## Solvability of equations 

### Charity

A1, 2015

Ten people sitting around a circular table decide to donate some money for charity. You are told that the amount donated by each person was the average of the money donated by the two persons sitting adjacent to him/her. One person donated Rs. $$500$$.

Choose the correct option for the following two questions.

Q1. What is the total amount donated by the 10 people?

(a) exactly Rs. 5000 (b) less than Rs. 5000 (c) more than Rs. 5000 (d) cannot decide.

Q2. What is the maximum amount donated by an individual?

(a) exactly Rs. 500 (b) less than Rs. 500 (c) more than Rs. 500 (d) cannot decide.

Solution

A1. Exactly Rs. $$5000$$.

A2. Exactly Rs. $$500$$.

Consider the person who donated Rs. 500. Suppose the neighbor to the left donates $$500+x$$. Then the one on the right donates $$500-x$$. But continuing leftward, the amounts donated are $$500+2 x, 500+3 x, \ldots,$$ forcing $$x$$ to be $$0,$$ since you come around to the neighbor to the right.

### Integers in a function range

A2, 2018

Consider the following function defined for all real numbers $$x$$ $f(x)=\frac{2018}{10+e^{x}}$ How many integers are there in the range of $$f$$?

Solution

There are 201 integers.

The function is strictly decreasing as $$x$$ goes from $$-\infty$$ to $$\infty$$. The range of the function is $$(0,201.8)$$. By continuity, all the integers in the range appear as values of $$f(x)$$.

### GDP vs Per-capita GDP

A2, 2016

A country's GDP grew by $$7.8 \%$$ within a period. During the same period the country's per-capita-GDP defined as the ratio of GDP to the total population, increased by $$10\%$$. By what percentage did the total population change during this period?

Solution

Notation.Let $$G$$ and $$P$$ denote the GDP and the population at the start of the period, respectively. Let $$x$$ be the percentage change of the population.

The new per-capita GDP is then given by $$\frac{1.078 G}{(1+x) P}$$. The percent increase in per-capita GDP is $$10 \%=0.1$$. So we have $$\frac{1.078}{1+x}=1.1 .$$ Solving for $$x$$ we get:

Per-capita GDP is $$\frac{\text { GDP }}{\text { population }}$$.

$1+x=\frac{1.078}{1.1}=0.98$ So population decreased by $$2 \%$$

### Monotonic functions

A1, 2021

Consider the following equations: \begin{align*} \log_{2021} a &= 2022 -a \\ 2021^b &= 2022-b \end{align*} (a) There is a unique $$a$$ that satisfies the first equation.
(b) There is a unique $$b$$ that satisfies the second equation.
(c) There exists a solution for both the equations such that $$a=b$$.
(d) There exists a solution for both the equations such that $$a+b$$ is an integer.

Solution
(a) True. $$f(a)=\log_{2021} a + a - 2022$$ is an increasing function in $$a$$. So $$a=2021$$ is the only solution.
(b) True. $$g(b)=2021^b + b - 2022$$ is an increasing function in $$b$$. So $$b=1$$ is the only solution.
(c) False. Follows from above.
(d) True. Since, $$2021+1$$ is an integer.

### Two equations

A7, 2021

$$a,b,c$$ are distinct positive constants. Consider the system of equations: \begin{align*} ax + by &= \sqrt{2}\\ bx + cy &= \sqrt{3} \end{align*} (a) There exist $$a,b,c$$ such that such that the system has infinite solutions.
(b) There exist $$a,b,c$$ such that such that the system has exactly one solution.
(c) If the system has no solution, it means that $$2b < a+c$$.
(d) If $$2b< a+c$$, then the system has no solution.

Solution (a) True. Pick $$a=2/3,b=\sqrt{2/3}$$ and $$c=1$$. Both the equations become equal.
(b) True. Pick $$a=2,b=1$$ and $$c=1$$.
(c) True. There is no solution if $$b^2-ac=0$$ and $$b/c\neq \sqrt{2/3}$$. The first condition implies that $$2b\leq a+c$$.
(d) False. Pick values given in (b).

### Solving a cubic root equation

A3, 2018

List all solutions of the following equation. Need not simplify the solutions. $\large \sqrt{x+4}-\sqrt{x}=1$

Solution

Put $$t=\sqrt{x}$$ to get: \begin{align} \left(t^{3}+4\right)&=(1+t)^{3} \\ 4&=t^{3}+3t^2+3t+1 \\ t^{2}+t-1&=0\\ \end{align} Solving for $$t$$ and then $$x$$ we get: \begin{align*} x &= \left( \frac{-1\pm\sqrt{5}}{2} \right)^{3} \\ &= -2\pm \sqrt{5} \end{align*}

### Solve: $$p(x)^{q(x)} = 1$$

B2a, 2018

Find all real solutions of the equation: $\left(x^{2}-2 x\right)^{x^{2}+x-6}=1$

Solution

The equation is of the form $$a^b=1$$. This can happen only in three cases:
(i) Either $$a=1$$, which means $$x^2-2x=1$$. So $$x=1\pm\sqrt{2}$$.
(ii) or $$a=-1$$ and $$b$$ is an even integer. So $$x=1$$ works.
(iii) or $$b=0$$ and $$a\neq 0$$. Which happens when $$x=-3$$ or $$1$$. The value $$x=2$$ makes $$a=b=0$$
So $$x=-3,1,1 \pm \sqrt{2}$$ are the only solutions.

### Solve: $$\sqrt{a}-\sqrt{b} = 1$$

B2b, 2018

The following expression is an integer. Find its value. $\sqrt{6 \sqrt{3}+10}-\sqrt{6 \sqrt{3}-10}$

Solution

Let $$a=\sqrt{6\sqrt{3}+10}$$ and $$b=\sqrt{6\sqrt{3}-10}$$. We want to find the value of $$a-b$$. We have the following: \begin{align} a^3 - b^3 &= 20 \\ ab &= \sqrt{ (6\sqrt{3})^2-10^2 } = \sqrt{108-100} = 2 \end{align} Substituting the values above in $$(a-b)^3$$ gives a cubic in $$(a-b)$$. \begin{align} (a-b)^3 &= a^3 - b^3 - 3ab(a-b) \\ (a-b)^{3}&=20-6(a-b) \end{align} This cubic has one real root $$a-b=2$$ and two complex roots. Hence, the value of the given expression is 2.

A similar problem was discussed on Quora in 2017.

### Symmetric log reciprocals

A7, 2010

If $$a, b, c$$ are real numbers $$> 1$$, then show that:

$\frac{1}{1+\log_{a^{2} b} \frac{c}{a}}+\frac{1}{1+\log_{b^{2} c} \frac{a}{b}}+\frac{1}{1+\log_{c^{2} a} \frac{b}{c}}=3$

Solution

We will make use of the identity: $$\log_{x} a+\log_{x} b=\log_{x} a b$$.

\begin{align} \frac{1}{1+\log_{a^{2} b}\left(\frac{c}{a}\right)}&=\frac{1}{\log_{a^{2} b}\left(a^{2} b\right)+\log_{a^{2} b}\left(\frac{c}{a}\right)} \\ &=\frac{1}{\log_{a^{2} b}\left(\frac{c}{a} \cdot a^{2} b\right)} \\ &=\frac{1}{\log_{a^{2} b}(a b c)}\\ &\\ &=\log_{a b c}\left(a^{2} b\right) \end{align}

Hence the given expression is:

$$\log_{a b c}\left(a^{2} b\right)+\log_{a b c}\left(b^{2} c\right)+\log_{a b c}\left(c^{2} a\right)$$ $$=\log_{a b c}(a b c)^{3}$$ $$=3$$

### Solutions to simultaneous equations

A8, 2017

State Yes or No for each of these questions.
Is it possible to find a $$2 \times 2$$ matrix $$A$$ for which the equation $$A \vec{x}=\vec{b}$$ has:

• (a) no solutions for some but not all $$\vec{b}$$; exactly one solution for all other $$\vec{b} ?$$
• (b) exactly one solution for some but not all $$\vec{b} ;$$ more than one solution for all other $$\vec{b} ?$$
• (c) no solutions for some but not all $$\vec{b}$$; more than one solution for all other $$\vec{b} ?$$
• (d) no solutions for some $$\vec{b},$$ exactly one solution for some $$\vec{b}$$ and more than one solution for some $$\vec{b} ?$$

Solution

No-No-Yes-No.
Suppose $$A$$ has nonzero determinant. Then for any $$\vec{b},$$ we see that $$A \vec{x}=\vec{b}$$ has the unique solution $$\vec{x}=A^{-1} \vec{b}$$. If determinant of $$A$$ is zero then we can make two cases:
(i) If $$A$$ is the zero matrix, then $$A \vec{x}=\vec{b}$$ has infinitely many solutions for $$\vec{b}=\overrightarrow{0}$$ and no solutions otherwise.

(ii) If $$A$$ is nonzero then it is easy to see that we are solving two equations in two variables whose left hand sides are proportional. So if the two right hand constants that make up $$\vec{b}$$ are in the same proportion, then we will have infinitely many solutions (because one of the variables can be arbitrary). If the constants are not in the same proportion, then the two equations will be inconsistent and we will have no solutions.

#### Problem credits The statements in the problem are basic facts in linear algebra (LA). It is a good idea to be familiar with LA although the syllabus does not mention it. Read the first three chapters of the book by Sheldon Axler. Another LA problem from CMI exam: B5 part(b) from 2014.

B2B, 2019

Find real numbers $$x$$ that satisfy: $\frac{8^{x}+27^{x}}{12^{x}+18^{x}}=\frac{7}{6}$

Solution

$\frac{\left(2^{x}\right)^{3}+\left(3^{x}\right)^{3}}{\left(2^{x}\right)^{2} \cdot 3^{x}+\left(3^{x}\right)^{2} \cdot 2^{x}}=\frac{7}{6}$

Putting $$a=2^x$$ and $$b=3^x$$ in the above equation, we get:

$\frac{a^3 + b^3}{a^2b+ab^2} = \frac{7}{6}$

$\frac{ (a+b)(a^2+b^2-ab) }{ab(a+b)} = \frac{7}{6}$

$\frac{ (a+b)(a^2+b^2-ab) }{ab(a+b)} = \frac{7}{6}$

$6a^2 - 13ab + 6b^2 = 0$

$6t^{2}-13t+6=0 \text{ where } t=a/b$

The above quadratic has $$t = 3/2$$ and $$t=2/3$$ as the roots. So $$x=\pm 1$$ satisfies the original equation.