Pre RMO (PRMO) practice questions, specially selected: PRMO 2018

The following set of problems tinkers, kindles some basic mathematical concepts as well as algebraic manipulations, so I suggest you try them out:

Try to decide by yourself which set of points are defined by these relations:

(a) |x| = |y|

(b) \frac{x}{|x|} = \frac{y}{|y|}

(c) |x| + x = |y| + y

(d) [x] = [y].

Note: The symbol [x] denotes the whole part of the number x, that is, the largest whole number not exceeding x. For example, [3.5] = 3, [5] = 5, [-2.5] = -3.

(e) x - [x] = y - [y]

(f) x - [x] > y - [y].

Good luck.

Nalin Pithwa.

Pre RMO August 2018: some practice problems selected

Question 1:

Can the product of 31256 and 8427 be 263395312? Give reasons (of course, brute force long calculation will not be counted as an answer ! :-)).

Solution 1:

Use the rule “casting out the nines”: a number divided by 9 will leave the same remainder as the sum of its digits divided by nine.

In this particular case, the sums of the digits of the multiplicand, multiplier, and product are 17, 21, and 34 respectively, again, the sums of the digits of these three numbers are 8, 3, and 7, hence, 8 times 3 is 24 and, which has 6 for the sum of the digits; thus, we have two different remainders, 6 and 7, and the multiplication is incorrect.

Question 2:

Prove that 4.41 is a square number in any scale of notation whose radix is greater than 4.

Solution 2:

Let r be the radix; then, 4.41 = 4 + \frac{4}{r} + \frac{1}{r^{2}}=(2 + \frac{1}{r})^{2};

thus, the given number is the square of 2.1

Question 3:

In what scale is the decimal number 2.4375 represented by 2.13?

Solution 3:

Let r be the radix; then, 2 + \frac{}{} + \frac{}{} = 2.4375= 2 \frac{7}{16}

hence, 7r^{2}-16r-48=0

that is, (7r+12)(r-4)=0.

Hence, the radix is 4.

More later,

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 !!! ๐Ÿ™‚ ๐Ÿ™‚ ๐Ÿ™‚

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<x<1 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. ๐Ÿ™‚ ๐Ÿ™‚

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! ๐Ÿ™‚ ๐Ÿ™‚ ๐Ÿ™‚

Nalin Pithwa.