Quiz Question

Number 214 of a series

Here is a calendar. It could be any calendar. I drew a 3×3 box around 9 dates. What is the sum of the numbers in the box? You are allowed five seconds.

Post your answer in the comments section below.

Update and Solution

Mike has the right approach. To solve this in under five seconds you need to know about the life of Carl Friedrich Gauss. As a young school boy he was assigned the problem of adding up a column of large numbers. He noticed the professor wrote the column of numbers as an arithmetic series. He quickly realized the sum was the product of the average and the number of numbers. The average would be the middle number (odd number of numbers) or half way between the two middle numbers. He turned in his answer so quickly the professor was astounded.

Days in a month are an arithmetic sequence. All this problem does is to chop the sequence into three , equally spaced, sequences of the same length. The sum will be 9 times the middle number.

180 (in under 5 seconds)

Quiz Question

Number 212 of a series

The circles represent four pennies arranged in a square. All right, they don’t look like pennies, but use your imagination. Also use your imagination to move two pennies to form another square smaller than the original.

Post your solution in the comments section below.

Update and Solution

Everybody who had a go at this one got it. Here’s the picture.

Quiz Question

Number 209 of a series

What’s the last digit of the following?

171999 + 111999 – 71999

Enter your answer in the comments section below.

Update and Solution

Mike saw through this one early, only his comment was so cryptic I missed it. See below. I didn’t see it until I started working through it, then I thought, “too easy.” Here is how it works.

A is the last digit of 171999. B is the last digit of  111999. C is the last digit of 71999.

It is obvious A = C, so the answer is B. And B is 1.

Quiz Question

Number 206 of a series

This one is from The Moscow Puzzles by Boris A. Kordemsky, page 44.

That’s supposed to be a chess board above, although I have not shaded in the black and white squares. It doesn’t matter, because this is not a real chess game. The black circles are black pawns, about twice as many as in a real chess game. The question is can you set a white knight on the board and proceed to capture all 16 pawns in 16 consecutive moves? Let’s assume you can, then what are the moves?

Post your answer in the comments section below.

Update and Solution

Mike comments this is too easy. Could be. Kordemsky has this to say:

Start by capturing one of the pawns not marked with a red dot in the diagram below.

Then the rest should be straight-forward.

Quiz Question

Number 205 of a series

I’m late posting this week’s Quiz Question. Monday was a busy day. But here it is. I came across an item posted on the Internet—The Subtle Art of the Mathematical Conjecture. It has some interesting stuff, including proved and yet to be proved conjectures in mathematics. There is a discussion of Fermat’s Conjecture, and there is this:

Counterexamples can lie far ashore, like the one found by Noam Elkies, a mathematician at Harvard University, disproving Euler’s conjecture, a variation on Fermat’s conjecture that states that a fourth power can never be written as a sum of three other fourth powers. Who would have guessed that the first counterexample involved a number of 30 digits?*

The asterisk points to a note:

20,615,6734 = 2,682,4404 + 15,365,6394 + 18,796,7604.

Am I the only one who sees something wrong with this equation? Post your comments below.

Update and solution

Mike and Elmo got this one right. I had it wrong. I tried to check the math in my head and miscounted the multiplications. Shit happens.

Quiz Question

Number 204 of a series

Math time again. Some time ago I had the idea to clear out shelf space, so I dumped all the copies of Schaum’s Outline books that are available on Kindle. This is from Schaum’s Vector Analysis, Chapter 2.

2.11. Show that the vectors A = − i + j, B = − ij – 2k, C = 2j + 2k form a right triangle.

i, j, and k are the unit vectors parallel to the 3D coordinate axes. To be clear, imagine each vector as a straight line oriented in space but movable. Can you move the three lines into position to form a right triangle? Give a mathematical proof.

The answer is in the book, but you know better than to look for the answer. Post your answer as a comment below.

Update and solution

First it is needed to demonstrate the three vectors satisfy the Pythagorean Theorem. Square the lengths of all vectors and see if the sums of two add up to the third. So we have:

|A|² = 2

|B|² = 6

|C|² =8

That works.

To form a closed polygon (triangle) the sum of the vectors must be zero.

Sum = (-1 -1)i (1 -1 +2)j (-2 +2)k, which is not zero. But if you reverse A, then

Sum = (1 -1)i (-1 -1 +2)j (-2 +2)k, which is zero.

Also, note the dot product of A and B is zero. The two vectors are orthogonal.

So the three vectors form a right triangle.

Quiz Question

Number 200 of a series

Here is a nice problem, not too difficult, pertinent to a current hot topic.

Hypothetical scenario: Nothing is adding carbon dioxide to the atmosphere. Carbon dioxide has a 100-year half life in the atmosphere. We crank up a contraption that pumps 100 million tons of carbon dioxide into the atmosphere each year. How much carbon dioxide is in the atmosphere when  a steady state is obtained?

Post your answer as a comment below. Extra points for describing the calculation.