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Solution: Math 101
Answer: POLYDACTYLY

##### Written by Jon Schneider and Josh Alman

We are presented with 11 multiple choice math problems (with choices A-E). Each question can have multiple correct choices (for example, both 12 and 60 can be the sum of the side lengths of a right triangle of integer length).

Reading each set of correct answers as 5-bit binary and converting to letters reveals the message “HEXADECIMAL”. This suggests redoing the puzzle, interpreting all numbers in hexadecimal this time around. Doing this directly, however, does not work (this is confirmed by none of 0x21, 0x31, 0x32, 0x45, or 0x50 being prime). We must now notice that each of the choices (A-E) are themselves digits in hexadecimal. Including these choices as digits in the numbers (i.e. reading **A** 21 as 0xA21 = 2593) and solving the puzzle in the same way (reading the correct answers as 5-bit binary and converting to letters) reveals the answer **POLYDACTYLY**.

Problem # | Decimal answers | Message 1 | Hexadecimal answers | Message 2 |
---|---|---|---|---|

1 | aBcde | H | Abcde | P |

2 | abCdE | E | aBCDE | O |

3 | ABcde | X | aBCde | L |

4 | abcdE | A | ABcdE | Y |

5 | abCde | D | abCde | D |

6 | abCdE | E | abcdE | A |

7 | abcDE | C | abcDE | C |

8 | aBcdE | I | AbCde | T |

9 | aBCdE | M | ABcdE | Y |

10 | abcdE | A | aBCde | L |

11 | aBCde | L | ABcdE | Y |

Many of the questions can be easily solved given Wolfram Alpha or a powerful calculator. Some notes on the trickier questions:

Question 2: To solve this, it helps to know that any Pythagorean triple can be written in the form (*m*^{2} - *n*^{2}, 2*m**n*, *m*^{2} + *n*^{2}), for some positive integers *m* and *n*, with *m* > *n*. The sum of such a triple is 2*m*(*m*+*n*). Therefore a number *N* is the sum of the elements of a Pythagorean triple if and only if *N* is even and *N*/2 can be written as the product *xy* where *y* < 2*x*.

Question 3: Legendre’s three-square theorem gives a characterization of all natural numbers representable as the sum of three squares (all numbers not of the form 4^{a}(8*b* + 7) for integers *a*, and *b*).

Question 7: All odd numbers can be written as the sum of two consecutive integers (2*k* + 1 = *k* + (*k* + 1)).

Question 9: Leap years are years that are multiples of 4, except for years which are multiples of 100 but not 400.

Question 10: Every number *n* larger than 2 appears at least twice in Pascal’s triangle: as C(*n*,1) and C(*n*, *n*-1). To check if a number appears at least three times, it suffices to check whether it appears as something of the form C(*n*,*m*) for 1 < *m* < *n*-1. Checking whether C(*n*, 2) = C(*n*, *n*-2) = *x* can be done by solving a quadratic equation in *n*; all other coefficients grow quickly enough that it is possible to check whether *x* can be written in that form by looking at the first 20 or so rows.

Question 11: Goldbach’s conjecture states that every even number can be written as the sum of two primes (and is verified up to very large numbers). Since 2 is the only even prime, an odd number *n* can be written as the sum of two primes if and only if *n* - 2 is prime.

##### Author’s Notes

This puzzle was inspired by the puzzle Multiple Choice, a puzzle which another Galactic Trendsetter (Ben Yang) wrote for BAPHL 11.