BMAD

I meant sixth graders came to bother Timmy not 8th graders.

Answers Timmy is much older and thus more physically mature than James.

very clever answers but not what i am looking for

There are three questions but the third one is meant to be the edge of the new stamp to the center of the previous circle.

I have a stamp in the shape of a circle with a diameter of 4 inches and no special design. I need to completely cover my 8.5 by 11 sheet of paper with this stamp completely. I can only use the stamp to cover the sheet, so no dipping the sheet into the ink , what is the least amount of times i could stamp this sheet and still completely cover it? can you model that? What is the most amount of times i could stamp the sheet if each successive stamp must be at least 1 inch away from the center of the previous stamp?

I have a different proof of the same, which gives even more insight into divisibility rules, but it would take longer to explain. We have another Divisibility puzzle going here: http://brainden.com/forum/index.php/topic/15583-more-divisibility/?p=328196]More divisibility Perm Russia actually

Another one

Six coins make the Triangle Man shown in the illustration. The coins form three identical equilateral triangles as shown with the dotted lines. Rearrange the coins into another shape with three equilateral triangles, each with a different size. You may only move one coin at a time so that in its new position it touches at least two other coins.

In this diagram 11 matches make 3 squares: Your challenge is to move 3 matches to show 2 squares. Note: you may not change the orientation of a match. If it is vertical, it must remain vertical. All matches must make polygons

Two blondes are sitting in a street cafe, talking about the children. One says that she has three daughters. The product of their ages equals 36 and the sum of the ages coincides with the number of the house across the street. The second blonde replies that this information is not enough to figure out the age of each child. The first agrees and adds that the oldest daughter has the beautiful blue eyes. Then the second solves the puzzle. What happened?

In Russia you get into a bus, take a ticket, and sometimes say : Wow, a lucky number! Bus tickets are numbered by 6-digit numbers, and a lucky ticket has the sum of 3 first digits being equal to the sum of 3 last digits. When we were in high school (guys from math school No. 7 might remember that ) we had to write a code that prints out all the lucky tickets' numbers; at least I did, to show my loyalty to the progammers' clan. Now, if you add up all the lucky tickets' numbers you will find out that 13 (the most unlucky number) is a divisor of the result. Can you prove it (without writing a code)?

A group of four people has to cross a bridge. It is dark, and they have to light the path with a flashlight. No more than two people can cross the bridge simultaneously, and the group has only one flashlight. It takes different time for the people in the group to cross the bridge: Annie crosses the bridge in 1 minute, Bob crosses the bridge in 2 minutes, Caren crosses the bridge in 5 minutes, Dorothy crosses the bridge in 10 minutes. How can the group cross the bridge in 17 minutes?

A better question would had been: how do you share 17 chocolates equally with your three friends?

But i just tried it now and see that everything is working fine. I only hope my posts yesterday weren't to redundant. If they were, i am sorry.

I used only the forum search engine. I tried key words of some classic riddles and math problems just to see if any of them have been posted before and each of them (all six) came up with no hits. I then, out of suspicion, tried searching out some of my own post titles and again came up empty. I then tried giving generic words like 'square' and 'rectangle' and again nothing in the search forums was returned.

nice. now how many different 'correct' answers are there? IN SECOND ROW SECOND COLUMN IT SHOULD BE 'KH'. I TRIED TO EDIT THE POST BUT EDITING FAILED.

Take from a pack of cards all the Aces, Kings, Queens and Jacks. Arrange them in a 4 × 4 square so that every row, column and diagonal contains one card of each value (A,J,Q,K) and one card of each suit (Heart, Spade, Diamond, Club).

5 people are standing in a queue for plane tickets in Germany; each one has a name, an age, a favorite TV program, where they live, a hairstyle and a destination. Names: Bob, Keeley, Rachael, Eilish and Amy TV programs: The Simpsons, Coronation Street ("Corrie"), Eastenders, Desperate Housewives and Neighbours. Destinations: France, Australia, England, Africa and Italy Ages: 14, 21, 46, 52 and 81 Hairstyle: Afro, long, straight, curly and bald Where they live: A town, a city, a village, a farm and a youth hostel 1. The person in the middle watches Desperate Housewives 2. Bob is the first in the queue 3. The person who watches the Simpsons is next to the person who lives in a youth hostel 4. The person going to Africa is behind Rachael 5. The person who lives in a village is 52 6. The person who is going to Australia has straight hair 7. The person travelling to Africa watches Desperate Housewives 8. The 14-year-old is at the end of the queue 9. Amy watches Eastenders 10. The person heading to Italy has long hair 11. Keeley lives in a village 12. The 46-year-old is bald 13. The fourth in the queue is going to England 14. The people who watch Desperate Housewives and Neighbours are standing next to each other 15. The person who watches Coronation Street stands next to the person with an afro 16. A person next to Rachael has an afro 17. The 21-year-old lives in a youth hostel 18. The person who watches Corrie has long hair 19. The 81-year-old lives on a farm 20. The person who is travelling to France lives in a town 21. Eilish is not next to the person with straight hair

I have tried and tried but cannot get the search engine to work even on problems that i created. So i have no idea if i am reposting problems you guys have once seen in past. If I am, sorry!

Bluebeard explains that his bunch of keys was strung upon an endless key ring and divided into three groups so that the first group multiplied by the second equaled the third! That was the secret by which he knew whether the keys had been tampered with and forbidden chambers had been entered. You see that 6910 multiplied by 7 does not amount to 83452, so the keys were not replaced properly in their groups. Can our clever puzzlists show how the keys must have been arranged in three groups so that the first group multiplied by the second makes the third?

good try but no