Michael Reid’s Happy New Year Puzzles, 2018

Belatedly posted by permission of Michael Reid. Enjoy!

Here are some New Year’s puzzles to help start out 2018.

1. Arrange the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 in the
expression a^b + c^d + e^f + g^h + i^j to make 2018.

2. (a) Express 2018 = p^q + r^s where p, q, r, s are primes.
(b) Express 2018 = p^q - r^s where p, q, r, s are primes.

3. (a) Is it possible to put the first 9 primes, 2, 3, 5, 7, 11, 13,
17, 19 and 23 into a 3×3 matrix that has determinant 2018?
(b) Is it possible to put the first 16 primes, 2, 3, 5, … , 53,
into a 4×4 matrix that has determinant 2018?

4. (a) Express 2018 = A / B using the fewest number of distinct
For example, the expression 7020622 / 3479 uses only seven
different digits. But it is possible to do better than this.
(b) Express 2018 = (A_1 \cdot A_2 \cdot ... \cdot A_m) / (B_1 \cdot B_2 \cdot ... \cdot B_n) using the fewest number of distinct digits.

Michael Reid’s Happy New Year Puzzle, 2017

Belatedly posted by permission of Michael Reid. Enjoy!

Here are some interesting puzzles to start the New Year; hopefully they are not too easy!

1. Express 2017 as a quotient of palindromes.

2. (a) Are there two positive integers whose sum is 2017 and whose product
is a palindrome?
(b) Are there two positive integers whose difference is 2017 and whose
product is a palindrome?

3. Is there a positive integer n such that both 2017 + n and
2017 n are palindromes?

4. What is the smallest possible sum of the decimal digits of 2017 n ,
where n is …
(a) … a positive integer?
(b) … a prime number?
(c) … a palindrome?

5. Consider the following two operations on a positive integer:
(i) replace a string of consecutive digits by its square,
(ii) if a string of consecutive digits is a perfect cube,
replace the string by its cube root.

Neither the string being replaced, nor its replacement, may have
have “leading zeros”. For example, from 31416 , we may change it to
319616 , by squaring the 14 . From 71253 , we may change it to
753 by taking the cube root of 125 .

(a) Starting from the number 2017 , what is the smallest number we
can reach with a sequence of these operations?
(b) What is the smallest number from which we can start, and reach
the number 2017 with a sequence of these operations?

6. Find a list of positive rational numbers, q_1 , q_2 , … , q_n
whose product is 1 , and whose sum is 2017 . Make your list as
short as possible.
Extra Credit: Prove that you have the shortest possible list.

Michael Reid’s Happy New Year Problems, 2020

Posted by permission of Michael Reid. Enjoy!

New Year’s Greetings!

Here are some fun puzzles to start the year.

1. Substitute the numbers 1, 2, … , 9 for the letters
a, b, … , i in the expression a^b + c^d + (e + f + g - h)^i
to get 2020.

2. Use the digits 1, 2, … , 9 in order, and any of the usual
arithmetic operations and parentheses to get a number that is
as close as possible to, but not exactly equal to 2020.

3. Express 2019/2020 as a sum of distinct Egyptian fractions,
i.e. 1 / n_1 + 1 / n_2 + ... + 1 / n_k for integers 0 < n_1 < n_2 < ...  n_k < 202049
(but 202049 is not square).

5. Make a 4×4 matrix of single-digit integers (0-9) with digits
2, 0, 2, 0 on the main diagonal, and having determinant 2020.
Is it possible to do it with a symmetric matrix?

If you liked this one, check out other puzzles ont this blog tagged with “Michael Reid”.

Problem of the Week, #121

A former colleague Bill Wardlaw (March 3, 1936-January 2, 2013) used to create a “Problem of the Week” for his students, giving a prize of a cookie if they could solve it. Here is one of them.

Problem 121

The Maryland “Big Game” lottery is played by selecting 5 different numbers in \{ 1,2,3,\dots, 50\} and then selecting one of the numbers in \{ 1,2,3,\dots, 36\}. The first section is an unordered selection without replacement (so, arrange them in increasing order if you like) but the second selection can repeat one of the 5 numbers initially picked.

How many ways can this be done?

Simple unsolved math problem, 6

If you know a little point-set topology, below is an unsolved math problem whose statement is relatively simple.

Probably everyone has at least seen the Mandelbrot set in some form, as it’s a popular object of mathematical artists. Here’s a picture from Wikipedia:


The formal definition is as follows. Let f_c (z)=z^2+c, where c\in \mathbb{C} is a complex number. The Mandelbrot set X is the complex plot of the set of complex numbers c for which the sequence of iterates

f_c (0), f_c (f_c (0)), f_c (f_c (f_c (0))), \dots,

remains bounded in absolute value.
We say X is locally connected if every point x\in X admits a neighborhood basis consisting entirely of open, connected sets.

Conjecture: The Mandelbrot set X is locally connected.

Simple unsolved math problem, 5

It seems everyone’s heard of Pascal’s triangle. However, if you haven’t then it is an infinite triangle of integers with 1‘s down each side and the inside numbers determined by adding the two numbers above it:


First 6 rows of Pascal’s triangle

The first 6 rows are depicted above. It turns out, these entries are the binomial coefficients that appear when you expand (x+y)^n and group the terms into like powers x^{n-k}y^k:


First 6 rows of Pascal’s triangle, as binomial coefficients.

The history of Pascal’s triangle pre-dates Pascal, a French mathematician from the 1600s, and was known to scholars in ancient Persia, China, and India.

Starting in the mid-to-late 1970s, British mathematician David Singmaster was known for his research on the mathematics of the Rubik’s cube. However, in the early 1970’s, Singmaster made the following conjecture [1].

Conjecture: If N(a) denotes the number of times the number a > 1 appears in Pascal’s triangle then N(a) \leq 12 for all a>1.

In fact, there are no known numbers a>1 with N(a)>8 and the only number greater than one with N(a)=8 is a=3003.


[1] Singmaster, D. “Research Problems: How often does an integer occur as a binomial coefficient?”, American Mathematical Monthly, 78(1971) 385–386.

Simple unsolved math problem, 3

A perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself. For example,  1 + 2 + 3 = 6 implies 6 is a perfect number.

Unsolved Problem: Are there any odd perfect numbers? 

The belief, by some, that there are none goes back over 500 years (wikipedia).

If you want to check out some recent research into this problem, see oddperfect.org.


(Another unsolved problem: Are there an infinite number of even perfect numbers?)

Simple unsolved math problem, 2

In 1911, Otto Toeplitz asked the following question.

Inscribed Square Problem: Does every plane simple closed curve contain all four vertices of some square?

This question, also known as the square peg problem or the Toeplitz’ conjecture, is still unsolved in general. (It is known in lots of special cases.)


Inscribed square, by Claudio Rocchini

Thanks to Mark Meyerson (“Equilateral triangles and continuous curves”,Fundamenta Mathematicae, 1980) and others, the analog for triangles is true. For any triangle T and Jordan curve C, there is a triangle similar to T and inscribed in C. (In particular, the triangle can be equilateral.) The survey page by Mark J. Nielsen has more information on this problem.

Added 2016-11-23: See also this recent post by T. Tao.

Added 2020-07-01: This has apparently been solved by Joshua Greene and Andrew Lobb! See their ArXiV paper (https://arxiv.org/abs/2005.09193).

Simple unsolved math problem, 1

In 1937 Lothar Collatz proposed the 3n+1 conjecture (known by a long list of aliases), is stated as follows.

First, we define the function f on the set of positive integers:

If the number n is even, divide it by two: f(n)=n/2.
If the number n is odd, triple it and add one: f(n)=3n+1.

In modular arithmetic notation, define the function f as follows:
f(n)=  {n/2},\  if \ n\equiv 0 \pmod 2, and f(n)=  {3n+1},\  if \ n\equiv 1 \pmod 2. Believe it or not, this is the restriction to the positive integers of the complex-valued map (2+7z-(2+5z)\cos(\pi z))/4.

The 3n+1 conjecture is: The sequence
n,\ f(n),\ f^2(n)=f(f(n)),\ f^3(n)=f(f^2(n)),\ \dots
will eventually reach the number 1, regardless of which positive integer n is chosen initially.

This is still unsolved, though a lot of people have worked on it. For a recent survey of results, see the paper by Chamberland.