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.

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.

Problem of the week, 161

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

The residue of an integer n modulo an integer d > 1 is the remainder r left when n is divided by d. That is, if n = dq + r for integers q and r with 0 < r < d, we write $r \equiv n \pmod d$ for the residue of n modulo d. Show that the residue modulo 7 of a (large) integer n can be found by separating the integer into 3-digit blocks $n = b(s)b(s-1)\dots b(1)$.(Note that b(s) may have 1, 2, or 3 digits, but every other block must have exactly three digits.) Then the residue modulo 7 of n is the same as the residue modulo 7 of $b(1) - b(2) + b(3) - b(4) + \dots \pm b(s)$. For example,
$n = 25,379,885,124,961,154,398,521,655 \pmod 7$
$\equiv 655 - 521 + 398 - 154 + 961 - 124 + 885 - 379 + 25 \pmod 7$ $\equiv 1746 \pmod 7$ $\equiv 746 - 1 \pmod 7$ $\equiv 745 \pmod 7 \equiv 3 \pmod 7$.
Explain why this works and show that the same trick works for residues modulo 13.

Problem of the week, 137

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

Chain addition is a technique employed in cryptography for extending a short sequence of digits, called the seed to a longer sequence of pseudorandom digits. Quoting David Kahn (in Kahn on Codes, MacMillan, New York, 1983, p. 154), “the first two digits of the [seed] are added together modulo 10 [which means they are added and the carry is neglected] and the result placed at the end of the [sequence], then the second and third digits are added and the sum placed at the end, and so forth, using also the newly generated digits when the [seed] is exhausted, until the desired length is obtained”. Thus, the seed 3964 yields the sequence 3964250675632195… .

Periodic pattern

a. Show that this sequence eventually repeats itself.
b. Show that the sequence begins repeating itself with “3964”.
c. EXTRA CREDIT: How many digits are there before the first repetition of “3964”?

Problem of the week, 148

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

Suppose p and q are each monic polynomials of degree 4 with real coefficients and the intersection of their graphs is {(1, 3), (5, 21)}. If p(3) – q(3) = 20, what is the area enclosed by their graphs?

A particularly simple puzzles on round pegs

I just thought of this simple (at least I think it is simple) puzzle.

Consider a long loop of string and a number (at least 2) of round pegs of radius 1 inch each, parallel to each other. Drape the string around the pegs and pull the pegs so that the string is tight, as in the picture (which has only 3 pegs). Notice some sections of the string are straight and some are curved (shown in red in the picture).

Why is the total length of the curved sections equal to $2\pi$?

This is related to a pulley puzzle of Harry Langman, published in 1949.