# Puzzles and Solutions 2016

All Puzzles and Solutions

### Mathematical RecreationsPUZZLE of the MONTH

1. Puzzles will be uploaded at the 1st day of each month.
2. Solutions will be published at the beginning of the following month. It is possible that puzzles have more than one correct solution.
3. You may send your solution or propose a new puzzle anytime.
4. All original puzzles are under the copyright of the APM Institute and their authors.

#### May 2016(by Kyriakos Kefalas)

R. Graham, D. Knuth and O. Patashnik ask if there exist positive integers k, n, such that:

$\forall&space;k>1,\exists&space;n>1:2^{n}\equiv&space;k\;&space;\left&space;(&space;mod\:&space;n&space;\right&space;)$

(Problem 4.71 in Concrete Mathematics, Addison Wesley, 1998). For, n = 2, 3, … , 32, … , we have:

k = 0, 2, 0, 2, 4, 2, 0, 8, 4, 2, 4, 2, 4, 8, 0, 2, 10, 2, 16, 8, 4, 2, 16, 7, 4, 26, 16, 2, 4, 2, 0, …

(Sequence A015910 in OEIS). We immediately see that odd k, are scarce. For, k = 1, there is no solution at all. For, k = 3, the smallest solution is, n = 4700063497. In spite of this, are there infinitely many n, for each k ? For some progress on this problem see Richard Guy, Unsolved Problems in Number Theory, Springer, 2004, problem F10.

Our puzzle of the month refers to a simpler but related problem: How common are odd quotients of, 2n / n ? For, n = 1, 2, 3, …, 17, …, the quotients are:

$\left&space;\lfloor&space;\frac{2^{n}}{n}&space;\right&space;\rfloor\rightarrow&space;2,2,2,4,6,10,18,32,56,102,186,341,630,1170,2184,4096,7710,...$

(Sequence A000799 in OEIS). We again see that odd quotients are scarce. We ask you to estimate (or even better compute) how many odd quotients appear in the set where n, takes the values, 1 ≤ n ≤ 214 . For help we give the number of odd quotients for similar smaller sets:

 n ≤ odd quotients 1 = 1 0 2 = 2 0 22 = 4 0 23 = 8 0 24 = 16 1 25 = 32 3 26 = 64 8 … … … 214 = 16384 ?

You may choose among three possible answers:    A:  69,  B:  369,  C:  5000.

Solutions (Available after 1st June 2016)

#### Apr 2016(by Kyriakos Kefalas)

Write 4 mathematical expressions that are equal to 666, using only the integers, 1, 11, 111, …, and simple mathematical operations, with the following restrictions:

1. Write one expression which uses in total 8 or less digits of ‘1’.
2. Write two different expressions which use in total exactly 9 digits of ‘1’.
3. Write one expression which uses in total exactly 13 digits of ‘1’.

Example

$1+11-111+\frac{1111-1}{11+\frac{1}{11}}+11^{11}-111111+-\cdots=\:&space;?$

OOOPS, something went wrong in this example I think, …

Solutions (Available after 1st May 2016)

Created: 21 Mar 2016