BMAD

Look inside the spoiler.

It won't let me mark solved for some reason.

It appears that an ingenious or eccentric teacher being desirous of bringing together a number of older pupils into a class he was forming, offered to give a prize each day to the side of boys or girls whose combined ages would prove to be the greatest. Well, on the first day there was only one boy and one girl in attendance, and, as the boy's age was just twice that of the girl's, the first day's prize went to the boy. The next day the girl brought her sister to school, and it was found that their combined ages were just twice that of the boy, so the two girls divided the prize. When school opened the next day, however, the boy had recruited one of his brothers, and it was found that the combined ages of the two boys were exactly twice as much as the ages of the two girls, so the boys carried off the honors of that day and divided the prizes between them. The battle waxed warm and on the fourth day the two girls appeared accompanied by their elder sister; so it was then the combined ages of the three girls against the two boys, and the girls won off course, once more bringing their ages up to just twice that of the boys'. The struggle went on until the class was filled up, but as our problem does not need to go further than this point, to tell the age of that first boy, provided that the last young lady joined the class on her twenty-first birthday. Now, guess the first boy's age.

What is the probability that the next person you meet has an above average number of arms?

What mathematical symbol can be put between 5 and 9, to get a number bigger than 5 and smaller than 9? (Caution: Trick)

this is irony.

7

Recently, Bonanova made a comment to one of my problems about Fibonacci numbers. In all honesty I am not too familiar with their work so I began studying it. In researching it I have found a reference to their congruum problem. But something about it has me stumped; I hope one or more of you can tell me what I am doing wrong: Quote from the article at the MacTutor History of Mathematics archive at St. Andrews University: http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Fibonacci.html "[Fibonacci] defined the concept of a congruum, a number of the form ab(a + b)(a - b), if a + b is even, and 4 times this if a + b is odd where a and b are integers. Fibonacci proved that a congruum must be divisible by 24 and he also showed that for x,c such that x^2 + c and x^2 - c are both squares, then c is a congruum. He also proved that a square cannot be a congruum." With x=15 and c=216, we get the two squares 441 and 9, meaning that 216 should be a congruum. Thus we should be able to find numbers a and b such that 216 = ab(a+b)(a-b) if a+b is even or 54 = ab(a+b)(a-b) if a+b is odd Now, b must be smaller than a and d:= ab(a+b)(a-b) is increasing in both a and b. Let's look at the possible scenarios. Say a+b is odd. We want d to equal 54. Our triples (b,a,d) give us (1,2,6) (1,4,60); thus b=1 is not possible (2,3,30) (2,5,210); thus b=2 is not possible (3,4,84); thus b = 3 and b > 3 is not possible (since d is increasing); thus a+b odd is not possible Say a+b is even. We want d to equal 216. Our triples (b,a,d) give us (1,5,120) (1,7,336); thus b=1 is not possible (2,4,96) (2,6,3849; thus b=2 is not possible (3,5,360); thus b = 3 and b > 3 is not possible (since d is increasing); thus a+b even is not possible So, because 15^2 + 216 = 21^2 and 15^2 - 216 = 3^2 we should have that 216 is a congruum, but we cannot put it in the form ab(a+b)(a-b). What is wrong with this argument?

For a given regular heptagon ABCDEFG prove that 1/AB = 1/AC + 1/AD.

You have an equilateral triangle ABC. Extend a line from vertex B to some point near the center of the triangle. The length of this line is 4. Now extend another line from vertex A to that point. The length of this side is 4. Now extend another point form vertex C to that same point. The length of this side is 5. What is the exact area of equilateral triangle ABC?

What would be the remainder of 3^4^5^6^7^... and so on to infinity is when divided by 17?

Place n distinct points on the circumference of a circle and draw all possible chords through pairs of these points. Assume that no three of these chords pass through the same point. Find and solve the recurrence relation for the number of interior intersection points formed inside the circle.

On the right track but minor mistake

I never know when to use Q.E.D. vs Without loss of generality, is there a difference or is it just a style thing?

Given four points A, B, C, D, and four directly similar quadrilaterals AP1P2B, CQ1Q2B, CR1R2D, AS1S2D with respective centroids P, Q, R, S, let K, L, M, and N be the midpoints of the segments P1Q2, Q1R2, R1S2 and S1P2 respectively (or of P1S2, S1R2, R1Q2 and Q1P2), and let V, W, X be the centroids of the quadrilaterals ABCD, PQRS, KLMN respectively. Then what is special about PQRS, KLMN, and point W?

See attached:

nice try, but no.

For every prime number (p) greater than 3, there exists a natural number (n) such that p^2 = 12(n)-11. Can you provide a counterexample? Else, can you prove it?

Is it too late to blame the calculator?

There is an old saying that one dislikes in others that they dislike in themselves. Assume this line to be true. Also suppose it references personality traits and that it's opposite (opposite defined by me ) is true. Opposite: you like others that do not have your disliked traits. A PhD student for there dissertation is seeking to replicate the above theory(ies) using an experiment. By asking several acquaintices to identify friends from people they dislike and to identify their personality traits. When asked to identify their personality traits in isolation however, the following six people labeled their personality traits using different systems (adjectives). But fortunaetly, everyone either liked or disliked someone. If the theories are true, what are the negative traits that are causing the people to dislike each other? (Which letters match up) Identified Negative Traits Jon: a, b Jacob: e, f Junior: h, I Heimer: l, m, n Smidth: q, r Sara: u, v Jon only hates Junior and Sara. Heimer likes everyone. Jacob only hates Sara and Smidth. Junior only hates Jon. Smidth hates Jacob and Sara and no one else. Sara hates Jon, jacob and Smidth. However, she hates Jacob and Smidth for the same reason. The hate between Jacob and Smidth is the strongest. V does not equal f or q. h=b. List of some of the corresponding adjectives: Unrealistic Ridicule Arrogant Quixotic Embarass Moan Hedonistic

Unless I made a computation mistake, I am getting 137 when n=6 not 132 using your formula.

I was on my junior-high chess team! One time, another seventh grader and I entered an eighth-grade tournament. Every player played every other player once (round-robin); a win counted as one point and a draw was 1/2 point. My friend and I got a total of 8 points, while all the 8th-graders got the same number of points (as each other). How many eighth-graders were in the tourney, and why? *Note there are two possible answers