Rap:
Hungry Hungry Carnivores
A catches B at the same time that B catches C, C catches D, and D catches A
At 1 cm/s and a initial square size of 4cm--4 seconds
A, B, C, & D all move 4 cm
They all meet proboscis to proboscis in a round robin eat off.
Poor robin
Three digit number
If abc|11 then some #pq*11=abc
pq*11=pq*10+pq=p(p+q)q
p(p+q)q=abc so p=a q=b & p+q=b=a+b
So any number abc with b=a+b is abc=11*ab and abc|11.
There is a restriction, this will only be true is a+b<10
Mark:
BEETLES
yes, 4 seconds, 4 cm, they converge in the center
The key is the fact that the one you're chasing is always moving orthogonally to you.
3-DIGIT NUMBER
Proof 1: A test for divisibility by 11 is to sum the digits in a +, -, +, - pattern (every other digit is subtracted). If the final total is a multiple of 11, then so is the number.
In this case we have a - (a+b) + b = 0.
Proof 2: a(a+b)b = 100a + 10(a+b) + b = 110a + 11b = 11(10a+b) = 11*ab
***In the above, adjacent letters are concatenated, not multiplied.
Damn, those guys sure know their beetles.
For the rest of you;
The problem is symmetrical so we can see straight away that whatever paths the beetles take they will always be at the four vertices of a square whose origin remains fixed.
We know that A's path must curve because B is also moving. From the direction of B's motion we can also see that A's path must curve towards the interior of the square. So the square is shrinking (as well as rotating clockwise).
Notice that the component of B's velocity in the direction of AB is always zero so the length of the side from A to B is shrinking at the speed at which A is moving towards B, 1cm/s. After 4s the square has shrunk to a point with all the beetles having spiralled into the centre.
To summarise: A takes 4s to catch B and travels 4cm in this time. As to what happens, that is left to your imagination!
Three digits.
What I meant to write was, "I don't doubt it"
When we write a three-digit number in base-ten we are really saying "so many hundreds", "so many tens" and "so many units". If our number is n and the digits are represented by x, y and z (reading from left to right) then we can write:
n = 100x + 10y + z
We are told that
y = x + z
Therefore
n = 100x + 10(x + z) + z
n = 110x + 11z
n = 11(10x + z)
Therefore n must be divisible by 11!
Trains.
Mark was on the right track when he said,
"TRAINS
72.25%
25.5%"
Rap was on time with:
Both are on time 72.25% of the time
One will be late 25.5% of the time
Both will be late 2.25% of the time
Armies.
Right theorem. I cannot see any avatars at the moment. I will return later.
Rap, thank you for the information on the bikes. Totally unsuitable for road surfaces of the time. A wonder how the idea got off the ground, never mind the rider.
The local recycling plant has just bought a new metal compactor that produces a smaller cube of scrap iron than does the older machine. Mark noticed however, that the combined volumes of one cube from each compactor was numerically the same as the combined lengths of all their edges.
What are the dimensions of the cubes, if you consider only integral solutions
The Valley High School Math Club was out on its field day. The teacher, Mr. Rap, assigned Mark and Whim the problem of finding a buried box. He told them that the box was buried at the fourth vertex of the parallelogram having three of its vertices at (-1,4), (1,1), and (3,5) on the Cartesian grid that was laid out on the field. Mark and Whim dug at (1,8), but the box was not there.
Why didn't they find the buried box
What would you do
Defend your decision, to your last breath.
A bag contains 50 red and 50 yellow balls. There are three boxes on a shelf. One is labeled RED; one is labeled YELLOW; and one is labeled MIXED.
Two balls at a time are taken from the bag. If both are yellow, they are placed in the YELLOW box; if both are red, they are placed in the RED box. If one of each is picked, both go into the box marked MIXED.
What is the probability that the box marked RED and the box marked YELLOW will have the same number of balls after all pairs of balls have been drawn from the bag
