Some comments on questions in Journal of Recreational Mathematics

The Journal of Recreational Mathematics (JRM) has now ceased publication, but here are some of my contributions to, or comments on, some of the contents of the Problems and Conjectures section of that journal. After 2015, it was followed by the Topics in Recreational Mathematics (TRM) edited by Charles Ashbacher and available from Amazon.

The questions are simply listed here with links to the individual documents where my comments and solutions are to be found, together with detailed references in each case to the relevant issue of the journal. All the links are to PDF files, except where otherwise indicated.

I’ve edited some for clarity or convenience, and added some extra comments where that seemed appropriate. My extra comments are usually enclosed in []. Sometimes I found time to extend the results further. I also fixed any editorial errors I found, but essentially these contributions are as I submitted them, if I did. Not all were published, and I have selected only those where the contribution appeared to add to the solution provided. The large gaps in the sequence is due to the fact that there were periods when I had no time to look at them to try to find solutions. And, of course, some simply did not interest me enough.

Friday the Thirteenth

Question 2534 in JRM 30(3). Answer in JRM 31(3).

Show that the thirteenth day of each month falls on a Friday more often than any other day of the week.

Solution

A Special Age

Question 2536 in JRM 30(3). Answer in JRM 31(3)

Four people of different ages have each told each other how old they are. One of them then said “If I multiply my age by any of your ages, the product is a permutation of the digits in the two ages.”
How old is everyone?

Solution and comment

Morra

Question 2537 in JRM 30(3). Answer in JRM 31(3)

The following description of Morra: A Game for Two appears in Math Puzzles and Games by M. Holt, Dorset Press, 1992.

This very old finger game comes from Italy. One player is called Morra. On a given signal — a nod of the head, for example — both players put up either one or two fingers, both at the same time. The payoff rules are, in summary form:

• If both players show the same number of fingers, Morra wins two pennies.
• If Morra shows two fingers and his opponent one, Morra loses one penny.
• If Morra shows one finger, and his opponent two, Morra loses three pennies.

In his analysis, Holt displays the payoff matrix, then simply states “Morra’s best strategy, to reduce his losses, is to show two fingers all the time, then he never loses more than one penny.”

Solution

Trisecting a Square

Question 2538 in JRM 30(3). Answer in JRM 31(3)

Determine all ways to dissect a square into three similar parts, two of which are congruent.
Editor’s note: Similar allows mirror images.

Solution and comments

Mr Smith’s Problem

Question 2541 in JRM 30(3). Answer in JRM 31(3)

This problem was brought to the attention of the diarist Samuel Pepys in 1693 by a “Mr. Smith”. Pepys knew a number of eminent scientists and mathematicians through his connections with the Royal Society, so on November 22, 1693, he sent with a letter of introduction to Isaac Newton. Newton found the question badly worded in its original form, but Smith was able to clear up the ambiguities. Three weeks later, Newton wrote back to Pepys giving the correct solution and some details of his working

The problem concerns three men, A, B, and C, playing the following dice game:

• A throws six dice simultaneously and tries to score at least one six.
• B throws twelve dice simultaneously and tries to score at least two sixes.
• C throws eighteen dice simultaneously and tries to score at least three sixes.

Who has the best chance of winning? More specifically:

1. Rank the probabilities of success of A, B and C in decreasing order.
2. How is the ranking changed if A, B and C have 7, 14 and 21 dice respectively, but still require the same number of sixes?
3. Same as (b), but with 8, 16 and 24 dice.

Solution

The Pandigital Society

Question 2543 in JRM 30(3). Answer in JRM 31(3)

George, his wife Georgina, and their daughter Georgette have joined the Pandigital Society. Each member has a Membership Number and a Secret Code Number used for encrypting messages. The Secret Code is always a pandigital number, and is always the square of the Membership Number, which is not very secret, but that’s the way it is.

The society defines a pandigital number as one containing all ten digits once each. Neither Membership Numbers nor Secret Codes may have leading zeros.

On checking their Membership Cards, Georgette discovered that her Membership Number contains all the digits which do not appear in either of her parents’ Membership Numbers.

I can tell you that the difference between George’s Secret Code and Georgina’s is 2048627457 — can you tell me Georgette’s Membership Number?

Solution and Comment

Square Dance

Question 2544 in JRM 30(3). Answer in JRM 31(3).
I found a better solution, and proved it minimal, which was published in JRM 35(1)

Dennis Shasha featured this interesting form of puzzle in Scientific American September 2001. My puzzle, unlike his, really is square!

A square dance floor is marked out on an 8 x 8 grid of squares. Each square is occupied by a dancer. Initially, thirty-two women (indicated by white circles in the first diagram) surround thirty-two men (black circles). The choreographer wants them to change places (creating the second diagram) using a minimum of steps.

At each step, all the dancers move simultaneously — each dancer may either “step on the spot” remaining on the same square, or step into an orthogonally adjacent square. Two adjacent dancers may not exchange places in one step, because they would collide in the process, which would definitely lose points for presentation! Also, only one dancer may occupy any one square following any step.

What is the minimum number of steps required to achieve the exchange of men and women?

Solution

One to Nine

Question 2558 in JRM 30(4). Answer in JRM 31(4).

In how many ways can we place the digits 1 to 9 into a 3 x 3 array so that each digit is smaller than the digit immediately to its right and is smaller than the digit immediately below it?

Solution

Chord of Triangle

Question 2561 in JRM 30(4). Answer in JRM 31(4).

Given a general triangle ABC, construct a chord DE parallel to AB such that DE = AD + EB (see diagram).

Solution

A Literal Sequence

Question 2562 in JRM 30(4). Answer in JRM 31(4).

1. Find the next number in this series: 1, 11, 21, 1112, 3112, 211213, 312213
2. Find the 1000th number in the series.

Solution

A colourful Cube

Question 2563 in JRM 30(4). Answer in JRM 31(4).

A square is divided into four triangles, as shown in the diagram. Not counting rotations as different, there are six ways to colour the triangles red, yellow, blue and green, such that each colour is used once in the square.

Is it possible to make a cube from a set of these six squares such that each edge borders two triangles having the same colour — if so, in how many ways can this be done?

Solution

A New Pentomino Tiling Problem

Question 2564 in JRM 30(4). Answer in JRM 31(4).

Select any one of the twelve pentominoes (see diagram) and place it on a 5 x 7 board in such a way that the rest of the board can be tiled with ten “right trominoes”.

There are several placements of the P pentomino, one of which is shown here.

You are asked to find (at least) three possible placements of each of the 12 pentominoes. The number of possible placements of trominoes in each case is irrelevant, and rearrangement of the trominoes only does not constitute a different solution.

Solution

Four Cuts

Question 2565 in JRM 30(4). Answer in JRM 31(4).

A 6 x 6 square has to be cut into eight pieces of any shape with areas 1, 2, 3, 4, 5, 6, 7, and 8 square units.

1. The figure shows how to achieve this with five cuts. Can you do it with four straight cuts? In how many ways?
2. How many straight cuts are needed if each of the eight pieces has to be a triangle? Please offer one solution.

Solution

Integral Sided Tetrahedron

Question 2566 in JRM 30(4). Answer in JRM 31(4).

1. Find a tetrahedron with integral volume and edges, n, n+1, ... n+5, where n is any integer.
2. Can you find more than one solution?

Comments

Square Cut

Question 2567 in JRM 30(4). Answer in JRM 31(4).

Dissect a unit square into a number of rectangles of different shapes but all having the same area.

Can you find the minimum number of rectangles, and the precise dimensions of each rectangle?

Solution and comment

Touching Circles

Question 2568 in JRM 30(4). Answer in JRM 31(4). My comment on the solution was accepted for publication in JRM 35(1).

Four mutually touching circles have radii in geometric progression, denoted here by 1, r, r², r³. What is the value of r?

The geometric progression can be continued indefinitely in each direction — the previous and next circles are shown in the diagram by the black dot and the shaded segment, which have radii 1/r and r⁴.

Can anyone determine the link between this puzzle and the study of isotopes?

Solution and comment

Ever Even Chance

Question 2581 in JRM 31(1). Answer in JRM 32(1).

A large sack contains a number (less than 10000) of balls of nine different colours. If I select two balls at random there is an exactly even chance that they will be of the same colour.

If, instead, I throw away all the balls of one colour, it is still true that if I select two balls at random there is an exactly even chance that they will be of the same colour.

How many balls were there originally in the sack

Solution

Athletics and Soccer

Question 2582 in JRM 31(1). Answer in JRM 32(1).

There are many stadia in Europe with a soccer pitch inside an athletics track (see diagram). can you find the area of the largest possible soccer pitch which can be contained within an athletics track?

For the purposes of this puzzle, assume only that the inside of the track comprises two parallel straights and two semi-circles with a total perimeter of 400 meters, and that the soccer pitch is rectangular. In real life there are other constraints, so don’t worry if your solution to the puzzle looks a little odd!

Solution

Egyptian Fractions

Question 2583 in JRM 31(1). Answer in JRM 32(1).

The ancient Egyptians used fractions in their calculations, but their notation did not allow them to write fractions such as 11/199, and they did not comprehend such fractions. They could only use unit numerators, so the result of dividing 11 by 199 might be written, used, and understood as 1/20 + 1/199 + 1/3980. There is a software package for PCs which transforms proper fractions into “Egyptian Fractions” — the sum of a series of reciprocals. For the fraction 50/89, it gives:
50/89 = 1/2 + 1/17 + 1/337 + 1/145681 + 1/29711989585 + 1/1471337208468868797457 + 1/6494499543074890436870241790813851000203090

Can you find something more manageable, with fewer terms and less horrendous denominators?

1. What is the smallest number of different reciprocals which sum to 50/89?
2. What is the smallest possible value for the largest denominator in an Egyptian fraction totalling 50/89?

Solution

Prime Square

Question 2584 in JRM 31(1). Answer in JRM 32(1).

In the diagram, the four rows, four columns, and two diagonals, each of which can be read both ways, show 20 different 4-digit numbers. Most of them are prime numbers.

Can you construct a four-by-four grid of digits in which the 20 4-digit numbers are all different and all prime?

Solution

Identity

Question 2585 in JRM 31(1). Answer in JRM 32(1).

An unidentified country has a 7-digit population — and everyone has been given a National ID Number, sequentially from one, allocated by no identifiable logic.

The Census Minister has chosen three names at random, and is finding their ID numbers on the computer. When the first number appears on the screen, the Government’s mathematical whizz-kid informs the Minister that there is precisely 50-50 chance that the other two numbers will both be less than the one just displayed.

What is the population, and what is the first number?

Solution

Red, White and Blue

Question 2586 in JRM 31(1). Answer in JRM 32(1).

I have a number of small wooden cubes, all the same size. I arrange them into a cuboid and paint the whole outer surface red. I then arrange the cubes into a different cuboid shape, with no red squares visible, and paint the outer surface white. Finally, I rearrange them again into a cuboid with no painted squares visible, and paint the outer surface blue. I now find that every face of every cube has been painted

If I have the smallest number of cubes, and use all of them each time, how many of them have only two different colours of paint?

Solution

Triplets

Question 2587 in JRM 31(1). Answer in JRM 32(1).

The four triplets of numbers (14, 50, 54), (15, 40, 63), (18, 30, 70), (21, 25, 72) each have the same sum — 118 — and the same product — 37800. Can you find five triplets with this property, with a common product under 100000?

[Misprint in JRM is here corrected.]

Solution

Semi-Prime Sequence

Question 2591 in JRM 31(1). Answer in JRM 32(1).

A semi-prime is a composite number which has only one pair of divisors. The first few are 4, 6, 9, 10, 14, 15, 21, 22, .... The first trio of consecutive numbers which are all semi-primes is 33—34—35. There are many more such trios, but there cannot be more than three consecutive numbers which are semi-prime, and there cannot be more than eight consecutive odd numbers which are all semi-prime.

If you can explain why those statements are true, that might help you tackle this problem. Can you find a sequence of four (or more) numbers in arithmetic progression which are all semi-prime?

Solution

Maximum Coverage by Triangles

Question 2604 in JRM 31(3). Answer in JRM 32(3).

Cover the maximum possible portion of the area of a square, using four equilateral triangles. The triangles must not overlap each other or extend beyond the sides of the square.

Solution

Diophantine Determination

Question 2605 in JRM 31(3). Answer in JRM 32(3).

Consider the equation a² + b² + c² = 10ⁿ − 1, where a, b, c, and n are positive integers. We may assume without loss of generality, that a ≤ b ≤ c. The simplest solution is (a, b, c, n) = (1, 2, 2, 1). Find all possible solutions and prove that your list is complete.

Solution

Tetromino Dissection

Question 2607 in JRM 31(3). Answer in JRM 32(3).

Dissect the five tetrominoes (see below) in the smallest number of pieces which can be reassembled to form a square.

Solution

An Exponential Equation

Question 2609 in JRM 31(3). Answer in JRM 32(3).

Determine the values of α for which the function f(x) = 3^x + ∣x∣ − α has

1. no zeros;
2. exactly one zero; and
3. exactly two zeros.

Solution

A New Pentomino Cross

Question 2625 in JRM 32(1). Answer in JRM 33(1).

The cross shown in the diagram below differs from those presented by Golomb as numbers 19, 71, 89, and 90 in his book Polyominoes (revised edition, Princeton, 1974).

Construct this cross from the 12 pentominoes.

Assuming that the centre of the X pentomino is in the upper left quadrant, it seemingly may occupy any of the 5 numbered positions. We seek solutions for each of these five possibilities.

Solution

Degree Two Equation

Question 2628 in JRM 32(1). Answer in JRM 33(1).

(Note: A misprint in the published question has here been corrected.)

Find all real solutions of the equation
4a² + 6b² + 2c² + d² = 2 (2ac + ad + bc − bd − cd)

Solution

Folded Triangle

Question 2629 in JRM 32(1). Answer in JRM 33(1).

An equilateral triangle ABC is folded along the line DE so that its apex A touches the base BC at point F, as shown here.

1. If the distances from A to points D and E are 91 and 65 respectively, what is the side of the original triangle ABC?
2. In general, let AD = x and AE = y and find a formula for the length BC in terms of x and y.

Solution

The Longest Reach

Question 2650 in JRM 32(3). Answer in JRM 33(3).

The entire surface of an L-shaped smørgåsbord table is covered with dishes of food. The ends of the table are two metres wide, while the outer sides are each 4 metres long: each interior side is thus 2 metres in length. If there is access to a dish from any point along the table perimeter, where is the dish which requires the longest reach and how long is that reach?

Solution

Nine Digits Make 2004

Question 2652 in JRM 32(3). Answer in JRM 33(3).

Use each of the digits, 1, 2, 3, 4, 5, 6, 7, 8, and 9 once, in ascending order from left to right, and any of the usual operations — addition, subtraction, multiplication, division and exponentiation — along with parentheses, to represent the number 2004.

[Comment: Almost all the solutions provided, also allowed digits to be simply catenated, so that the number twelve could be used as simply 12.
I also found a solution to this using the digits in reverse order, from 9 to 1.

Solution

Lattice Triangles

Question 2701 in JRM 33(4). Answer in JRM 34(4).

Define a lattice triangle as one whose vertices lie on points in the unit square lattice.

1. Which lattice triangle of area 2004 has the smallest perimeter?
2. Which lattice triangle of area 2004 and with the smallest possible number of lattice points on its edge has the smallest perimeter?

Solution

Equal Distances

Question 2707, originally posed (but with an error) in JRM 33(4), and corrected in JRM 34(4). It was answered in 35(2)

Triangle ABC is inscribed in a circle. The tangents to the circle at points B and C intersect at D. What is the relation between the sides of the triangle when A and D are at the same distance from line BC? What possible values can be assumed by angle A?

Solution

Hardware Logarithms

Question 2723 in JRM 34(2). Answer in JRM 35(2).

In the 1690s, it was determined that a heavy, hanging chain assumes the shape of the catenary curve, which has the equation
y = a cosh(x/a) = (a/2)(e^(x/a) + e^(−x/a)).

L. Hodgkin in his A History of Mathematics from Mesopotamia to Modernity (Oxford 2005) reports that Leibniz proposed that ships at sea should have a chain suspended among their instruments, so that, should they lose their tables of logarithms, they could deduce them by measurements on the curve of the chain.

How could they have done this?

[Note that the original question quoted directly from Hodgkin’s work, which I have paraphrased here.]

Solution

Pieces of Eight Challenge

Question 2724 in JRM 34(2). Answer in JRM 35(2).

For each of the ten decimal digits d, find an expression using only the digits 2, 7, and d exactly once each, and having value 8. Only these three digits and the operators for addition, subtraction, multiplication, division, and exponentiation as well as decimal points and parentheses may be used. For example, 7 – 2 + 3 = 8 is one known solution for d = 3.

Solution

Magic Square Relation

Question 2727 in JRM 34(2). Answer in JRM 35(2).

If the two numbers B and C are placed in the positions shown in the 3×3 array in the diagram, there are still an infinite number of ways in which seven additional numbers can be placed so that the sums of the three numbers in each row, column, and diagonal are equal. Nevertheless, given B and C as shown, what must be the value of A?

Solution

Football Results

Question 2728 in JRM 34(2). Answer in JRM 35(2).

In a certain year, the standings in the first round of the FIFA world cup, Group B, were as shown in the table below:

Knowing that each team plays each of the others exactly once, use the given information to determine the score of each match.

Solution

Triangle Sequence

Question 2731 in JRM 34(3). Answer in JRM 35(3).

In the figure A₀BC is equilateral and D is midpoint of BC. Let E be any point on the segment BD.

We define a sequence of points A_n on A₀D by A_nB = A_n−1E. Find the measure of angle A₃ED.

Solution

Fibonacci Sum Sequence

Question 2732 in JRM 34(3). Answer in JRM 35(3).

Arrange the integers 1 through 34 in a sequence such that the sum of every pair of consecutive terms is a Fibonacci number.

Solution

Consecutive Integer Perimeters

Question 2733 in JRM 34(3). Answer in JRM 35(3).

1. Find two n-gons, all of whose vertices are lattice points in the Cartesian plane and whose sides measure, in order, 1, 2, 3, ..., n units.
2. Are there infinitely many n for which such an n-gon exists?

Solution

Maximal Convex Hull

Question 2734 in JRM 34(3). Answer in JRM 35(3).

Let S be a set of points in the Euclidean plane such that each point has at least one neighbouring point at distance 1 unit or less. For each fixed n, determine the maximum possible area of the convex hull of S. The convex hull is the minimum convex set containing all points of S.

Solution

A Pair of Ages

Question 2735 in JRM 34(3). Answer in JRM 35(3).

“Amy, how old are you?” asked Bill. Amy replied “Ladies do not like to tell their age, but I will tell you that my age is the smallest integer that can be expressed as the sum of two distinct squares in two different ways.” “How interesting,” said Bill. “My age is the next larger number having the same property.” How old are Amy and Bill?

Solution

Semi-Magic Square Product

Question 2737 in JRM 34(3). Answer in JRM 35(3).

A semi-magic square is a square matrix in which the sums of all rows and of all columns are equal. Prove or disprove the following: If A and B are two 3 × 3 semi-magic squares with magic sums S_A and S_B respectively, then the matrix product AB is semi-magic with magic sum S_A × S_B.

Solution

Integer Distances to Vertices

Question 2738 in JRM 34(3). Answer in JRM 35(3).

1. (From a posting on alt.math.recreational) Place a point P in a square so that the distances from P to the vertices are distinct integers. If this is possible, what is the smallest such square?
2. Solve the same problem for an equilateral triangle rather than a square.

Solution

Cubes of Primes

Question 2739 in JRM 34(3). Answer in JRM 35(3).

Prove or disprove: If p is a prime greater than 3, then either p³ + 1 or p³ − 1 is divisible by 18.

Solution

Asymmetric Dice Game

Question 2740 in JRM 34(4). Answer in JRM 35(4).

A game is played between Angelica and Basil with a pair of standard cubical unbiased dice. On each throw of the dice, if a double is thrown, Angelica wins. Otherwise, if the total number of pips repeats a total previously thrown in the game, then Basil wins. Otherwise the dice are thrown again. Who has the advantage, and by how much?

Solution

Handy Tips

Question 2742 in JRM 34(4). Answer in JRM 35(4).

A clock has an hour hand and a minute hand, the latter being longer than the former. At 25 minutes past 2, the distance between the tip of the hour hand and the tip of the minute hand is exactly 161 mm. At 25 minutes to 4, this distance is exactly 199 mm. How far apart are the tips of the hands at 9 o’clock?

Solution

Sum of Sixth Powers

Question 2743 in JRM 34(4). Answer in JRM 35(4).

Find an integer, greater than 1, which is the sum of the sixth powers of its digits.

Solution

Rectangular Spiral

Question 2745 in JRM 34(4). Answer in JRM 35(4).

Find the area and perimeter of the shaded spiral.

Solution

Shoelace Clocks

Question 2746 in JRM 34(4). Answer in JRM 35(4).

You have two shoelaces and some matches. You are told that each shoelace takes exactly one hour to burn (in fuse-like fashion, starting by lighting one end) but that the burning is irregular and possibly different for the two laces, some parts of each lace burning faster than other parts. Explain how to use these laces and matches to time exactly 45 minutes.

Solution

L-shaped Pentominoes Redux

Question 2748 in JRM 34(4). Answer in JRM 35(4).

We define an L-shaped polyomino as one consisting of two “arms” projecting at a right angle from a common unit square, each arm being one unit in width. The set of L-shaped polyominoes through size 7 is shown below.

These nine pieces have a total of 50 unit squares. Use the full set to cover the 11 by 5 rectangle with a 5-unit central hole, and the 7 by 8 rectangle with a 6-unit central hole, which are shown in the diagram below.

Solution

Unit Interval Probability

Question 2749 in JRM 34(4). Answer in JRM 35(4).

Two points are picked at random in the interval (0,1). What is the probability that one is more then twice the other?

Solution

MathDice

Question 2755 in JRM 35(1). Answer in JRM 36(1).

For each of the following, you may use addition, subtraction, multiplication, division, powers,decimal points, concatenation, and parentheses, but no roots, factorials, or other notations. Find as many solutions as possible for each part.

1. Use each of the digits 2, 3, and 4 exactly once to form an expression that equals 16.
2. Use each of the digits 4, 6, and 8 exactly once to form an expression that equals 8.

Solution

Expected Length

Question 2758 in JRM 35(1). Answer in JRM 36(1).

A bag initially contains 10 black balls. We repeatedly draw a ball and if it is black, we replace it with a white ball, while if it is white, we return it to the bag. How many repetitions will there be, on average, before all the balls in the bag are white?

Solution

Jigsaw Assemblies

Question 2760 in JRM 35(2). Answer in JRM 36(2).

Here are the set of all 24 possible four-sided jigsaw-puzzle pieces where each side is either straight, has one lobe or one indentation.

Arrange these pieces to form a three by eight rectangle. Pieces may be rotated but not flipped over.

Solution

Fibonacci Seventh Powers

Question 2761 in JRM 35(2). Answer in JRM 36(2).

Generalize the pattern illustrated by the four sums given below.

2⁷ + 1⁷ + 1⁷ = 3 × 2⁶ − 1 × 1 × 2 × 31
3⁷ + 2⁷ + 1⁷ + 1⁷ = 5 × 3⁶ − 1 × 1 × 2 × 31 − 1 × 2 × 3 × 211
5⁷ +3⁷ + 2⁷ + 1⁷ + 1⁷ = 8 × 5⁶ − 1 × 1 × 2 × 31 − 1 × 2 × 3 × 211 − 2 × 3 × 5 × 1441
8⁷ + 5⁷ +3⁷ + 2⁷ + 1⁷ + 1⁷ = 13 × 8⁶ − 1 × 1 × 2 × 31 − 1 × 2 × 3 × 211 − 2 × 3 × 5 × 144 − 3 × 5 × 8 × 9881

In particular, find a general formula for the last number (31, 211, 1441, etc.) in successive equalities.

Solution

Jigsaw Assemblies Again

Question 2770 in JRM 35(3). Answer in JRM 36(3).

Here are the set of all 24 possible four-sided jigsaw-puzzle pieces where each side is either straight, has one lobe or one indentation.

Arrange one complete set of these pieces to form two separate rectangles. Pieces may be rotated but not flipped over.

Solution

Digit Extension of Primes

Question 2771 in JRM 35(3). Answer in JRM 36(3).

For any given natural number n, ending in the digit 1, 3, or 7, is there a prime number which “ends in n”, i.e. whose rightmost digits are the number n?

Solution

Palindromic Composite Numbers

Question 2772 in JRM 35(3). Answer in JRM 36(3).

Prove that any palindrome, other than 11, having an even number of digits is not prime.

Solution

Smarandache Prime Parts

Question 2773 in JRM 35(3). Answer in JRM 36(3).

For an integer n≥2, we define the Inferior Smarandache Prime part, ISPP(n), as the largest prime less than or equal to n. Similarly, the Superior Smarandache prime Part, SSPP(n), is the smallest prime greater than or equal to n. Solve the Diophantine equation

ISPP(x) + SSPP(x) = k, where k is a positive integer.

Solution

Generalized Smarandache Palindromes

Question 2774 in JRM 35(3). Answer in JRM 36(3).

A Generalized Smarandache Palindrome (GSP) is an integer having either of the concatenated forms a₁a₂...a_na_n...a₂a₁ or a₁a₂...a_n−1a_na_n−1...a₂a₁, where the a_i are positive integers. We note that if all a_i are single-digit integers, then the GSP is palindromic in the usual sense. We also note that every integer is trivially a GSP by letting n = 1 in the second pattern of the definition; we hereby rule out this possibility. Find the number of four digit GSP which are not palindromic.

Solution

Digit Cube Sum

Question 2781 in JRM 35(4). Answer in JRM 36(4).

For a positive integer n, let f(n) compute the sum of cubes of digits of n, for example, f(325) = 3³ + 2³ + 5³ = 160. Find a solution, other than 1, of f(f(n)) = n.

Solution

Explicit Inversion

Question 2783 in JRM 35(4). Answer in JRM 36(4).

It is easy to show that for rational numbers a and b, not both zero, we have (a + b√2)⁻¹ = a / (a² − 2b²) + (b√2) / (a² − 2b²). Find an analogous formula for (a + b·2^⅓ + c·2^⅔)⁻¹, where a, b, and c are rational numbers, not all zero. Prove that your formula is valid whenever a, b and c are not all zero.

Solution

Fibonacci Identity

Question 2792 in JRM 36(1). Answer in JRM 37(1).

Let F_n denote the n^th Fibonacci number. and U(x) denote the units digit of x in base ten. Find k such that the following equality is true for all n.

U(F_n+6 F_n+10 F_n+11 F_n+13) = U(F_n F²_n+1 F_n+k)

Solution

Double Duty Pentomino Shapes

Question 2793 in JRM 36(1). Answer in JRM 37(1).

Use one full set of pentominoes (below)

to construct two shapes which can in turn be used to construct either a 6 x 10 rectangle or a 5 x 13 rectangle which encloses a unit square hole and two domino holes (see below).

Solution

A Special Age bis

Question 2794 in JRM 36(1). Answer in JRM 37(1).

Sixteen inhabitants of Mars announce their ages, all of which are different. After a moment of mental calculation one of them notes that if his age is multiplied by that of any of the others, the product is a permutation of the digits in the two ages which were multiplied. We know that Martians use base ten arithmetic and that they can live up to 1000 years. What are the sixteen ages? This problem is offered with an appropriate nod to Problem 2536 (JRM, 30:3, p. 224).

Solution

Pythagorean Pyramids

Question 2795 in JRM 36(1). Answer in JRM 37(1).

A pyramid has a rectangular base and four Pythagorean triangle faces. Find such a pyramid in which the four Pythagorean triangles all have different underlying primitive Pythagorean triples.

Solution

Circle-Squaring-Near-Miss

Question 2796 in JRM 36(1). Answer in JRM 37(1).

Adam Kochansky, a Jesuit mathematician, gave the construction shown below in 1685.

Line RS is tangent at A to the unit circle with centre O. Then with A as centre, draw an arc with unit radius cutting the circle at C. With C as centre, draw another arc with unit radius, cutting the first arc at D. Draw OD intersecting RS at E. Construct EF = 3. Now use the standard square root construction (not shown) to construct a square with side √BF.

1. What is the area of the constructed square, that is, what is the length of BF?
2. This construction amounts to an approximation of π to what accuracy?

Reference: B. Bold, Famous Problems of Geometry and How to Solve Them Dover Publications Inc., New York, 1969.

Solution

Shadow Coincidence

Question 2797 in JRM 36(1). Answer in JRM 37(1).

Early on a fall morning, I was walking between buildings on the Mount Mercy campus. The sun was directly at my back as I started down a set of stairs. I noticed that as I raised and lowered my feet to move from one step to the next, the shadows cast by my feet on the flat ground at the bottom of the flight of stairs were in the identical position each time. Assuming that all of the steps are identical and that there are n of them, determine the relation between the dimensions of a step and the angle at which the sun’s rays strike the sidewalk.

Solution

Smarandache Prime Base

Question 2799 in JRM 36(1). Answer in JRM 37(1).

The infinite Smarandache Prime Base is {1, 2, 3, 5, 7, 11, 13, 17, ...} (i.e., 1 and all the primes). We express a number in this base making a “greedy” selection of base elements. For example,

16 = 1×13 + 0×11 + 0×7 + 0×5 + 1×3 + 0×2 + 0×1 = 1000100_SPB

1. Prove that any integer can be expressed in this base using only digits 1 and 0.
2. What is the integer with the largest number of digits 1 in this base?

Solution

Smarandache Ratio Theorem

Question 2824 in JRM 36(4). Answer in JRM 37(4).

Prove: If the points A₁, B₁, C₁ divide the sides ||BC|| = a, ||CA|| = b and ||AB|| = c, respectively, of triangle ABC in the ratio k>0, then

||AA₁||² + ||BB₁||² + ||CC₁||² ≥ ¾(a² + b² + c²).

Solution

Generating Sequences

Question 2826 in JRM 36(4). Answer in JRM 37(4).

Each member of a group of people generates a sequence of numbers by the following procedure. First, each person chooses any real number as the initial term of his sequence. Then he calculates the sum of the first terms in all other sequences (not including his own) and uses that sum as his second term. This procedure is repeated for each term, i.e. the (i+1)^th term in each sequence is the sum of the i^th terms of all other sequences. One person generates a sequence whose first three terms are 1, 3 and 13.

1. What are the next three numbers in this sequence?
2. How many people are in the group?

Solution

Classification by N Plus Digit Sum

Question 2830 in JRM 37(1). Answer in JRM 38(1).

For a positive integer n, let f(n) be the sum of n and its individual digits, e.g. f(25) = 25 + 2 + 5 = 32. We can generate an increasing sequence by iteration of this function, e.g. 14, 19, 29, 40, 44, ... Then we can partition the set of positive integers according to the number of iterations needed to reach a palindrome. Class 0 consists of the palindromes, Class 1 contains 10, 20, 30, 40, ... etc.

1. Is there a non-empty Class n for each n?
2. Is there an n for which Class n is finite?
3. Are these classes as chaotic as they seem at first glance or are there underlying patterns?

Comment

Making Twelve

Question 2834 in JRM 37(1). Answer in JRM 38(1).

You may choose to throw any number of standard, unbiased, cubical dice. You win if the total of the numbers rolled on all of the dice is 12. How many dice should you use?

Solution

Diophantine Delight

Question 2836 in JRM 37(1). Answer in JRM 38(1).

Consider the Diophantine equation a^m + a^k = b². For which values in the range 1 ≤ a ≤ 20 does the equation have a solution? In problem 27 of J. Cofman’s What to solve? (Clarendon, 1990), it is shown that if, for a given value of a there is one solution, then there are, in fact, infinitely many solutions.

Solution

Tangent Circles

Question 2855 in JRM 37(3). Answer in TRM:2.

In the diagram, a circle of radius a is inscribed in a circle of radius 2a so that it is tangent to the larger circle at the point (0, 2a). A circle with centre B is tangent to the larger and smaller circle as well as the x-axis. A circle with centre C is tangent to the circle centred at B and circle centred at (0, a) as well as to the x-axis.

What are the coordinates of the points B and C? This problem was inspired by a wooden decoration at the front of some meeting rooms in the Sheraton Hotel and Towers in Chicago.

Solution

Matrix Square roots

Question 2863 in JRM 37(4). Answer in TRM:3.

Find all solutions of the matrix equation .

Solution

Quadratic Function and Arithmetic Sequence

Question 2866 in JRM 37(4). Answer in TRM:3.

For the quadratic equation f(x) = ax² + bx + c, prove that there is a constant C and an arithmetic sequence (s_n) such that for positive integer values n, f(n) is equal to C plus the nth partial sum S_n of the arithmetic sequence.

Solution

Box of Balls

Question 2869 in JRM 37(4). Answer in TRM:3.

In the diagram, if the radius of each inscribed circle is 1, what are the dimensions of the bounding rectangle?

Solution

Pentomino Doublers

Question 2871 in JRM 38(1). Answer in TRM:4.

The figure below, a solution to problem 2822 (JRM 36(4), p. 359), shows that the Z-pentomino and three others can be used to construct a double scale version of itself.

We readily see that the X-pentomino does not share this property. What about the other ten pentominoes?

Solution

Digital Fraction Sum

Question 2874 in JRM 38(1). Answer in TRM:4.

We can express the number 1 in several ways as the sum of two fractions which together use each of the nine non-zero digits exactly once. The fractions are not necessarily in lowest terms. For example, 1 = ⁴⁄₁₂ + ⁶³⁸⁄₉₅₇.

1. What is the smallest possible value for any of the fractions?
2. If we include zero, but not as the first digit in any term, then what is the smallest possible value for any of the fractions?

Solution

Integral Powers of an Irrational Number

Question 2875 in JRM 38(1). Answer in TRM:4.

Show that every positive integral power of √2 - 1 can be expressed in the form √m - √(m-1).

Solution

Dudeney’s Century Puzzle

Question 2876 in JRM 38(1). Answer in TRM:4.

H. E. Dudeney in his Amusements in Mathematics, question 90, “The Century Puzzle”, asks us to represent the number 100 as a “mixed fraction”, that is of the form A + ^B⁄_C, using each of the digits 1 – 9, once only. For example 100 = 82 + ³⁵⁴⁶⁄₁₉₇. He says that Edouard Lucas had found seven solutions, but in fact there are 11, and in that Dudeney was correct. There are 11 solutions, which he supplies in his answers.

1. What about other powers of ten? Can the numbers 10, 1000, 10000, etc. be so represented, and if so, in how many ways?
2. What about using all ten digits, not allowing zero as the first digit in any number?
3. In question 91, “More Mixed Fractions”, he states in his answers that, for the number 26, he had “recorded no fewer than 25 different arrangements” using the nine digits. In fact there are more. How many?

Solution

Recurring 3’s

Question 2877 in JRM 38(1). Answer in TRM:4.

In his book, Mathematics Galore (MAA, Washington DC), James Tranton poses the problem of finding a number N such that all of the multiples N, 2N, ..., 10N contain the digit 3.

1. Tanton provides the solution N = 19507893 which “more than” solves this problem. Find the first positive integer m, such that mN does not contain a digit 3.
2. Find a value of N which improves on the one given in part a. That is, find N such that all the multiples N, 2N, ..., kN contain the digit 3, and k≥m.
3. Is there an upper bound on the value of k?

Solution

Linear Combination of Irrationals

Question 2879 in JRM 38(1). Answer in TRM:4.

We consider the inequality |aπ + be + cφ| < 10^-6 where π, e, and φ (golden ratio) are mathematical constants, while a, b, and c are integers. Find solutions where abc ≠ 0, and

1. a + b + c = 0
2. |a| + |b| + |c| is minimum.

Solution

Prime Search

Question 2880 in JRM 38(2). Answer in TRM:8.

The ordered set of odd numbers beginning with the integer 3 is a list that contains all the prime numbers greater than 2. It has half as many integers as does the ordered set of all integers beginning at 3. One process for generating the elements of the list of odd numbers is to begin with the number 3 and repeatedly add 2.

Define a similar process that generates a list of integers that contains all the prime numbers beginning with prime P and has fewer than one-fourth as many integers as the ordered set of all integers beginning at P. As in the example given above, your list may contain integers which are not prime.

Solution

Clock Progression

Question 2883 in JRM 38(2). Answer in TRM:8.

A circular clock face has radius 12. The hour hand and minute hand have zero thickness and move continuously at a constant speed, not in discrete jumps. At 12:00 the hands coincide, of course. At how many minutes after 12:00 will they be positioned so that a disk of radius 2 can be tangent to each of them and to the circumference of the clock?

Solution

Close Encounters of the n-th Kind

Question 2884 in JRM 38(2). Answer in TRM:8.

1. Find a solution for the inequality |xⁿ + yⁿ - zⁿ| < |1 + 2ⁿ - 3ⁿ|, where x, y, z are distinct integers and n is an integer greater than 3.
2. Find another solution subject to the same conditions specified in part a.

Solution

Math Dice (2)

Question 2885 in JRM 38(2). Answer in TRM:8.

For each of the following, you may use addition, subtraction, multiplication, division, powers, decimal points, concatenation, and parentheses, but no roots, factorials, or other notations. Find as many solutions as possible for each part.

1. Use each of the digits 2, 3, and 4 exactly once to form an expression which equals 4.
2. Use each of the digits 2, 3, and 6 exactly once to form an expression which equals 8.

Solution

Digital Fractions

Question 2886 in JRM 38(2). Answer in TRM:8.

1. We can represent the number 2012 as ⁵⁶¹³⁴⁸⁄₂₇₉, using each of the digits 1 to 9 exactly once. What is the next higher number that can be so represented?
2. If we wish to find such representations using all ten decimal digits, then 2012 cannot be so represented. The next lower number with this representation is 2006 = ^1293870⁄₆₄₅. Which is the next higher number with such a representation?

Solution

Prime Power Expansion

Question 2887 in JRM 38(2). Answer in TRM:8.

1. Find the smallest prime which can be expressed in five different ways as a sum 2ⁿ + p, where n is an integer and p is prime.
2. Find the smallest prime which can be expressed in five different ways as a sum 2ⁿ + p^m, where n and m are integers and p is prime, and m > 1 for at least one of these expressions.

Solution

Pancake Flipping

Question 2889 in JRM 38(2). Answer in TRM:8.

The pancake flipping problem was mentioned in a presentation by Ivars Peterson at the Iowa section meeting of the Mathematical Association of America. It has a rich history and has applications in computer network theory.

A cook is incapable of cooking a batch of pancakes where all of them are the same size. Once a stack is complete, they will be stacked in a random order and the waiter will reorganize the stack so that they go from the largest to the smallest from the bottom to the top. This is accomplished by performing a flip, which is placing a spatula somewhere in the pile and flipping the entire stack of pancakes above the spatula over.

Problem: Assuming that it takes at most k flips to orient a stack of n pancakes, prove that it will take at most k+2 flips to orient a stack of n+1 pancakes.

Solution

Infinite Sequence of Integral Square roots

Question 1 in TRM:1. Answer in TRM:5.

Prove that, for a suitably chosen integer x₁, the recursive formula x_n+1 = x_n + 4 + 4√(x_n - a), where a is a positive integer, produces an infinite sequence of integers.

Solution

Binary and Dyadic Numerals

Question 2 in TRM:1. Answer in TRM:5.

On page 188 of The Gödelian Puzzle Book: Puzzles, Paradoxes & Proofs, Raymond M. Smullyan describes dyadic notation, where all integers greater than zero can be expressed using a string of 1‘s and 2‘s. Using the alphabet {1,2} all such integers can be expressed in the form
2ⁿd_n + 2^n-1d_n-1 + ... + 2¹d₁ + 2⁰d₀
For example, 4, 5 and 6 are 12, 21, and 22 in dyadic form.

Shortly after the description of dyadic notation there is the sentence: “It is just that for any given number in decimal notation, the digits of the dyadic and binary expressions of the number are different.”

1. Determine an infinite family of integers whose binary and dyadic expressions are identical.
2. Prove that no number other than those in the family of part (a) can have identical binary and dyadic expressions.

Solution

Circle Triplet (1)

Question 3 in TRM:1. Answer in TRM:5.

In the figure, a circle of radius a is inscribed in a circle with radius 2a and centre (0,0), so that it is tangent to the larger circle at the point (2a, 0). A circle with centre B(b, d) is tangent to the other two circles as well as to the x-axis. Finally, a circle with centre C(c, e) is tangent to all three circles mentioned previously.

Find the centre and radius of this last circle.

Solution

Circle Triplet (2)

Question 4 in TRM:1. Answer in TRM:5.

In the figure, a circle with centre A(a, 0) is inscribed in a circle radius 2a and centre (0, 0), so that it is tangent to the larger circle at point (2a, 0). A circle with centre B(b,d) is tangent to these other two circles as well as to the x-axis. Finally, a circle with centre C is tangent to the x-axis and to the two circles with centres A and B.

Find the centre and radius of this last circle.

Solution

Catch Twenty-Two

Question 9 in TRM:1. Answer in TRM:5.

Four students (A, B, C and D) enter a three-day mathematics competition and each is assigned two integers to be used throughout. A is assigned 2 and 9, B gets 3 and 8, C gets 4 and 7, while D has 5 and 6. Each day they are paired in teams of two and the pairings are never repeated. Each day the team’s task is to use their four digits to form expressions which equal 22. They may concatenate digits as well as using the operators +, –, ×, ÷ and √. Each expression must use all four of their digits and operators may be used more than once in any expression. The radical symbol may not have an index, so it indicates a square root only.

1. Determine the number of possible solutions for each team.
2. Determine the total number of possible solutions by each student over the three days.

Solution

Pythagorean Dissection

Question 12 in TRM:1. Answer in TRM:5.

Find a four-piece dissection of a 7 × 7 square and a 24 × 24 square such that the pieces can be re-assembled to form a 25 × 25 square. Pieces may be turned over but cuts may only be made along boundaries of the unit squares which make up the larger squares.

Solution

Point of Concurrency in a Square

Question 4 in TRM:3. Answer in TRM:8.

Let ABCD be a square. Let E be the midpoint of BC, F the midpoint of CD and G the midpoint of BE. Prove that the lines AE, BF and DG are concurrent.

Solution