Since developing a love for mathematics, I have always tried to solve any problem on my own. In this pursuit, I have solved several problems, including those from my textbooks and from the preparation stages for the Mathematics Olympiad. I have tried to include all the problems that I have solved on my own approach up to this day.
There are two diagonals of the given cube which are \(AE\) and \(BD\). If \(\angle ACB = \cos^{-1}(x)\), then what is the value of \(x\)?
Let's consider the diagonal as \( d \) and the side of the cube as \( m \). As we know, the \( d \) of a cube is \( \sqrt{3}a = \sqrt{3}m \). Here we have to remember that the two diagonals where they have been inscribed are the middle point of the cube because both sides and each diagonal are the same since it’s a cube. Therefore, \( \frac{d}{2} = \frac{\sqrt{3}m}{2} \). We also have \( \angle ACB = \cos^{-1}(x) \). To solve this problem, we must apply the cosine rule for \( \angle C \), which is \( c^2 = a^2 + b^2 - 2bc \cos C \).
Now from the cosine rule,
\( c^2 = a^2 + b^2 - 2bc \cos C \)
\(\implies \cos C = \frac{a^2 + b^2 - c^2}{2bc} \)
\(\implies \cos C = \frac{(\frac{\sqrt{3}m}{2})^2 + (\frac{\sqrt{3}m}{2})^2 - m^2}{2(\frac{\sqrt{3}m}{2})(\frac{\sqrt{3}m}{2})}\)
\(\implies \cos C = \frac{\frac{3m^2}{4} + \frac{3m^2}{4} - m^2}{\frac{6m^2}{4}}\)
\(\implies \cos C = \frac{3m^2 + 3m^2 - 4m^2}{4} \times \frac{4}{6m^2}\)
\(\implies \cos C = \frac{2m^2}{4} \times \frac{4}{6m^2}\)
\(\implies \cos \cos^{-1}(x) = \frac{1}{3}\)
\(\implies x = \frac{1}{3}\)
Therefore the value of \(x\) is \(\frac{1}{3}\)
A \(20\) meters long ladder is placed vertically against a wall. How far must the base of the ladder be moved away from the wall so that the top of the ladder slides down \(4\) meters?
Let's consider the base as \(CD = x\). Before moving away, \(AD\) was \(20m\). Since moving away from the wall so that top of the ladder slides down \(4m\), \(BD\) has become (\(20-4) = 16m\). The hypotenuse \(BC = 20m\). Therefore, we can apply \( \sin \theta \) to evaluate the value of \( \angle C \) so that we can use it to determine the length of \(x\) which is \(CD\)
Now,
\( \sin \theta = \frac{BD}{BC}\)
\(\implies \sin \theta = \frac{16}{20}\)
\(\implies \theta = \sin^{-1} \frac{4}{5}\)
\(\implies \theta = 53.13^\circ\)
So we have got the value of \( \angle C \) and now we will use \( \tan \theta \) to determine the length of \(CD\).
Again,
\( \tan \theta = \frac{BD}{CD}\)
\(\implies \tan \theta = \frac{16}{x}\)
\(\implies \tan 53.13^\circ\ = \frac{16}{x}\)
\(\implies 1.33 = \frac{16}{x}\)
\(\implies x = \frac{16}{1.33}\)
\(\implies x = 12.03\)
Therefore, the it should be moved away \(12m\).
This problem can also be solved by pythagorean theorem which is mostly preferable for this kind of problems.
Prove that \( \sqrt{2} \) is an irrational number.
To solve this problem, we will use proof by contradiction.
So, let's consider \( \sqrt{2} \) is a rational number.
As we know we can express a rational number by \(\frac{p}{q}\) where \(p\) and \(q\) are coprime to each other.
Therefore,
\( \sqrt{2} = \frac{4}{5}\)
\(\implies 2 = \frac{p^2}{q^2}\)
Here,
\(p ≡ m \pmod{q}\)
\(\implies p^2 ≡ n \pmod{q^2}\)
Since \( p^2 ≡ n \pmod{q^2} ≠ 2\) Therefore, \( \sqrt{2} \) is an irrational number. ■
Prove that the square of any odd perfect number is an odd number.
To solve this problem, we will use proof by contradiction as problem 3.
Let's consider the odd number as \(p\).
Then,
\(p ≡ 1 \pmod{2}\)
\(\implies p^2 ≡ 1^2 \pmod{2}\)
But it is impossible to have any remainder of an even number if we divide by \(2\)
Therefore, it can be said that the square of any odd perfect number is an odd number. ■
Prove that the product of two consecutive even numbers is divided by \(8\)
Let's consider two consecutive even numbers as \(2n\) and \(2n + 2\)
Their product,
\(p = 2n × (2n + 2) \)
\(\implies p = 2n × 2(n + 1)\)
\(\implies p = 4n(n + 1)\)
Here, \(8\) is divided by \(4\) which means \(4\) is a factor of \(8\). Among two consecutive natural numbers, there is always a number divisible by \(2\). \(p\) will be divisible by \(2\) if \(n(n + 1)\) is a multiple of \(2\). As a result, \(4n(n + 1)\) must be divisible by \(8\).
Therefore, \(8 | 4n(n + 1)\) ■
If \(n = 2x - 1\), where \(x \in \mathbb{N}\). Then show that, if \(n^2\) is divided by \(8\), the remainder will always be \(1\).
According to the question, we have to show \(n^2 ≡ 1 \pmod{8}\)
Now,
\(8 | (n^2 - 1)\)
\(\implies 8 | (2x - 1)^2 - 1\)
\(\implies 8 | (2x - 1)(2x - 1) - 1\)
\(\implies 8 | (4x^2 - 4x + 1) - 1\)
\(\implies 8 | 4x(x - 1)\)
Therefore, when \(4x(x - 1) + 1\) is divided by \(8\), the remainder will be \(1\)
So, \(4x(x - 1) ≡ 1 \pmod{8}\) ■
If \(2N + 127\) is divisible by \(13\), then what is the smallest integer value of \(N\)?
According to the divisibility rule, we know that if \(x | y\) and \(x | (y \pm z)\), then \(x | z\)
Now,
\(13 | (2N + 127)\)
\(\implies 13 | (2N + 127 + 3 - 3)\)
\(\implies 13 | (127 + 3) + (2N - 3)\)
Since, \((127 + 3)\) is divisible by \(13\), then (\(2N - 3)\) must be divisible by \(13\)
So,
\(13 | (2N - 3)\)
\(\implies 2N - 3 = 13\)
\(\implies 2N = 16\)
\(\implies N = 8\)
We are called \(127 + 2N\) is divisible by \(13\). Since only integer \(127\) is not divisible by \(13\), we can simply modify to make divisible. We can add \(3\) and then subtract. What is left now? It is \(2N - 3\). Since \(13\) divides \(127 + 3\), then \(2N - 3\) should be divisible also. The place where \(2N - 3\) is written reffers which smallest integer should be divisible by \(13\). The smallest integer is \(0\), but this will make negative value. Then what is next smallest integer value which is divided by \(13\)? Yes, \(13\) itself! That is why we have written \(2N - 3 = 13\) and we get \(N = 8\).
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Two numbers are to be inserted into a magic box. If the second number is multiplied by any other number to get the first number then the box lights up green. If not, the red light will flash. For example, if you put in \(16\) and \(2\), then the green light will turn on, because \(2 × 2 × 2 × 2 = 16\), but if you put in \(18\) and \(4\), then the red light will turn on. If the first and second numbers are the same, the green light will still light up. Suppose you enter \(256\) as the first number. How many different numbers are there which when entered as the second number will light the green light?
If we look at the problem statement sharply, we can see that the green light only flashes when a divisor of first number is multiplied with that.
So, we have to determine all the divisors of \(256\) and the total number of the divisors will be the answer.
Then,
\(256\)
\(= 1 × 256\)
\(= 2 × 128\)
\(= 4 × 64\)
\(= 8 × 32\)
\(= 16 × 16\)
As we can see, total number of divisors is \(9\). So, the answer is \(9\).
If \(xy + x + y = 19\), then determine all the positive integers of \(x\), \(y\).
Let's solve the equation,
\(xy + x + y = 19\)
Since, \(19\) is a prime number, so there is no other divisors except \(19\) itself and \(1\) which will divide \(19\). So either the value of \(y\) will be \(19\) or \(1\)
For \(y = 19\),
\(2x + 19 = \frac{19}{19}\)
\(\implies 2x = 1 - 19\)
\(\implies x = \frac {-18}{2}\)
\(\implies x = -9\)
For \(y = 1\),
\(2x + 1 = \frac{19}{1}\)
\(\implies 2x = 19 - 1\)
\(\implies x = \frac {18}{2}\)
\(\implies x = 9\)
Since, we are called to determine only positive integers, therefore, \(x = 9\) and \(y = 1\).
\(2520\) is the smallest number that can be divided by each of the numbers from \(1\) to \(10\) without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from \(1\) to \(20\)?
The numbers from \(1\) to \(10\) are the factors of \(2520\). So, \(2520\) is the common multiple of the numbers from \(1\) to \(10\). More precisely Least Common Multiple (LCM). So we need to figure out the Least Common Multiple (LCM).
By doing prime factorizing,
\(1 = 1\)
\(2 = 2\)
\(3 = 3\)
\(4 = 2^2\)
\(5 = 5\)
\(6 = 2 \times 3\)
\(7 = 7\)
\(8 = 2^4 \)
\(9 = 3^2\)
\(10 = 2 \times 5\)
\(11 = 11\)
\(12 = 2^2 \times 3\)
\(13 = 13\)
\(14 = 2 \times 7\)
\(15 = 3 × 5\)
\(16 = 2^4\)
\(17 = 17\)
\(18 = 2 \times 3^2\)
\(19 = 19\)
\(20 = 2^2 \times 5\)
By taking and multiplying the highest powers of all the prime factors to get Least Common Multiple (LCM),
\(2^4 \times 3^2 \times 5 \times 7 \times 11 \times 17 \times 19\)
\(= 232792560\)
So, the smallest integer is \(232792560\).
If the \(m\) th term of an arithmetic series is \(n\) and \(n\) th term is \(m\), then what is the
(\(m + n\)) th term?
Let the
\(m\) th term be \(a + (n - 1)d = m...(i)\)
\(n\) th term be \(a + (m - 1)d = n...(ii)\)
Subtracting from equation (\(ii\)) to (\(i\)),
\(n - m = (m - 1)d - (n - 1)d\)
\(\implies n - m = d(m - 1 - n + 1)\)
\(\implies n - m = d(m - n)\)
\(\implies -(m - n) = d(m - n)\)
\(\implies d = \frac {-(m - n)}{(m - n)}\)
\(\implies d = -1\)
Replacing the value of \(d\) to equation \(i\),
\(a + (m - 1)d = n\)
\(\implies a - (m - 1)d = n\)
\(\implies a - m + 1 = n\)
\(\implies m + n = a + 1...(iii)\)
So, \((m + n)\) th term is,
\(a + (m + n - 1)d\)
\(\implies a - (m + n - 1)d = a - (a + 1 - 1)\)
\(\implies a - a = 0\)
If \((ax - cy, a^2 - c^2) = (0, ay - cx)\), then determine \((x, y)\).
According to the condition of ordered pair,
\((ax - cy) = 0\)
\(\implies ax - cy = 0\)
\(\implies cy = ax\)
\(\implies y = \frac {ax}{c}...(i)\)
And,
\((a^2 - c^2) = (ay - cx)\)
\(\implies a^2 - c^2 = ay - cx\)
\(\implies y = \frac {a^2 - c^2 + cx}{a}...(ii)\)
Replacing the value of \((i)\) to \((ii)\),
\(\frac {ax}{c} = \frac {a^2 - c^2 + cx}{a}\)
\(\implies a^2x = c(a^2 - c^2 + cx)\)
\(\implies a^2x = a^2c - c^3 + c^2x\)
\(\implies a^2x + c^2x = a^2c - c^3\)
\(\implies x(a^2 - c^2) = c(a^2 - c^2)\)
\(\implies x = c\)
Replacing the value of \(x\) to \((ii)\),
\(y = \frac {a^2 - c^2 + c^2}{a}\)
\(\implies y = \frac {a^2}{a}\)
\(\implies y = a\)
Therefore, \((x, y) = (c, a)\)
Prove that the sum of the degrees of all nodes in any graph is twice the total number of edges.
Let the nodes be \( n\).
So, edges will be \(n - 1\).
Therefore,
total degree \(= n(n - 1)...(i)\)
But, if the degrees of all nodes are summed, each edge will be counted twice—once from each of its endpoints.
For that case,
total degree \(= 2n(n - 1)...(ii)\)
Dividing equation \((ii)\) by \((i)\),
\(\frac {2n(n - 1)}{n(n - 1)} = 2\)
Therefore, the sum of the degrees of all nodes will be twice the total number of edges in the graph. ■
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Prove that, \(A(a, 0)\), \(B(0, b)\) and \(C(1, 1)\) will be collinear, if \(\frac {1}{a} + \frac {1}{b} = 1\)
As we are given \(A(a, 0)\), \(B(0, b)\) and \(C(1, 1)\). We need to prove that
\(\frac {1}{a} + \frac {1}{b} = 1\)
We know that if the lines \(AB\) and \(BC\) are collinear, then their slopes will be equal.
Mathematically, if points \(A(x_1, y_2)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) are collinear, then: \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_2}{x_3 - x_2}\)
Then,
slope of line \(AB\) is \(m_{AB}\)
\(= \frac{y_2 - y_1}{x_2 - x_1}\)
\(= \frac{b - 0}{0 - a}\)
\(= \frac {b}{-a}...(i)\)
slope of line \(BC\) is \(m_{BC}\)
\(= \frac{y_3 - y_2}{x_3 - x_2}\)
\(= \frac{1 - b}{1 - 0}\)
\(= 1 - b...(ii)\)
According to the law of colinear points and replacing \(i\) and \(ii\),
\(m_{AB} = m_{BC}\)
\(\implies\frac {b}{-a} = 1 - b\)
\(\implies\frac {b}{-a} = -(b - 1)\)
\(\implies\frac {b}{a} = b - 1\)
\(\implies a(b - 1) = b\)
\(\implies a = \frac {b}{b - 1}...(iii)\)
\(\implies \frac {a}{b} = \frac {1}{b - 1}\)
\(\implies \frac {1}{a} = \frac {b - 1}{b}\)
\(\implies \frac {1}{a} = 1 - \frac {1}{b}\)
\(\implies \frac {1}{a} + \frac {1}{b} = 1\) ■
Alternate Approach-
We could prove this problem through an easy way and that's how alternate solution comes from.
We could take \(iii\) and replace it instead of \(a\) inside our given statement:
\(\frac {1}{a} + \frac {1}{b}\)
\(= \frac{\frac{1}{b}}{b - 1} + \frac {1}{b}\)
\(= \frac {b - 1}{b} + \frac {1}{b}\)
\(= \frac {b - 1 + 1}{b}\)
\(= 1\) ■
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if \(ab - c = 3\), \(abc = 18\) and \(a\), \(b\), \(c \) \(\in \mathbb{N}\) then \(\frac {ab}{c} =\) ?
Given,
\(ab - c = 3...(i)\) and \(abc = 18...(ii)\)
From \((i)\), we get \(ab = 3 + c\)
Now, replacing the value of \(ab\) in \((ii)\) we get-
\((3 + c)c = 18\)
\(\implies 3c + c^2 = 18\)
\(\implies c^2 + 3c - 18 = 0\)
\(\implies c^2 + 6c - 3c - 18 = 0\)
\(\implies c(c + 6) - 3(c + 6) = 0\)
\(\implies (c - 3)(c + 6) = 0\)
Since, we are called \(a\), \(b\), \(c \) \(\in \mathbb{N}\) and \((c + 6) = 0 \not\in \mathbb{N}\) so the value of \(c\) is \(3\).
Now, replacing the value of \(c\) in \((i)\) we get \(ab = 6\)
Therefore, \(\frac {ab}{c} = \frac {6}{3} = 2\)