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"Tha" Riddles - Next 10 of 947.
Riddle:
The land was white the seed was black It'll take a good scholar to riddle me that. What am I?
Answer: An eye or an eyeball
Riddle:
An eye in a blue face saw an eye in a green face. "That eye is like to this eye," Said the first eye, "But in low place not in a high place." What is it?
Answer: The sun on the daisies.
Riddle:
A soccer fan, upset by the defeat of his favorite team, slept restlessly. In his dream a goalkeeper was practicing in a large unfurnished room, tossing a soccer ball against the wall and then catching it.
But the goalkeeper grew smaller and smaller and then changed into a ping-pong ball while the soccer ball was swelled up into a huge cast-iron ball. The iron ball circled round madly, trying to crush the ping-pong ball, how did the ping-pong find safety whithout leaving the floor?
Answer: If the ping-pong ball rolls flush against the wall, the cast-iron ball cannot crush it.
Those who know geometry can determine that if the diameter of a large ball is at least 5.83 (3+2(square root of 2) times as large as the diameter of a little ball, then the little ball will be safe if it hugs the wall.
A cast-iron ball that is larger than a soccer ball is more than 4.83 times as large in diameter as a ping-pong ball.
Riddle:
When the celebrated German mathematician Karl Friedrich Gauss (1777-1855) was nine he was asked to add all the integers from 1 through 100. He quickly added 1 to 100, 2 to 99, and so on for 50 pairs of numbers each adding to 101.
Answer: 50 X 101=5,050.
What is the sum of all the digits in integers from 1 through 1,000,000,000? (That's all the digits in all the numbers, not all the numbers themselves.)
Answer: The numbers can be grouped by pairs:
999,999,999 and 0;
999,999,998 and 1'
999,999,997 and 2;
and so on....
There are half a billion pairs, and the sum of the digits in each pair is 81. The digits in the unpaired number, 1,000,000,000, add to 1. Then:
(500,000,000 X 81) + 1= 40,500,000,001.
Riddle:
You want to send a valuable object to a friend. You have a box which is more than large enough to contain the object. You have several locks with keys. The box has a locking ring which is more than large enough to have a lock attached. But your friend does not have the key to any lock that you have. How do you do it? Note that you cannot send a key in an unlocked box, since it might be copied.
Answer: Attach a lock to the ring. Send it to her. She attaches her own lock and sends it back. You remove your lock and send it back to her. She removes her lock.
Riddle:
A camel travels a certain distance each day. Strangely enough, two of its legs travel 30 miles each day and the other two legs travel nearly 31 miles. It would seem that two of the camel's legs must be one mile ahead of the other two legs, but of course this can't be true.
Since the camel is normal, how is this situation possible?
Answer: The camel operates a mill and travels in a circular clockwise direction. The two outside legs will travel a greater distance than the two inside legs.
Riddle:
A boy presses a side of a blue pencil to a side of a yellow pencil, holding both pencils vertically. One inch of the pressed side of the blue pencil, measuring from its lower end, is smeared with paint. The yellow pencil is held steady while the boy slides the blue pencil down 1 inch, continueing to press it against the yellow one. He returns the blue pencil to its former position, then again it slides down 1 inch. He continues until he has lowered the blue pencil 5 times and raised it 5 times- 10 moves in all.
Supposed that during this time the paint neither dries nor diminishes in quantity. How many inches of each pencil will be sneared with paint after the tenth move?
Answer: At the start, 1inch of the yellow pencil gets smeared with wet paint. As the blue pencil is moved downward, a second inch of the blue pencils smears a second inch of the yellow pencil.
Each pair of down and up movesof the blue pencil smears 1 more inch of each pencil. 5 pairs of moves will smear 5 inches. This together with the initial inch, makes 6 inches for each pencil.
Riddle:
There were five men going to church and it started to rain. The four that ran got wet and the one that stood still stayed dry. How did the one stay dry?
Answer: It was a body in a coffin with the bearers.
Riddle:
The king dies and two men, the true heir and an impostor, both claim to be his long-lost son. Both fit the description of the rightful heir: about the right age, height, coloring and general appearance. Finally, one of the elders proposes a test to identify the true heir. One man agrees to the test while the other flatly re-fuses. The one who agreed is immediately sent on his way, and the one who re-fused is correctly identified as the rightful heir. Can you figure out why?
Answer: The test was a blood test. The elder remembered that the true prince was a hemophiliac.
Riddle:
There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, ...). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, ...). This continues until all 100 people have passed through the room. What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?
Answer: First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........ That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it's original state. So why aren't all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.
