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Past Problems

2009
71 (Nov 10) 70 (Sep 29) 69 (May 14) 68 (Apr 27)
67 (Apr 6) 66 (Mar 16) 65 (Feb 11)
2008
64 (Dec 20) 63 (Nov 17) 62 (Oct 22) 61 (May 7)
60 (Mar 26) 59 (Feb 15) 58 (Jan 25)
2007
57 (Nov 20) 56 (Oct 19) 55 (Sep 21) 54 (May 11)
53 (Apr 20) 52 (Apr 4) 51 (Mar 13) 50 (Feb 21)
2006
49 (Dec 5) 48 (Nov 14) 47 (Oct 30) 46 (Oct 9)
45 (Sep 19) 44 (Apr 28) 43 (Apr 7) 42 (Mar 20)
41 (Feb 15)
2005
40 (Dec 11) 39 (Nov 11) 38 (Oct 27) 37 (Oct 5)
36 (Sep 16) 35 (May 18) 34 (Apr 29) 33 (Apr 8)
32 (Mar 9) 31 (Feb 11)
2004
30 (Dec 15) 29 (Nov 22) 28 (Nov 3) 27 (Oct 13)
26 (Sep 24) 25 (May 3) 24 (Apr 16) 23 (Mar 24)
22 (Feb 25)
2003
21 (Dec 10) 20 (Nov 14) 19 (Oct 22) 18 (Oct 1)
17 (May 5) 16 (Apr 25) 15 (Mar 28) 14 (Mar 7)
13 (Feb 7)
2002
12 (Dec 20) 11 (Dec 2) 10 (Nov 11) 9 (Oct 21)
8 (Oct 4) 7 (Sep 18) 6 (May 8) 5 (Apr 22)
4 (Mar 27) 3 (Mar 11) 2 (Feb 20) 1 (Feb 6)

Problems in 2009



Problem Number 71

November 10, 2009

Winners:
Math solutions: Rosa Lemus (IV) (most elegant)
William Amidan (IV)
Jecsi Astorgan (IV)
Quang Bach
Jeff Burt (IV)
Jorge Dominguez (IV)
Jose Fuentes (IV)
Eva Garcia (IV)
Laura Gonzalez (IV)
Jacquelyn Mart

Computer Solution:
Michael Haddad

Consider the function f(x) = x^3 + 6*x^2 -15*x + k. Find all values of k such that the absolute maximum value and the absolute minimum value of f on the interval [-10,2] have the same absolute value.


Problem Number 70

September 29, 2009

Winners:
undergrad math solutions:
Amir Ruhi (SD)
William Amidon (IVC)
Jecsi Astroga (IVC)
Jeff Burt (IVC)
Jorge Dominguez (IVC)

graduate math solution:
Jonathan Boiser (SD)
Kameryn Denaro (SD)

Twenty people are sitting around a round table. A committee of three people is chosen randomly from among them. What is the probability that at least two of them were sitting next to each other?


Problem Number 69

May 14, 2009

Winners:
Undergraduate, blend of computer and math solutions Ryan Rosenbaum
http://www-rohan.sdsu.edu/%7Epsalamon/ProblemOfTheFortnight/solutions/POF69Rosenbaum.pdf

Among all triangles whose three sides are consecutive positive integers n, n+1, and n+2,
find the smallest sides for which the following three conditions hold:

1) two sides are odd numbers
2) at least one side is a prime number
3) the triangle's area is divisible by 20

Hint: The formula for the area A of a triangle with sides a, b, c, is
A = sqrt(s(s-a)(s-b)(s-c)), where s=(a+b+c)/2.


Problem Number 68

April 27, 2009

Winners:
Undergraduate math solutions:
Ryan Rosenbaum (with proof)
Laura Gonzalez (IV campus)
Rosa Lemus (IV campus)
Maxwell Handy (Bio major)
Andre Williams
Jeff Burt

Graduate solutions:
Brandon Fahy (Math & Computer solutions)
Vince Dayes

Determine the last two digits of

2^20000009 + 6^20000009 + 7^20000009.


Problem Number 67

April 6, 2009

Winners:
Daniel McGrath (undergraduate, combined computer and math solution)

Define a positive integer to be a "factorial-number" if when expressed with no leading zeros in base 10, it equals the sum of the factorials of its digits. For example,

1 is a factorial-number because 1 = 1!
2 is a factorial-number because 2 = 2!
145 is a factorial-number becase 145 = 1 + 24 + 120 = 1! + 4! + 5!

Prove or disprove: exactly four different positive integers are factorial-numbers.


Problem Number 66

March 16, 2009

Winners:
Nick Whaley (computer solution).

Find all polynomials of degree n<11 with all real roots
and with each of their coefficients equal to 1 or -1.


Problem Number 65

February 11, 2009

Winners:
Undergraduate Math Winners: Philip Tabares (most elegant), Jeff Burt, Rodolfo Nunez, and Procorro Gonzalez (partial solution)

Undergraduate Computer Winners: Nick Whaley (most elegant), Michael Lee, and Michael Haddad.

My odometer has six digits which can indicate a number miles up to 99999.9. The sum of the digits it is currently showing is as large as it has ever been. This sum was the same 900 miles ago. How far do I have to drive before the sum of the digits will be larger than it is now?


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