Monday, December 29, 2014

100

Find at least four ways of writing the number 100, each time using only one digit repeated five times.
For example: (999 / 9) - 9 = 102 (but you must get 100, not 102!!!)

Good luck!

100 Puzzle Solution

  • 111 - 11 = 100
  • (3 * 33) + (3 / 3) = 100
  • (5 * 5 * 5) - (5 * 5) = 100
  • (5 + 5 + 5 + 5) * 5 = 100
  • (11 - 1) ^ ( 1 + 1) = 100 [Thanks to Steven Renich for that one!]
  • ((2 * 2 * 2) + 2) ^ 2 = 100 [Thanks to David Cohen for this other one!]
  • ((99 * 9) + 9) / 9 = 100 [Thanks to Taylor Lowry for this other one!]
  • ((22 - 2) / 2) ^ 2 = 100 [Thanks to Karen D. Miller for this other one!]
  • 6! / 6 - 6! / (6 * 6) = 100 [Thanks to Karen D. Miller for this other one!]
  • 5! - 5 - 5 - 5 - 5 = 100 [Thanks to Karen D. Miller for this other one!]
  • 5! - (5 + 5 + 5 + 5) = 100 [Thanks to Jim St. Clair for this other one!]
  • 5 * 5 * (5 - 5 / 5) = 100 [Thanks to Rishi Mohan Sanwal for this one!]
  • 4! + 4! + 4! + 4! + 4 = 100 [Thanks Karen D. Miller and Saurabh Gupta!]
  • 99 + 9 ^ (9 - 9) = 100 [Thanks Gopalakrishnan Thirumurthy for this one!]
  • 5! - 5 * (5 - 5 / 5) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • (4! + 4 ^ (4 - 4)) * 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • 4! * 4 + 4 - 4 + 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • (4! * 4) + (4 * 4 / 4) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • 5 * 5 * (5 - (5 - 5)!) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • ((9 - 9) */ 9)! + 99 = 0! + 99 = 1 + 99 = 100 [Thanks to Bala Neerumalla!]
  • (3 - 3)! + 33 * 3 = 100 [Thanks Bala Neerumalla for this other one!]
  • (5 + 5) ^ ((5 + 5) / 5) = 100 [Thanks Bala Neerumalla for this one!]
  • ((2 ^ 2) * 2 + 2) ^ 2 = 100 [Thanks Bala Neerumalla for this one!]
  • (4! + ((4 - 4) */ 4)!) * 4 = 100 [Thanks Bala Neerumalla for this one!]
Forum member Bealzbob reminds us that, whenever we have a subtraction like A - B, we can rewrite it as an addition with a negative number, like A + (-B). In the case of the first solution, 111 - 11 = 111 + (-11).

Glen Parnell points out that, in any solution in which the digit 2 is used, it could be replaced by 2!, whenever the 2 is not only a digit but also a number, as 2 = 2! (but of course, 22 is not equal to 22!).
Bala Neerumalla has suggested some alternative solutions using trigonometric functions Sin() and Cos():
  • Sin(99 - 9) + 99 = Sin(90) + 99 = 1 + 99 = 100
  • Cos((9 - 9) */ 9) + 99 = Cos(0) + 99 = 1 + 99 = 100
  • Cos(3 - 3) + 33 * 3 = 100
  • (4! + (Sin(4-4)*4)!)*4 = 100
Gopalakrishnan Thirumurthy has intelligently played around with cubic root functions. If we were to represent this function with a name like, say, CR(), the following alternative solutions could be achieved:
  • 88 + 8 + CR(8) + CR(8) = 100
  • (8 + CR(8)) ^ (CR(8) * (8 / 8)) = 100
  • ((CR(8) * CR(8) * CR(8)) + CR(8)) ^ CR(8) = 100
  • ((8 * (8 / 8)) + CR(8)) ^ CR(8) = 100
However! If the cubic root function is represented as it should be, ie ³Ö, then there is the extra "3" which invalidates the solution. These alternative solutions using CR(), have therefore been given as possible examples that will not be accepted in the future. Unless, of course, the "3" in ³Ö is actually one of the five digits used in a solution. Same goes with any other root or power. This also applies to any usage of the symbol "%", as it really involves the number 100.
Bala Neerumalla smartly noticed that he can produce number 100 by using a different number base than the normal base 10 (b10). The following example is in base 5 (b5):
  • 444 - 44 = (4 * 5^2 + 4 * 5^1 + 4 * 5^0) - (4 * 5^1 + 4 * 5^0) = 100
However! The above equation mixes up two different number bases: the left hand side is b5, while the right hand side is b10 (100 b5 is equal to 25 b10). This is the reason why we will not be posting anymore solutions that mix different number bases. We will, however, accept solutions that use number bases different from 10, as long as they are consistent on both sides of the equation. Therefore, we will accept:
  • 111 - 11 = 100 [Wolfgang Solfrank pointed out that this works in any number base, thanks!]
Here are some more examples of equations using different number bases:
  • b16 4 ^ 4 +- 4 * (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
  • b16 (2 * 2 * 2 * 2) ^ 2 = 100 [Thanks to Glen Parnell for this one!]
  • b8 4 * 4 * 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
  • b8 (4 ^ 4) / 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
  • b8 4 * 4 * 4 * (4 / 4) = 100 [Thanks to Glen Parnell for this one!]
  • b8 (2 ^ (2 * 2)) * 2 * 2 = 100 [Thanks to Glen Parnell for this one!]
  • b2 11 - 1 + 1 + 1 = 100 [Thanks to Glen Parnell for this one!]
Furthermore, Glen Parnell noticed that there exists a general rule that applies to equations using any number base:
  • bX+1 ((X - X) */ X)! + XX = 100
Here are some examples of it:
  • b16 ((F - F) */ F)! + FF = 100
  • b12 ((B - B) */ B)! + BB = 100
  • b10 ((9 - 9) */ 9)! + 99 = 100 [As already shown by Bala Neerumalla]
  • b8 ((7 - 7) */ 7)! + 77 = 100
  • b4 ((3 - 3) */ 3)! + 33 = 100
  • b2 ((1 - 1) */ 1)! + 11 = 100
Jeff Shall cleverly discovered that he can produce 100 using the Roman numeral 'L', which corresponds to number 50:

  • ((L / L) + (L / L)) * L = ((50 / 50) + (50 / 50)) * 50 = 100
Unashamedly based on the above solution by Jeff Shall, we found that we can also use the Roman numeral 'C', which corresponds to number 100:

  • (C / C) * (C / C) * C = (100 / 100) * (100 / 100) * 100 = 100
  • (C / C) - (C / C) + C = (100 / 100) - (100 / 100) + 100 = 100
  • CCC - CC = 300 - 200 = 100
  • CC - CC + C = 200 - 200 + 100 = 100
Gopalakrishnan Thirumurthy has discovered a lot of new combinations with Roman numerals, including numeral 'X', which is number 10:

  • (L + L) * L ^ (L - L) = (50 + 50) * 50 ^ (50 - 50) = 100
  • (L + L) / L ^ (L - L) = (50 + 50) / 50 ^ (50 - 50) = 100
  • XX * X - X * X = 20 * 10 - 10 * 10 = 100
  • (X + (X / X)) * X - X = (10 + (10 / 10)) * 10 - 10= 100
  • X * X * (X ^ (X - X)) = 10 * 10 * (10 ^ (10 - 10) = 100
  • X * X / (X ^ (X - X)) = 10 * 10 / (10 ^ (10 - 10) = 100
  • C * C * C / C * C = 100 * 100 * 100 / 100 * 100 = 100
  • (C - C) + (C - C) + C = (100 - 100) + (100 - 100) + 100 = 100
  • L * (L - L) + L + L = 50 * (50 - 50) + 50 + 50 = 100
  • C * (CC / C) - C = 100 * (200 / 100) - 100 = 100
  • C * (C / C) + (C - C) = 100 * (100 / 100) + (100 - 100) = 100
Gopalakrishnan Thirumurthy has also noticed that all the solutions involving five times the number 5, could be rewritten by substituting it with Roman numeral 'V'. This is also true if, instead of number 5 and numeral 'V', we were talking about number 1 and Roman numeral 'I'.
Now, how about some solution using Roman numerals 'D' (500), or 'M' (1000)? If you find any alternative solutions, get in touch with us by email!

Easy!

Wednesday, December 24, 2014

A Law-Abiding Citizen

"Where do you think you're going with that thing?" asked the bus driver.

"Where do you think I'm going? On this bus, of course. Why, can't I?" replied the electrician.

"No, of course you can't," said the bus driver in a very patronising way. "It is forbidden to bring any object of length, width, or height greater than one metre on any bus. That thing you're carrying is longer than one metre."

"It's got nothing to do with a ticket," screeched the driver. "You could buy a dozen tickets, and I still would not let you ride on this bus!"

Irritation grew rapidly within the electrician. "Listen! I need to take this neon light tube to a ceremony. I don't have a car. The cabbies are on strike. And it's raining. What do you expect me to do!"

"I don't know, and I don't care anyway. You ain't gonna come on this bus with that tube. End of story."

Quickly, the electrician dashed into a shop next to the bus stop and came out with a package containing the neon tube. Smugly, with all thirty-two teeth on display, he showed the package to the bus driver. "Now can I come on the bus?"

With a snort, the bus driver pulled out a folding rule and performed a precise measurement. Scowling, he waved in the smug commuter.

How did the electrician manage to pack a 1.2 metre neon tube into a package less than one metre?

A Law-Abiding Citizen Puzzle Solution

The electrician packed the tube diagonally into a flat-ish squared package, with sides of less than one metre. More precisely, the sides were about 0.85 metres long, because [squareroot(1.2² / 2) = 0.84852...]

Friday, December 19, 2014

Tamerlano's Trap II

Alvise Moschin, a Venetian merchant, was dragged into the Hazel Room of Samarcanda Palacethe by a pair of soldiers. Although fairly worried, Moschin felt some confidence due to his knowledge of the East. He knew, through tales heard in wine bars, what was waiting for him and how he should react. For a start, he would find himself in front of two doors guarded by two soldiers, a liar and and truth-teller. That would not be a problem.

The street-smart Venetian was thrown onto the rug before a throne. Despite his predicament, he could not contain a grin, which only widened when he saw Tamerlano enter the room and take a seat upon the throne before him.

"Get up, merchant!" barked the conqueror. "There are two doors behind you--"

"Behind one of them there's a horse, and behind the other there are crocodiles, am I right?" interrupted the merchant.

Tamerlano leaned back. "You are smart and well informed, christian," he said. "However, this time we'll have a slight variation. You will not find two guards, but one. He will be the one to whom you may ask the single question. From that, you must decide which door will lead you to certain death and which to freedom. Also, you will not know whether he always lies or always tells the truth."

With his face pale, as if he had seen a ghost, Moschin turned around and saw that between the two doors, there was indeed only one guard. The guard bore a satanic grin, his piercing eyes staring. Moschin approached the guard slowly, his mind working frantically...

What question must Alvise Moschin ask to determine which is the door to freedom?

Tamerlano's Trap II Puzzle Solution

Moschin asked the guard: "If I had asked you which door leads to freedom, which door would you have pointed me to?"

If the guard was truthful, he would have shown the right door. If he lies, he would have again shown the right door, because he would have given the merchant the opposite answer to what he would have given if he was asked a direct question.

Sunday, December 14, 2014

Tamerlano's Trap

"Now be careful!" Tamerlano warned his prisoner. "You can see that in this room, there are two doors guarded by two soldiers. You can tell by their clothes that they come from two different clans. One of the doors leads to a pool of crocodiles; the other one leads to a healthy horse and a sack of gold. To determine which door leads to certain death and which leads to freedom and wealth, you may ask a single question to one of these soldiers. From that answer, you must make your decision. One more warning; one of these soldiers always tells the truth, and the other one always lies."

The prisoner, an intelligent Greek merchant, meditated for a while, bowed to the great conqueror, and with a grin on his face, approached one of the soldiers.

What question would open the door of freedom?

Tamerlano's Trap Puzzle Solution

The merchant asked one of the soldiers, it didn't matter which one: "If I had asked your colleague which door leads to freedom, which of the two doors would he have pointed me to?"

If the interrogated soldier was the one that tells the truth, he would have pointed him to the door that leads to death, because that's the door that the liar would have showed. But even if the same question was asked the liar, the same door (the one that leads to death) would have been the one pointed at, ie the door opposite the one that would have been shown by the truthful soldier.

Once obtained the answer, the merchant went to the door he was NOT pointed at, and enjoyed his freedom.

Tuesday, December 9, 2014

Drama Galore!

It was a beautiful day, perfect for a stroll. After leaving the cars at the edge of the woods, four couples moved towards the river, reaching its bank after a two-mile walk. The restaurant where they intended to dine was on the other side, partially hidden by trees.

But even on a perfectly planned day, evil could come and spoil it. During the stroll, Albert quietly told Amanda that she shouldn't have dressed quite so promiscuously, whilst she replied that he could have done a better job in refraining from making his oh-so-kind compliments to the other three girls. Bernard whispered menacing words at Barbara The Easy Flirt (as he called her then), and Barbara told him that his relationship with the other girls was of dubious morality. Simillar sorts of exchanges happened between Charles and Corinna, and between Douglas and, err, Diana. Reaching the river and seeing the flowing water did little to tame the souls. On the contrary, when the eight friends noticed that instead of the large boat that would carry them all over to the opposite bank, there was only a little boat that would carry no more than two persons at once, the irritation grew to the point that everybody started arguing with everybody else.

The river was about one hundred yards wide, with a small island in the middle. None of the four men were keen on leaving his girlfriend alone with one or more of his other male friends. On the other hand, the women found out that they could only agree on one point: none of their boyfriends should be alone on the boat when one of the girls, excluding his girlfriend, was all alone on any of the two banks or on the island.

Once the tempers calmed, Bernard and Douglas forumlated a plan involving many trips.

How many trips would it take to ferry everyone across whilst still adhering to the wishes of all the people?

 

Notes:
  • In case you haven't guessed, the couples are, Albert and Amanda, Bernard and Barbara, Charles and Corinna, and Douglas and Diana.
  • No woman should stay on one of the banks, on the island, or on the boat, in the company of one or more other men and without her boyfriend
  • No man should be alone on the boat when one of the girls, except his girlfriend, was all alone on one of the two banks or on the island

Drama Galore Puzzle Solution

They needed seventeen trips, frenquently using the island in the middle of the river as temporary destination. If we denote the names of the men with their initial in upper case, and the names of the women with their initial in lower case (by complete chance, the initials of the men are the same as the initials of their girlfriends), we would get the following table:
Trip # Departure bank Direction Island Direction Arrival bank
1 ABCDcd -
=>
ab
2 ABCDbcd
<=
- a
3 ABCDd
=>
bc a
4 ABCDcd
<=
b a
5 CDcd b
=>
ABa
6 BCDcd
<=
b Aa
7 BCD
=>
bcd Aa
8 BCDd
<=
bc Aa
9 Dd bc
=>
ABCa
10 Dd abc
<=
ABC
11 Dd b
=>
ABCac
12 BDd
<=
b ACac
13 d b
=>
ABCDac
14 d bc
<=
ABCDa
15 d -
=>
ABCDabc
16 cd
<=
- ABCDab
17 - -
=>
ABCDabcd

Thursday, December 4, 2014

Hydra


hydra jigsaw puzzle

A team of four girls and six boys put together a 2200-piece jigsaw puzzle in 4 hours. The same jigsaw puzzle was put together in 8 hours by a team of two boys and five girls.

Who are better at putting jigsaw puzzles together, boys or girls?

Hydra Puzzle Solution

If 6 boys and 4 girls spend 4 hours to complete the jigsaw puzzle, then 3 boys and 2 girls will need 8 hours, which is also the amount of time needed by the team of 2 boys and 5 girls. The reader can see that the input of 1 extra boy on the first team equals the input of 3 extra girls on the other team. The conclusion deduced is that the input from 1 boy is worth as much as the input from 3 girls.

To be more precise, it is possible to demonstate that each boy can put together, in one hour, 75 pieces of the puzzle, compared to the 25 for each girl. This is because, if we say that 6 boys and 4 girls (ie team 1) are equal to 22 girls ((6 * 3) + 4), and since that team puts together 550 pieces per hour (2200 / 4), then the workforce of each team member equals 25 pieces/hour (550 / 22).

Sorry ladies about the "non politically correct" nature of this puzzle, but I'm just translating. I was actually thinking of using Martians and Venusians instead, but then I thought "Who cares!" - Mickey.