

Problem 5 of 20

If set $X$ consists of three elements then the number of elements in the power set of power set of $X$ is

Solution:

From the concept of power set explained , the cardinality of the power set of a set with $n$ elements is $2^n$

Therefore, the power set of a $3$ element set is $2^3 = 8$.

The power set of this power set is will have $2^8$ elements.

Problem 6 of 20

In an $n$-sided regular polygon, the radius of the circum-circle is equal in length to the shortest diagonal. The number of values of $n \lt 60$ for which this can happen is

Solution:

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It is clear that the shortest diagonal of a regular polygon can be drawn by joining any two alternate vertices.

So, let us draw the figure as shown below, where $A,\ B,\ C$ are three consecutive vertices of a polygon, and let $O$ be the circumcentre. Therefore, $OC$ is one of the shortest diagonals.

Let us join the two radii $OA$ and $OC$

Since, we know that $AC = r$

$\triangle AOC$ is an equilateral triangle, and $\angle AOC = 60^\circ$

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Therefore, reflex $\angle AOC = 360 - 60 = 300^\circ$

Therefore, the angle subtended by the same arc at the circumference $\angle ABC = \dfrac{1}{2} \times 300^\circ = 150^\circ$

Therefore, each angle of the polygon is $150^\circ$

Let the number of sides of the polygon be $n$.

$(n - 2)180 = 150 \times n$

$\Rightarrow 180n - 150n = 360$

$\Rightarrow 30n = 360$

$\Rightarrow n = 12$

Therefore, a $12$-sided regular polygon with each of its angle $= 150^\circ$, is the only polygon that will satisfy the given condition.

Problem 7 of 20

Number of integers $n$ such that the number $1 + n$ is a divisor of the number $1 + n^2$ is

Solution:

Based on the given condition we can write,

$1 + n^2 \equiv 0\ (mod\ 1 + n)$

Substituting $n +1 = m$ we get:

$1 + (m - 1)^2 \equiv 0\ (mod\ m)$

$\Rightarrow 1 + m^2 - 2m + 1 \equiv 0\ (mod\ m)$

$\Rightarrow 2 \equiv 0\ (mod\ m)$

This is only possible for $4$ non-zero values of $m$, which are $-2,\ -1,\ 1,\ 2$

The corresponding values of $n$ are $-3,\ -2,\ 0,\ 1$

Therefore, there are $4$ possible integer values for which the given condition is satisfied.

Problem 8 of 20

If for $x,y \gt 0$ we have $\dfrac{1}{x} + \dfrac{1}{y} = 2$ then the minimum value of $xy$ is

Solution:

We know that for any two non-negative numbers $x$, $y$ their $G.M. \ge H.M.$

The geometric mean of $x,\ y$ is $\sqrt{xy}$ and their harmonic mean is $\dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}}$

Therefore,

$\sqrt{xy} \ge \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}}$

Substituting $\dfrac{1}{x} + \dfrac{1}{y} = 2$ we get:

$\sqrt{xy} \ge \dfrac{2}{2}$

$\Rightarrow \sqrt{xy} \ge 1$

$\Rightarrow xy \ge 1$

Therefore, the minimum value of $xy$ is $1$.

Problem 9 of 20

Tenth term in the sequence $12,\ 18,\ 20,\ 28\ ...$ is

Solution:

Let us express each term of the given series in terms of their prime factors and count the number of divisors for each of them using the method described .

$12 = 2^2 \times 3$ Number of divisors $= 3 \times 2 = 6$

$18 = 3^2 \times 2$ Number of divisors $= 3 \times 2 = 6$

$20 = 2^2 \times 5$ Number of divisors $= 3 \times 2 = 6$

$28 = 2^2 \times 7$ Number of divisors $= 3 \times 2 = 6$

So we can see that the series consists of integers that have $6$ divisors, in increasing order.

We know that for a number to have $6$ divisors it should be of the form $p^5$ or $p_1^2 \times p_2^1$ where $p$, $p_1$ and $p_2$ are prime numbers.

So continuing this given series we get:

$32 = 2^5$ Number of divisors $= 6$

$44 = 2^2 \times 11$ Number of divisors $= 3 \times 2 = 6$

$45 = 3^2 \times 5$ Number of divisors $= 3 \times 2 = 6$

$52 = 2^2 \times 13$ Number of divisors $= 3 \times 2 = 6$

$63 = 3^2 \times 7$ Number of divisors $= 3 \times 2 = 6$

$68 = 2^2 \times 17$ Number of divisors $= 3 \times 2 = 6$

$76 = 2^2 \times 19$ Number of divisors $= 3 \times 2 = 6$

Therefore, the $10\xasuper{th}$ number in this series is $68$

Problem 10 of 20

The probability of a point within an equilateral triangle with side $1$-unit lying outside its in-circle (inscribed circle) is

Solution:

Let us take the side of the equilateral triangle as $a$.

Therefore, the altitude is $\dfrac{\sqrt{3}}{2}a$ and the area is:

$\dfrac{\sqrt{3}a^2}{4}$

Using the relationship between the inradius, area and semi-perimeter from ,

$r \times \dfrac{3a}{2} = \dfrac{\sqrt{3}a^2}{4}$

$\Rightarrow r = \dfrac{a}{2\sqrt{3}}$

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Area of the incircle,

$\dfrac{\pi a^2}{12}$

Area outside the incircle:

$\dfrac{\sqrt{3}a^2}{4} - \dfrac{\pi a^2}{12}$

Probability of a randomly chosen point to be outside the incircle is:

$\dfrac{\dfrac{\sqrt{3}a^2}{4} - \dfrac{\pi a^2}{12}}{\dfrac{\sqrt{3}a^2}{4}}$

$= 1 - \dfrac{\pi a^2}{12} \times \dfrac{4}{\sqrt{3}a^2}$

$= 1 - \dfrac{\pi}{3\sqrt{3}}$

Problem 11 of 20

A triangle has perimeter $316$ and its sides are of integer length. The maximum possible area for such a triangle is achieved for

Solution:

We know that for a given perimeter $p$, a triangle with maximum area can be formed by making the triangle an equilateral triangle, with each side equal to $\dfrac{p}{3}$

In this case that is not possible since the sides are of integer lengths and perimeter $316$ is not divisible by $3$.

In this case the sides of the equilateral triangle will be $105\dfrac{1}{3}$

Therefore, to maximise the area, we can form an isosceles triangle with sides as close as possible to the dimensions of the equilateral triangle.

We can do so by making two sides as $105$ and the third side as $106$

Therefore, there is one triangle which will give the maximum area.

Problem 12 of 20

Number of numbers less than $40$ having exactly four divisors is

Solution:

We know from , that a number $N = p_1^{e_1} \times p_2^{e_2} \times ... \times p_n^{e_n}$ where $p_1,\ p_2, ..., p_n$ are prime numbers, has exactly $(e_1 + 1)(e_2 + 1)..(e_n + 1)$ divisors.

Since we need to find the numbers that have exactly $4$ divisors, that is possible under two conditions:

- The number is the of the for $p^3$ where $p$ is a prime number

- The number is of the form ${p_1}^1 \times {p_2}^1$ where $p_1$ and $p_2$ are prime numbers.

For the first case we have:

$2^3 \lt 40$

$3^3 \lt 40$

For the second case, we have:

$19 \times 2 \lt 40$

$17 \times 2 \lt 40$

$13 \times 2 \lt 40$

$13 \times 3 \lt 40$

$11 \times 2 \lt 40$

$11 \times 3 \lt 40$

$7 \times 2 \lt 40$

$7 \times 3 \lt 40$

$7 \times 5 \lt 40$

$5 \times 2 \lt 40$

$5 \times 3 \lt 40$

$3 \times 2 \lt 40$

Therefore, the count of numbers less than $40$ with exactly four divisors is $14$.

Problem 13 of 20

The number $3^8(3^{10} + 6^5) + 2^3(2^{12} + 6^7)$ is

Solution:

We can write the given expression as:

$3^8(3^{10} + 6^5) + 2^3(2^{12} + 6^7)$

$= 3^{18} + 2^5.3^{13} + 2^{15} + 2^{10}3^7$

$= (3^6)^3 + 3.(3^6)^2.2^5 + 3.3^6.(2^5)^2 + (2^5)^3$

$= (3^6 + 2^5)^3$

Now,

$3^6 + 2^5$

$= 729 + 32$

$= 761$

$761$ is not a perfect square.

Therefore, the given expression is a perfect cube, but not a perfect square.

Problem 14 of 20

Let the number of rectangles formed by $6$ horizontal and $4$ vertical lines be $n$ and those formed by $5$ vertical and $5$ horizontal lines be $m$ then we have

Solution:

From a given set of horizontal and vertical lines, a rectangle can be formed by choosing any two horizontal and and any two vertical lines.

Using the principles of choosing, as explained ,

for $6$, $4$ lines set it can be done in $n = \xacomb{6}{2} \times \xacomb{4}{2} = 15 \times 6 = 90$ ways.

for $5$, $5$ lines set it can be done in $m = \xacomb{5}{2} \times \xacomb{5}{2} = 10 \times 10 = 100$ ways

Therefore, $m \gt n + 5$ is the correct statement.

Problem 15 of 20

If $ABCD$ is a rhombus and $\angle ABC = 60^\circ$ then

Solution:

If one angle of a rhombus is $60^\circ$ then the two adjacent vertices are $120^\circ$ each and the opposite vertex is $60^\circ$

We can divide the rhombus into two identical equilateral triangles by joining the $120^\circ$

Each equilateral triangle will have an area of

$\dfrac{\sqrt{3}}{4}AB^2$

Therefore, the area of the rhombus is:

$2 \times \dfrac{\sqrt{3}}{4}AB^2 = \dfrac{\sqrt{3}}{2}AB^2$

Problem 16 of 20

If $a,b \gt 0$ then

Solution:

Since $a$ and $b$ are positive, both $\sqrt{a}$ and $\sqrt{b}$ are real numbers.

We can say that:

$(\sqrt{a} - \sqrt{b})^2$ is a non-negative quanitity, therefore,

$(\sqrt{a} - \sqrt{b})^2 \ge 0$

$\Rightarrow a + b - 2\sqrt{ab} \ge 0$

$\Rightarrow a + b \ge 2\sqrt{ab}$

Problem 17 of 20

The statement $\unicode{0x201C}a\ is\ not\ less\ than\ 4\unicode{0x201D}$ is correctly represented by

Solution:

The given statement can be rewritten as:

$a$ is either equal to or greater than $4$.

Therefore, the correct relationship is:

$a \ge 4$

Problem 18 of 20

Two friends $A$ and $B$ watched a car from the top of their buildings. Angle of depression for A was $10^\circ$ more than angle of depression for $B$, then

Solution:

Since angle of depression of the car for each person, is dependent on the height of the building and the distance of the car from the building, without knowing the distance of the car from each of the buildings, it is not possible to compare the heights of the building.

Problem 19 of 20

Total surface area of a sphere $S$ with radius $\sqrt{2} + \sqrt{3}\ cm$ is

Solution:

The surface area of the given sphere using the concept from , is:

$4\pi(\sqrt{2} + \sqrt{3})^2\ cm^2$

$= 4 \pi (2 + 3 + 2\sqrt{6})\ cm^2$

$= 4 \pi (5 + 2\sqrt{6})\ cm^2$

$= 400 \pi (5 + 2\sqrt{6})\ mm^2$

Problem 20 of 20

A craft teacher reshapes the wax from a cylinder of candle with section diameter $6$ cm and the height $6$ cm into a sphere. The radius of this sphere will be

Solution:

Let the radius of the sphere be $r$.

Since the volume of the sphere will be equal to that of the cylinder, using the formulae for sphere and cylinder from , we have:

$\dfrac{4}{3} \pi r^3 = 6 \times \pi 3^2$

$\Rightarrow r^3 = \dfrac{3}{2} \times 3^3$

$\therefore r = 3\sqrt[3]{\dfrac{3}{2}}$