December 1 Problem. My creativity has run dry. So, here is a problem sent to me by Ginny Nunemaker (originally from her son Ken). Computers usually represent color using a hexadecimal value for each primary color (red, green and blue). The range for each color is “00″ to “FF” ;. In this system, how many color combinations are there?
Answer by Steve Shantz. Short answer…. FFFFFF possibilities (for those amongst us who are not so ‘normal,’ and can comprehend large numbers in Hexadecimal notation). As most ‘normal’ people are better able to comprehend numbers when presented in base 10, Hex F is equivalent to 16. Hex FF is therefore equal to 16 x 16 = 256. This means that each pixel can have 255 possible shades of red, green, or blue, but each color can also have a value of 0 (black), so this makes 256. Because each color is independent of the others, there are 256 x 256 x 256 color possibilities, giving a total of 16,777,216 different colors, including ‘black’.
November 9 Problem. Abby’s GPA was higher than Brian’s GPA in the fall and spring semesters and neither took a May term course. Yet, Brian’s annual GPA was higher than Abby’s annual GPA. Show how such a scenario is possible.
Answer by Jim Miller. It could happen if one semester they both had a higher GPA than the other semester, and Brian took more credit hours that semester. Or, alternatively, it could happen if Abby took more hours the semester that they both had lower GPAs. And, of course, a combination of the above could occur.
For example if Abby: Abby takes 15 hours in the fall and has a 2.8 GPA, and Brian takes 12 hours and has a 2.75 GPA. Second semester, Abby takes 12 hours and has a 3.8 GPA, and Brian takes 15 hours and has a 3.75 GPA.
Abby has a higher GPA both semesters. But if you add up the quality points:
Abby 2.8*15 + 3.8*12 = 87.6 quality points
Brian 2.75*12 + 3.75*15 = 89.25 quality points
They have both taken 27 hours for the year, so Abby’s GPA is 87.6/27 = 3.244, and Brian’s GPA is 89.25/27 = 3.306. Or, to 2 decimal points, 3.24 and 3.31, respectively.
October 30 Problem. A common algebra mistake takes the form (a+b)2 = a2 + b2 . For example, claiming (2+3)2 = 22 + 32 is clearly a mistake. Note, however, that (2+3)2 = (2+1)2 + (3+1)2 . Determine all values for m and n such that (m+n)2 = (m+1)2 + (n+1)2 .
Answer by Rusty Emery. Go to here to seen the solution.
October 23 Problem. How many four-digit numbers are composed of three distinct digits (with no leading 0s), such that one digit is the average of the other three?
Answer. 462, but no one submitted a correct solution. So, this problem will be left for others to ponder.
October 7 Problem. Marie sells widgets for $29.95. It costs her $12.14 to produce a widget. During one month of producing and selling widgets, Marie had a net revenue of $6,453.17. How many widgets did Marie sell and how many widgets did Marie produce during that month?
Answer by Art Smucker. I think the smallest amounts are 512 produced and 423 sold. I did it the hard way, using a spread sheet, I set up cells so that x would equal 6453.17-29.95y. I then put in integer values in the y cell, and saw what x was. I quickly established that y had to be greater than 360. So I started increasing y by 1, and soon noted that when I got a value for x nearing integer, i.e. with decimal .9 or greater, that increasing y by 15 got the x closer to integer. So it was easy and quick to get to the final numbers, correct to at least 10 decimals.
Yes, I know that’s an old fashioned approach, using a newer tool (spreadsheet). I’d have written a little program to do that with a loop, but the only programing language I ever really learned was BASIC, and it’s too long since I used it.
October 1 Problem. A Walk-A-Thon raises money for a worthy cause by having donors pledge an amount of money per mile walked. A Hair-A-Thon would raise money for a worthy cause by having donors pledge an amount of money per mile of spliced together human hairs — piece by piece. If we could get every student, staff, and faculty member of Goshen College to donate all of the hairs on their heads, how many miles of hair would be produced?
Answer by Zach Yordy. Assuming that the average length of a males hair is 2 in and 12 inches for a females hair, and that we have a 60-40 female-male ratio, the average length of hair on campus would be 8 inches. Factoring in that we have 900 students and a 13:1 student-faculty ratio, we would have around 70 faculty members, which means 970 people in total. Considering there are around 100,000 hairs on the average head, I get around 12250 miles of human hair with the hair joined end-to-end.
September 15 Problem. Alan Turing was one of the 20th century’s greatest mathematicians, a founder of the field of computer science, and instrumental in breaking the German Enigma codes – critical to Allied success during WWII. Recently, British Prime Minister Gordon Brown issued an apology to Turing for the appalling way he was treated after the war (see http://www.number10.gov.uk/Page20571) . In honor of Turing, your challenge is to decypher PCLIMEMAXWX0ZJWANFS (blanks have been removed). A shift-increment cypher was used. For example, an initial shift of -3 and an increment of +2 would encypher MATH as JZUK because J is 3 letters before M, Z is 1 letter before A, U is 1 letter after T, and K is 3 letters after H. MATH can be recovered from JZUK with an initial shift of +3 and an increment of -2. You have to figure out what initial shift and increment I used.
Answer by Chaim Hodges. Based upon this encryption a shift initially of -1 and an increment of +3 comes out with the answer “Making Pecce With Math.” The issue with this answer however is that unless you intentionally spelled “peace” wrong then the solution is incorrect. I assigned each letter a number value and figured it out.
p(16) = x(shift)
c(3) = x + c(increment)
L(12) = x + 2c
And so on.