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Riddle:
An old parchment describes the location of buried treasure: "On the island there are only two trees, A and B, and the remains of a gallows. Start at the gallows and count the steps required to walk in a straight line to tree A. At the tree turn 90 degrees to the left and then walk forward the same number of steps. At the point where you top drive a spike into the ground. Now return to the gallows and walk in a straight line, counting your steps, to tree B. When you reach the tree, turn 90 degrees to the right and take the same number of steps forward, placing another spike at the point where you stop. Dig at the point exactly halfway between the spikes and you will find the treasure." However, our hero when he gets to the island finds the gallows missing. Is there any way he can still get to the treasure?
Answer: A simple experiment with a ruler and paper shows that any position for the gallows leads to the same point.
Riddle:
What is the 4 digit number in which the first digit is one-fifth of the last, and the second and third digits are the last digit multiplied by 3?
Riddle:
This number added to it's square, and the digits of that summation added together, bring it back to itself. What is it?
Answer: 3. 3 squared is 9. 3+9=12. Take the two digits of the summation and add them; 1+2=3. In short; 3-9-12-3. This can also work with the number 0 and 9.
Riddle:
If you like pretty gems that sparkle and shine, I invite you to dig in my virtual mine. My first is purple, fit for a king, My second is green where Dorothy did her thing. My third is red, July's birthstone as well, My fourth is seen in strings and is found inside a shell. My fifth is hard, pure Carbon, and expensive to buy, My sixth is Crocidolite, striped like the big cat's eye. Seventh is two words, a man-made fake of April's stone, Eighth is very dark and found at Lightning Ridge alone. Now take from each gem, one letter in its turn, And you will find the stuff for which even the gods yearn. What am I?
Answer: Answer: Ambrosia The gems are: Amethyst Emerald ( Emerald City in the Wizard of Oz ) Ruby Pearl Diamond Tiger's eye Cubic Zirconium (Fake diamond) Black Opal Taking the first letter of Amethyst, second of Emerald, etc.. gives Ambrosia - The food of the gods.
Riddle:
Julie is going on an extended trip for three weeks. She lives in a remote area where there are frequent electrical power outages which can last up to three or four days. Julie has quite a bit of food in her freezer which would go bad if it thawed and then re-froze. She does have digital clock and a VCR which would flash 12:00 if the power went out. Unfortunately the clock and VCR flash even if the power only goes out for a few seconds. What can Julie do so that when she returns home she will be able to determine whether the power was out long enough to thaw her food? Asking a neighbor whether the power was out, isn't a reliable option because the nearest house is half a mile away, and one house may have power, while another house may have no power. She won?t be able to have a neighbor check on her house every day, and has no one to house sit.
Answer: One thing Julie could do is freeze a tray of ice-cubes, and turn the tray of ice upside down in her freezer. When she comes home, she should check the tray. If the ice cubes are still in the tray, the food is safe to eat. If the trays are empty, it's time to clean out the freezer. She will have to make a judgment call if the ice-cubes are only slightly thawed.
Riddle:
(A) Use two digits to make the smallest possible positive integer.
(B) Five 3s can express 37: 37=33+3+3/3
Find another way to do it.
(C) Use six identical digits to make 100. (Several solutions are possible.)
(D) Use five 4s to make 55.
(E) Use four 9s to make 20.
Answer: (A) 1 X 1; 1/1;2/2;ect....;1-0;2-1;and many others.
(B) 37=333/3X3; 37=3 X 3 X 3 + 3/.3
(C) 99 + 99/99; 55+55- 5- 5; (666-66)/6
(D) 44 + 44/4=55.
(E) 9 + 99/9=20.
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.
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