# Pre RMO more algebra problems for practice

Question I:

I rode one third of a journey at 10kmph, one third more at 9, and the rest at 8 kmph; if I had ridden half the journey at 10kmph, and the other half at 8 kmph, I should have been half a minute longer on the way: what distance did I ride?

Question 2:

The express train leaves Bristol at 3pm and reaches London at 6pm; the ordinary train leaves London at 1:30pm and arrives at Bristol at 6pm. If both trains travel uniformly, find the time when they will meet.

Question 3:

Solve (a) $0.\dot{6}x + 0.75x-0.1\dot{6} = x - 0.58\dot{3}x+5$

Solve (b) $\frac{37}{x^{2}-5x+6} + \frac{4}{x-2} = \frac{7}{3-x}$

Question 4:

Simplify: $(1+x)^{2} \div \{ 1 + \frac{x}{1-x+ \frac{x}{1+x+x^{2}}}\}$

Question 5:

Find the square root of $\frac{4a^{2}-12ab-6bc+4ac+9b^{2}+c^{2}}{4a^{2}+9c^{2}-12ac}$

Question 6:

Find the square root of $4a^{4}+9(a^{2}+\frac{1}{a^{2}})+12a(a^{2}+1)+18$

Question 7:

Solve the following system of equations:

$\frac{1}{3}(x+y)+2z=21$

$3x - \frac{1}{2}(y+z) = 65$

$x + \frac{1}{2}(x+y-z) = 38$

Question 8:

A number consists of three digits, the right hand one being zero. If the left hand and middle digits be interchanged the number is diminished by 180; if the left hand digit be halved and the other two digits are interchanged, the number is diminished by 336; find the number.

Question 9:

$\frac{2}{x^{2}+xy+y^{2}}$, $\frac{-4x}{x^{3}-y^{3}}$, $\frac{x^{2}}{y^{2}(x-y)^{2}}$, and $\frac{-x^{2}}{x^{3}y-y^{4}}$

Question 10:

Simplify:

$\frac{a^{3}+b^{3}}{a^{4}-b^{4}} - \frac{a+b}{a^{2}-b^{2}} -\frac{1}{2} \{ \frac{a-b}{a^{2}+b^{2}} - \frac{1}{a-b} \}$

More later,
Nalin Pithwa

# Pre RMO Algebra problems for practice

Question 1:

Find the square root of $49x^{4}+\frac{1051x^{2}}{25} - \frac{14x^{3}}{5} - \frac{6x}{5} + 9$

Question 2:

The surface area of a circular cone is given by $A= {\pi}r^{2}+{\pi}rs$, where s cm is the slant height, r cm is the radius of the base and $\pi$ is $\frac{22}{7}$. Find the radius of the base if a cone of surface area 93.5 square cm has a slant height of 5 cm.

Question 3:

Solve:

$\frac{a+x}{a^{2}+ax+x^{2}} +\frac{a-x}{a^{2}-ax+ x^{2}} = \frac{3a}{x(a^{4}+a^{2}x^{2}+x^{4})}$

Question 4:

Subtract $\frac{x+3}{x^{2}+x-12} + \frac{x+4}{x^{2}-x-12}$ and divide the difference by $1 + \frac{2(x^{2}-12)}{x^{2}+7x+12}$

Question 5:

Solve the following for the unknown x:

$\frac{x}{2(x+3)} - \frac{53}{24} = \frac{x^{2}}{x^{2}-9} - \frac{8x-1}{4(x-3)}$

Question 6:

Find the square root of the following:

$a^{6}+ \frac{1}{a^{6}} -6(a^{4}+\frac{1}{a^{4}}) +12(a^{2}+\frac{1}{a^{2}})-20$; also, cube the result.

More later,
Nalin Pithwa.

# Practice questions based on combinatorics for RMO Training and IITJEE Mathematics

Question 1:

Prove that if n is an even integer, then

$\frac{1}{(1!)(n-1)!} + \frac{1}{3! (n-3)!} + \frac{1}{5! (n-5)!} + \ldots + \frac{1}{(n-1)! 1!} = \frac{2^{n-1}}{n!}$

Question 2:

If ${n \choose 0}$, ${n \choose 1}$, ${n \choose 2}$, ….${n \choose n}$ are the coefficients in the expansion of $(1+x)^{n}$, when n is a positive integer, prove that

(a) ${n \choose 0} - {n \choose 1}ย + {n \choose 2} - {n \choose 3} + \ldots + (-1)^{r}{n \choose r} = (-1)^{r}\frac{(n-1)!}{r! (n-r-1)!}$

(b) ${n \choose 0} - 2{n \choose 1} + 3{n \choose 2} - 4{n \choose 3} + \ldots + (-1)^{n}(n+1){n \choose n}=0$

(c) ${n \choose 0}^{2} - {n \choose 1}^{2} + {n \choose 2}^{2} - {n \choose 3}^{2} + \ldots + (-1)^{n}{n \choose n}^{2}=0$, or $(-1)^{\frac{n}{2}}{n \choose {n/2}}$, according as n is odd or even.

Question 3:

If $s_{n}$ denotes the sum of the first n natural numbers, prove that

(a) $(1-x)^{-3}=s_{1}+s_{2}x+s_{3}x^{2}+\ldots + s_{n}x^{n-1}+\ldots$

(b) $2(s_{1}s_{2m} + s_{2}s_{2n-1} + \ldots + s_{n}s_{n+1}) = \frac{2n+4}{5! (2n-1)!}$

Question 4:

If $q_{n}=\frac{1.3.5.7...(2n-1)}{2.4.6.8...2n}$, prove that

(a) $q_{2n+1}+q_{1}q_{2n}+ q_{2}q_{2n-1} + \ldots + q_{n-1}q_{n+2} + q_{n}q_{n+1}= \frac{1}{2}$

(b) $2(q_{2n}-q_{1}q_{2m-1}+q_{2}q_{2m-2}+\ldots + (-1)^{m}q_{m-1}q_{m+1}) = q_{n} + (-1)^{n-1}{q_{n}}^{2}$.

Question 5:

Find the sum of the products, two at a time, of the coefficients in the expansion of $(1+x)^{n}$, where n is a positive integer.

Question 6:

If $(7+4\sqrt{3})^{n} = p + \beta$, where n and p are positive integers, and $\beta$, a proper fraction, show that $(1-\beta)(p+\beta)=1$.

Question 7:

If ${n \choose 0}$, ${n \choose 1}$, ${n \choose 2}$, …,, ${n \choose n}$ are the coefficients in the expansion of $(1+x)^{n}$, where n is a positive integer, show that

${n \choose 1} - \frac{{n \choose 2}}{2} + \frac{{n \choose 3}}{3} - \ldots + \frac{(-1)^{n-1}{n \choose n}}{n} = 1 + \frac{1}{2} + \frac {1}{3} + \frac{1}{4} + \ldots + \frac{1}{n}$.

That’s all for today, folks!

Nalin Pithwa.

# Combinatorics for RMO : some basics and examples: homogeneous products of r dimensions

Question:

Find the number of homogeneous products of r dimensions that can be formed out of the n letters a, b, c ….and their powers.

Solution:

By division, or by the binomial theorem, we have:

$\frac{1}{1-ax} = 1 + ax + a^{2}x^{2} + a^{3}x^{3} + \ldots$

$\frac{1}{1-bx} = 1+ bx + b^{2}x^{2} + a^{3}x^{3} + \ldots$

$\frac{1}{1-cx} = 1 + cx + c^{2}x^{2} + c^{3}x^{3} + \ldots$

Hence, by multiplication,

$\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \times \ldots$

$= (1+ax + a^{2}x^{2}+a^{3}x^{3}+ \ldots)(1+bx + b^{2}x^{2} + b^{3}x^{3}+ \ldots)(1+cx + c^{2}x^{2} + c^{3}x^{3}+ \ldots)\ldots$

$= 1 + x(a + b + c + \ldots) +x^{2}(a^{2}+ab+ac+b^{2}+bc + c^{2} + \ldots) + \ldots$

$= 1 + S_{1}x + S_{2}x^{2} + S_{3}x^{3} + \ldots$ suppose;

where $S_{1}$, $S_{2}$, $S_{3}$, $\ldots$ are the sums of the homogeneous products of one, two, three, … dimensions that can be formed of a, b, c, …and their powers.

To obtain the number of these products, put a, b, c, …each equal to 1; each term in $S_{1}$, $S_{2}$, $S_{3}$, …now becomes 1, and the values of $S_{1}$, $S_{2}$, $S_{3}$, …so obtained give the number of the homogeneous products of one, two, three, ….dimensions.

Also,

$\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \ldots$

becomes $\frac{1}{(1-x)^{n}}$, or $(1-x)^{-n}$

Hence, $S_{r} =$ the coefficient of $x^{r}$ in the expansion of $(1-x)^{-n}$

$= \frac{n(n+1)(n+2)(n+3)\ldots (n+r-1)}{r!}= \frac{(n+r-1)!}{r!(n-1)!}$

Question:

Find the number of terms in the expansion of any multinomial when the index is a positive integer.

In the expansion of $(a_{1}+ a_{2} + a_{3} + \ldots + a_{r})^{n}$

every term is of n dimensions; therefore, the number of terms is the same as the number of homogeneous products of n dimensions that can be formed out of the r quantities $a_{1}$, $a_{2}$, $a_{3}$, …$a_{r}$, and their powers; and therefore by the preceding question and solution, this is equal to

$\frac{(r+n-1)!}{n! (r-1)!}$

A theorem in combinatorics:

From the previous discussion in this blog article, we can deduce a theorem relating to the number of combinations of n things.

Consider n letters a, b, c, d, ….; then, if we were to write down all the homogeneous products of r dimensions, which can be formed of these letters and their powers, every such product would represent one of the combinations, r at a time, of the n letters, when any one of the letters might occur once, twice, thrice, …up to r times.

Therefore, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of homogeneous products of r dimensions which can be formed out of n letters, and therefore equal to $\frac{(n+r-1)!}{r!(n-1)!}$, or ${{n+r-1} \choose r}$.

That is, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of combinations of $n+r-1$ things r at a time when repetitions are NOT allowed.

Example 1:

Find the coefficient of $x^{r}$ in the expansion of $\frac{(1-2x)^{2}}{(1+x)^{3}}$

Solution 1:

The expression $= (1-4x+4x^{2})(1+p_{1}x+p_{2}x^{2}+ \ldots + p_{r}x^{r}+ \ldots)$, suppose.

The coefficients of $x^{r}$ will be obtained by multiplying $p_{r}$, $p_{r-1}$, $p_{r-2}$ by 1, -4, and 4 respectively, and adding the results; hence,

the required coefficient is $p_{r} - 4p_{r-1}+4p_{r-2}$

But, with a little work, we can show that $p_{r} = (-1)^{r}\frac{(r+1)(r+2)}{2}$.

Hence, the required coefficient is

$= (-1)^{r}\frac{(r+1)(r+2)}{2} - 4(-1)^{r-1}\frac{r(r+1)}{2} + 4 (-1)^{r-2}\frac{r(r-1)}{2}$

$= \frac{(-1)^{r}}{2}\times ((r+1)(r+2) + 4r(r+1) + 4r(r-1))$

$= \frac{(-1)^{r}}{2}(9r^{2}+3r+2)$

Example 2:

Find the value of the series

$2 + \frac{5}{(2!).3} + \frac{5.7}{3^{2}.(3!)} + \frac{5.7.9}{3^{3}.(4!)} + \ldots$

Solution 2:

The expression is equal to

$2 + \frac{3.5}{2!}\times \frac{1}{3^{2}} + \frac{3.5.7}{3!}\times \frac{1}{3^{3}} + \frac{3.5.7.9}{4!}\times \frac{1}{3^{4}} + \ldots$

$= 2 + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times \frac{2^{2}}{3^{2}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times \frac{2^{3}}{3^{3}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times \frac{2^{4}}{3^{4}} + \ldots$

$= 1 + \frac{\frac{3}{2}}{1} \times \frac{2}{3} + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times (\frac{2}{3})^{2} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times (\frac{2}{3})^{3} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times (\frac{2}{3})^{4} + \ldots$

$= (1-\frac{2}{3})^{\frac{-3}{2}} = (\frac{1}{3})^{-\frac{3}{2}} = 3^{\frac{3}{2}} = 3 \sqrt{3}$.

Example 3:

If n is any positive integer, show that the integral part of $(3+\sqrt{7})^{n}$ is an odd number.

Solution 3:

Suppose I to denote the integral and f the fractional part of $(3+\sqrt{7})^{n}$.

Then, $I + f = 3^{n} + {n \choose 1}3^{n-1}\sqrt{7} + {n \choose 2}3^{n-2}.7 + {n \choose 3}3^{n-3}.(\sqrt{7})^{3}+ \ldots$…call this relation 1.

Now, $3 - \sqrt{7}$ is positive and less than 1, therefore $(3-\sqrt{7})^{n}$ is a proper fraction; denote it by $f^{'}$;

Hence, $f^{'} = 3^{n} - {n \choose 1}.3^{n-1}.\sqrt{7} + {n \choose 2}.3^{n-2}.7 - {n \choose 3}.3^{n-3}.(\sqrt{7})^{3}+ \ldots$…call this as relation 2.

Add together relations 1 and 2; the irrational terms disappear, and we have

$I + f + f^{'} = 2(3^{n} + {n \choose 2}.3^{n-2}.7+ \ldots ) = an even integer$

But, since f and $f^{'}$ are proper fractions their sum must be 1;

Hence, I is an odd integer.

Nalin Pithwa.

# Pre-RMO or RMO algebra practice problem: infinite product

Find the product of the following infinite number of terms:

$\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \ldots = \prod_{m=2}^{\infty}\frac{m^{3}-1}{m^{3}+1}$

$m^{3}-1=(m-1)(m^{2}+m+1)$, and also, $m^{3}+1=(m+1)(m^{2}-m+1)=(m-1+2)((m-1)^{2}+(m-1)+1)$

Hence, we get $P_{m}=\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \ldots \times \frac{m^{3}-1}{m^{3}+1}$, which in turn, equals

$(\frac{1}{3} \times \frac{7}{3}) \times (\frac{2}{4} \times \frac{13}{7}) \times (\frac{3}{5} \times \frac{21}{13})\times \ldots (\frac{m-1}{m+1} \times \frac{m^{2}+m+1}{m^{2}-m+1})$, that is, in turn equal to

$\frac{2}{3} \times \frac{m^{2}+m+1}{m(m+1)}$, that is, in turn equal to

$\frac{}{} \times (1+ \frac{1}{m(m+1)})$, so that when $m \rightarrow \infty$, and then $P_{m} \rightarrow 2/3$.

personal comment: I did not find this solution within my imagination !!! ๐ ๐ ๐

The credit for the solution goes to “Popular Problems and Puzzles in Mathematics” by Asok Kumar Mallik, IISc Press, Foundation Books. Thanks Prof. Mallik !!

Cheers,

Nalin Pithwa

# Algebra Training for RMO and Pre-RMO: let’s continue: Fibonacci Problem and solution

Fibonacci Problem:

Leonardo of Pisa (famous as Fibonacci) (1173) wrote a book “Liber Abaci” (1202), wherein he introduced Hindu-Arabic numerals in Europe. In 1225, Frederick II declared him as the greatest mathematician in Europe when he posed the following problem to defeat his opponents.

Question:

Determine the rational numbers x, y and z to satisfy the following equations:

$x^{2}+5=y^{2}$ and $x^{2}-5=z^{2}$.

Solution:

Definition: Euler defined a congruent number to be a rational number that is the area of a right-angled triangle, which has rational sides. With p, q, and r as a Pythagorean triplet such that $r^{2}=p^{2}+q^{2}$, then $\frac{pq}{2}$ is a congruent number.

It can be shown that square of a rational number cannot be a congruent number. In other words, there is no right-angled triangle with rational sides, which has an area as 1, or 4, or $\frac{1}{4}$, and so on.

Characteristics of a congruent number: A positive rational number n is a congruent number, if and only if there exists a rational number u such that $u^{2}-n$ and $u^{2}+n$ are the squares of rational numbers. (Thus, the puzzle will be solved if we can show that 5 is a congruent number and we can determine the rational number $u(=x)$). First, let us prove the characterisitic mentioned above.

Necessity: Suppose n is a congruent number. Then, for some rational number p, q, and r, we have $r^{2}=p^{2}+q^{2}$ and $\frac{pq}{2}=n$. In that case,

$\frac{p+q}{2}, \frac{p-q}{2}$ and n are rational numbers and we have

$(\frac{p+q}{2})^{2}=\frac{p^{2}+q^{2}}{4}+\frac{pq}{2}=(\frac{r}{2})^{2}+n$ and similarly,

$(\frac{p-q}{2})^{2}=(\frac{r}{2})^{2}-n$.

Setting $u=\frac{r}{2}$, we get $u^{2}-n$ and $u^{2}+n$ are squares of rational numbers.

Sufficiency:

Suppose n and u are rational numbers such that $\sqrt{u^{2}-n}$ and $\sqrt{u^{2}+n}$ are rational, when

$p=\sqrt{u^{2}+n}+\sqrt{u^{2}-n}$ and $q=\sqrt{u^{2}+n}-\sqrt{u^{2}-n}$

and 2n are rational numbers satisfying $p^{2}+q^{2}=(2n)^{2}$ is a rational square and also $\frac{pq}{2}=n$, a rational number which is a congruent number.

So, we see that the Pythagorean triplets can lead our search for a congruent number. Sometimes a Pythagorean triplet can lead to more than one congruent number as can be seen with $(9,40,41)$. This set obviously gives 180 as a congruent number. But, as $180=5 \times 36=5 \times 6^{2}$, we can also consider a rational Pythagorean triplet $(\frac{9}{6}, \frac{40}{6}, \frac{41}{6})$, which gives a congruent number 5 (we were searching for this congruent number in this puzzle!). We also determine the corresponding $u=\frac{r}{2}=\frac{41}{12}$.

The puzzle/problem is now solved with $x=\frac{41}{12}$, which gives $y=\frac{49}{12}$, and $z=\frac{31}{12}$.

One can further show that if we take three rational squares in AP, $u^{2}-n$, and $u^{2}$, and $u^{2}+n$, with their product defined as a rational square $v^{2}$ and n as a congruent number, then $x=u^{2}$, $y=v$ is a rational point on the elliptic curve $y^{2}=x^{3}-n^{2}x$.

Reference:

1) Popular Problems and Puzzles in Mathematics: Asok Kumar Mallik, IISc Press, Foundation Books, Amazon India link:

https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_1?s=books&ie=UTF8&qid=1525099935&sr=1-1&keywords=popular+problems+and+puzzles+in+mathematics

2) Use the internet, or just Wikipedia to explore more information on Fibonacci Numbers, Golden Section, Golden Angle, Golden Rectangle and Golden spiral. You will be overjoyed to find relationships amongst all the mentioned “stuff”.

Cheers,

Nalin Pithwa

# Solution to a RMO algebra practice question

Refer to question posted on blog of Mar 13 2018, reproduced here for your convenience. Compare your solution with the one given here:

Question:

Show that the following expression: $[4-3x+\sqrt{16+9x^{2}-24x-x^{3}}]^{1/3}+[4-3x-\sqrt{16+9x^{2}-24x-x^{3}}]^{1/3}$ remains constant in the interval $0 \leq x \leq 1$. Find this constant value.

Solution/proof:

Let y equal the given expression for x in the prescribed interval. Then, taking cube of both sides, we write

$y^{3}=8-6x+3xy$

$y^{3}-8-3x(y-2)=0$

$(y-2)(y^{2}+2y+4-3x)=0$

The only real value of $y=2$, a constant. The roots of the quadratic equation for $0 are complex. It is easy to check that $y=2$ for both $x=0$ and 1.

Alternately, derive the square roots of the expression within the radical; you can use the method of undetermined coefficients for this.

Cheers,

Nalin Pithwa.

# Solutions to two algebra problems for RMO practice

Problem 1.

If a, b, c are non-negative real numbers such that $(1+a)(1+b)(1+c)=8$, then prove that the product abc cannot exceed 1.

Solution I:

Given that $a \geq 0$, $b \geq 0$, $c \geq 0$, so certainly $abc>0$, $ab>0$, $bc>0$, and $ac>0$.

Now, $(1+a)(1+b) = 1 + a + b + ab$ and hence, $(1+a)(1+b)(1+c) = (1+a+b+ab)(1+c)= 1+a+b+ab+c +ac + bc + abc=8$, hence we get:

$a+b+c+ab+bc+ca+abc=7$.ย Clearly, the presence ofย $a+b+c$ and $abc$ reminds us of the AM-GM inequality.

Here it is $AM \geq GM$.

So, $\frac{a+b+c}{3} \geq (abc)^{1/3}$.

Also, we can say: $\frac{ab+bc+ca}{3} \geq (ab.bc.ca)^{1/3}$. Now, let $x=(abc)^{1/3}$.

So, $8 \geq 1+3x+3x^{2}+x^{3}$

that is, $8 \geq (1+x)^{3}$, or $2 \geq 1+x$, that is, $x \leq 1$.ย So, this is a beautiful application of arithmetic mean-geometric mean inequality twice. ๐ ๐

Problem 2:

If a, b, c are three rational numbers, then prove that :$\frac{1}{(a-b)^{2}} + \frac{1}{(b-c)^{2}} + \frac{1}{(c-a)^{2}}$ is always the square of a rational number.

Solution 2:

Let $x=\frac{1}{a-b}$, $y=\frac{1}{b-c}$, $z=\frac{1}{c-a}$. It can be very easily shown that $\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} =0$, or $xy+yz+zx=0$. So, the given expression $x^{2}+y^{2}+z^{2}=(x+y+z)^{2}$ is a perfect square !!!ย BINGO! ๐ ๐ ๐

Nalin Pithwa.

# RMO type algebra practice question

Question:

Show that the following expression

$[4-3x+ \sqrt{16+9x^{2}-24x-x^{3}}]^{1/3}+ [4-3x-\sqrt{16+9x^{2}-24x-x^{3}}]^{1/3}$

remains constant in the interval $0 \leq x \leq 1$. Find this constant value.