# 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, $(7r+12)(r-4)=0$.

More later,

Nalin Pithwa

# Solution: Intel Pentium P5 floating point unit error (1994): RMO problem !!!

Finally, the much awaited solution is here:

(I re-state the problem from a previous blog, almost a month old):

Two number theorists bored in a chemistry lab, played a game with a large flask containing 2 litres of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that $(p+2)$ is also a prime number. Then, the first number theorist would pipette out $1/p$ litres of chemical and the second $\frac{1}{p+2}$ litres. How many times do they have to play this game to empty the flask completely?

Solution:

It is easy to play this game initially even for ordinary people : one could guess p to be 3 because 5 is a prime number, then 5 and 7, 11 and 13, 17 and 19, 29 and 31, and so on. These are called twin primes. Number theorists need to be there to recall large twin primes. The emptied amount of liquid in litres is given by the twin prime harmonic series $H_{P}^{TP}$:

$H_{P}^{TP} = (\frac{1}{3} + \frac{1}{5}) + (\frac{1}{5} + \frac{1}{7}) + (\frac{1}{11} + \frac{1}{13}) + (\frac{1}{17}+\frac{1}{19}) + \ldots$

This series is known to converge to 1.902160583104…which is known as Brun’s constant, named after Viggo Brun, who proved it in 1919. It is a curious result because it is not known if infinitely many twin primes exist, refer for example,

http://www.math.sjsu.edu/~goldston/twinprimes.pdf

even though it is known that infinitely many primes exist (a result proved by Euclid in 300 BC!) and the harmonic series diverges (a result proved by Euler in the eighteenth century). Had the series $H_{P}^{TP}$ diverged, then one could say that infinite twin primes exist. But, as the series converges (must converge with finitely many twin primes or may converge even with infinitely many twin primes), the question of infinitude of twin primes is still an open one. (there is a recent famous result of Prof. Yitang Zhang also regarding this). Anyway, the point is that the two number theorists would not be able to empty 2 litres even if they play the game for infinitely long period. So, they are not bored and can keep themselves busy in the chemistry lab forever.

Another curious fact about Brun’s constant is that its computation in a computer revealed a floating point division arithmetic error in Intel’s Pentium P5 Floating Point Unit in 1994. This bug was discovered by Thomas Nicely while evaluating the reciprocals of twin primes 824633702441 and 824633702443. Consequently, Intel incurred USD 475 million to fix this bug. For a while in 1995, number theory and Brun’s constant took the centre stage in popular media.

For curious minds, there also exist prime triplets, prime quadruples etc. If four number theorists play the game, they will not be able to empty even 1 litre because the harmonic series of prime quadruples is estimated to be around 0.8705883800.

Reference:

Popular Problems and Puzzles in Mathematics, Asok Kumar Mallik, IISc Press, Foundation Books:

https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_1?s=books&ie=UTF8&qid=1530628680&sr=1-1&keywords=popular+problems+and+puzzles+in+mathematics

# AMS Menger Awards 2018

(shared from the AMS website for motivational purposes)

The AMS presented the Karl Menger Memorial Awards at the 2018 Intel International Science and Engineering Fair (Intel ISEF), May 13-18, 2018 in Pittsburgh, PA. The First Place Award of US$2000 was given to Ryusei Sakai, Sota Kojima, and Yuta Yokohama, Shiga Prefectural Hikone Higashi High School, Japan, for “Extension of Soddy’s Hexlet: Number of Spheres Generated by Nested Hexlets.” [Photo: bottom row (left to right): Dr. Keith Conrad (committee chair), Rachana Madhukara, Yuta Yokohama, Sota Kojima, Ryusei Sakai; top row (left to right): Chavdar Lalov, Gianfranco Cortes-Arroyo, Gopal Goel, Savelii Novikov, Boris Baranov. Not pictured: Muhammad Abdulla. Photo by the Society for Science & the Public.] The Menger Awards Committee also presented the following awards: • Second Award of$1,000: Gopal Krishna Goel (Krishna Homeschool, OR), “Discrete Derivatives of Random Matrix Models and the Gaussian Free Field” and Rachana Madhukara, Canyon Crest Academy, CA, “Asymptotics of Character Sums”
• Third Award of \$500: Chavdar Tsvetanov Lalov, Geo Milev High School of Mathematics, Bulgaria, “Generating Functions of the Free Generators of Some Submagmas of the Free Omega Magma and Planar Trees”; Gianfranco Cortes-Arroyo, West Port High School, FL, Generalized Persistence Parameters for Analyzing Stratified Pseudomanifolds”; Muhammad Ugur Oglu Abdulla, West Shore Junior/Senior High School, FL, “A Fine Classification of Second Minimal Odd Orbits”; Boris Borisovich Baranov and Savelii Novikov, School 564, St. Petersburg, Russian Federation, “On Two Letter Identities in Lie Rings”
• Certificate of Honorable Mention: Dmitrii Mikhailovskii, School 564, St. Petersburg, Russian Federation, “New Explicit Solution to the N-Queens Problem and the Millennium Problem”; Chi-Lung Chiang and Kai Wang, The Affiliated Senior High School of National Taiwan Normal University, Chinese Taipei, “’Equal Powers Turn Out’ – Conics, Quadrics, and Beyond”; Kayson Taka Hansen, Twin Falls High School, ID, “From Lucas Sequences to Lucas Groups”; Gustavo Xavier Santiago-Reyes and Omar Alejandro Santiago-Reyes, Escuela Secundaria Especializada en Ciencias, Matematicas y Tecnología, Puerto Rico, “Mathematics of Gene Regulation: Control Theory for Ternary Monomial Dynamical Systems”; Karthik Yegnesh, Methacton High School, PA, “Braid Groups on Triangulated Surfaces and Singular Homology”

A booklet on Karl Menger was also given to each winner. This is the 28th year of the presentation of the Karl Menger Memorial Awards. The Society’s participation in the Intel ISEF is supported in part by income from the Karl Menger Fund, which was established by the family of the late Karl Menger. For more information about this program or to make contributions to this fund, contact the AMS Development Office.

Cheers to the winners,

Nalin Pithwa.

# Solution to a “nice analysis question for RMO practice”

The question from a previous blog is re-written here for your convenience.

Question:

How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?

Solution:

The figure below shows how two and three cards can be stacked so that the mass of cards is equal on either side of the vertical line passing through the corner of table’s edge in order to just balance them under gravity:

the set of first two cards are arranged as follows (the horizontal lines represents the cards):

$xxxxxxxxxxxxxxxx\line(5,0){170}$

$\line(5,0){150}xxxxxxxxxxxxxxxxxxx$

the set of three cards are arranged as follows:

$xxxxxxxxxxxxxxxxxxxxxx\line(5,0){150}$

$xxxxxxxxxxxxx\line(5,0){150}$

$\line(5,0){150}xxxxxxxxxxxxxxxxxxxxxx$

We can see that the length of the overhand is a harmonic series of even numbers multiplied by the length of one card, L.

Overhand distance is $(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{52 \times 2})L$ for 52 cards.

It may be noted that the series if continued to infinity leads to $H_{\infty}^{E}$.

That is, $H_{\infty}^{E}=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots$

This series is known to diverge as proved below:

First consider, $H_{\infty}=1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}+ \ldots$, which is, greater than

$1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}+ \ldots$, which is greater than $1+ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots$. Hence, $H_{\infty}$ diverges as we go on adding 1/2 indefinitely.

Now, let $H_{E}=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots = \frac{1}{2}(1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots)=\frac{1}{2}H_{\infty}$

Since $H_{\infty}$ diverges, $H_{E}$also diverges.

Hence, the “overhang series” also diverges.

This means that the cards can be stacked indefinitely and the overhang distance can reach infinity. However, this will happen very slowly as shown in the table below:

$\begin{array}{cc} n^{E} & H_{n}^{E}\\ 2 & 0.5 \\ 10 & 1.46 \\ 100 & 2.59 \\ 1000 & 3.74 \\ 10000 & 4.89 \\ 100000 & 6.05 \end{array}$

Computing the number of cards that completely overhang off the table needs information about the overhang distance for different number of cards. As shown below in the figure, four cards are required to have one card completely away from the edge of the table. This is because $(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8}=1.0417 >1)$.

(the set of four cards are arranged as follows:)

$xxxxxxxxxxxxxxxxxxxxxxxxxxxx\line(5,0){150}$

$xxxxxxxxxxxxxxxx\line(5,0){150}$

$xxxxxxxxxx\line(5,0){150}$

$xxxxx\line(5,0){150}$

We can see that the length of the overhang is a harmonic series of even numbers multiplied by the length of one card, L:

Overhang distance = $(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{52 \times 2})L$ for 52 cards

It may be noted that the series if continued to infinity, leads to $H_{\infty}^{E}$

$H_{\infty}^{E} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots$

This series is known to diverge. This means that the cards can be stacked indefinitely and the overhang can reach infinity. However, this will happen very slowly as shown in the table below:

$\begin{array}{cc} n^{E} & H_{\infty}^{E} \\ 2 & 0.5 \\ 10 & 1.46 \\ 100 & 2.59 \\ 1000 & 3.74 \\ 10000 & 4.89 \\ 100000 & 6.05 \end{array}$

Computing the number of cards that completely overhang off the table needs information about the overhang distance for different numbers of cards. As shown in the above schematic figures of cards with overhangs, four cards are required to have one card completely away from the edge of the table. This is because

$(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8})=1.0417>1$

For the second card to overhang completely, leaving the first card (and hence one half) that is already completely overhung, it is now necessary that

$(\frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n})>1$, or

$(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n} )>1+ \frac{1}{2}$

where n needs to be found out. By generating some more data, we can find the value of n to be 11.

For third overhanging card, we need

$(\frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{2n})>1$ or

$(\frac{1}{2} + \frac{1}{4} + \frac{1}{6}+\frac{1}{8}+ \ldots + \frac{1}{2n})>1+\frac{1}{2} + \frac{1}{4}$

Thus, for m completely overhanging cards, we find n such that $H_{2n}^{E} > 1+ H_{2(m-1)}^{E}$

The table below shows these values wherein we see an approximate pattern of arithmetic progression by 7.

$\begin{array}{cccc} m & n & m & n \\ 1 & 4 & 11 & 78 \\ 2 & 11 & 12 & 85 \\ 3 & 19 & 13 & 92 \\ 4 & 26 & 14 & 100 \\ 5 & 33 & 15 & 107 \\ 6 & 41 & 16 & 115 \\ 7 & 48 & 17 & 122 \\ 8 & 55 & 18 & 129 \\ 9 & 63 & 19 & 137 \\ 10 & 70 & 20 & 144 \end{array}$

By examining the pattern in the table, we can get a simple rule to estimate the number of completely overhanging number of cards m, with an error of utmost one, for n cards stacked.

$m = round(\frac{n}{7.4})=round(\frac{10n}{74})$.

Reference:

Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books.

Hope you enjoyed the detailed analysis…

More later,

Nalin Pithwa