Difficult riddles

logic

Unusual paragraph

This is an unusual paragraph. I’m curious as to just how quickly you can find out what is so unusual about it. It looks so ordinary and plain that you would think nothing was wrong with it. In fact, nothing is wrong with it! It is highly unusual though. Study it and think about it, but you still may not find anything odd. But if you work at it a bit, you might find out. Try to do so without any coaching.
The letter "e", which is the most common letter in the English language, does not appear once in the long paragraph.
86.92 %
54 votes

cleaninterviewmath

Euro 2012 Football

What should be the value of the bottom row in our. Euro 2012 Football themed grid?
115. Each shirt is worth 40, each ball is worth 25 and each scarf is worth 10
86.81 %
31 votes

logicshort

Sequence of numbers

Finish the sequence: 7 8 5 5 3 4 4 ?
6 - the number of letters in the month august; (January has 7 letters, February has 8 etc.)
86.62 %
42 votes

logicmath

Camel and Banana

The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels. What is the most bananas you can bring over to your destination?
First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A. So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again. Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between. When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward. In the first part, P1, to shift the bananas by 1Km, the Camel will have to Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana Note: After point 5 the Camel does not need to return to point A again. So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas. After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas. Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km. Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Note: After point 3 the Camel does not need to return to the starting point of P2. So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas. After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas. The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example. Hence the length of part P2 is 333 Km. Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas. He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.
85.53 %
58 votes

cleanlogicwhat am I

Silver-tongued, but never lie

I’m teary-eyed but never cry, silver-tongued, but never lie. double-winged, but never fly, air-cooled, but never dry. What am I?
Mercury. The element looks shiny, silver, and is wet. The god Mercury has two wings but only uses them to run.
84.83 %
64 votes

logic

Three Childrens' Ages

A deliveryman comes to a house to drop off a package. He asks the woman who lives there how many children she has. "Three," she says. "And I bet you can't guess their ages." "Ok, give me a hint," the deliveryman says. "Well, if you multiply their ages together, you get 36," she says. "And if you add their ages together, the sum is equal to our house number." The deliveryman looks at the house number nailed to the front of her house. "I need another hint," he says. The woman thinks for a moment. "My youngest son will have a lot to learn from his older brothers," she says. The deliveryman's eyes light up and he tells her the ages of her three children. What are their ages?
Their ages are 1, 6, and 6. We can figure this out as follows: Given that their ages multiply out to 36, the possible ages for the children are: 1, 1, 36 (sum = 38) 1, 2, 18 (sum = 21) 1, 3, 12 (sum = 16) 1, 4, 9 (sum = 14) 1, 6, 6 (sum = 13) 2, 2, 9 (sum = 13) 2, 3, 6 (sum = 11) 3, 3, 4 (sum = 10) When the woman tells the deliveryman that the children's ages add up to her street number, he still doesn't know their ages. The only way this could happen is that there is more than one possible way for the children's ages to add up to the number on the house (or else he would have known their ages when he looked at the house number). Looking back at the possible values for the children's ages, you can see that there is only one situation in which there are multiple possible values for the children's ages that add up to the same sum, and that is if their ages are either 1, 6, and 6 (sums up to 13), or 2, 2, and 9 (also sums up to 13). So these are now the only possible values for their ages. When the woman then tells him that her youngest son has two older brothers (who we can tell are clearly a number of years older), the only possible situation is that their ages are 1, 6, and 6.
83.96 %
52 votes

logicmathshort

Really hard algebra puzzle

2+3=8, 3+7=27, 4+5=32, 5+8=60, 6+7=72, 7+8=? Solve it?
98 2+3=2*[3+(2-1)]=8 3+7=3*[7+(3-1)]=27 4+5=4*[5+(4-1)]=32 5+8=5*[8+(5-1)]=60 6+7=6*[7+(6-1)]=72 therefore 7+8=7*[8+(7-1)]=98 x+y=x[y+(x-1)]=x^2+xy-x
80.86 %
43 votes

logicshort

The smallest number

Which is the smallest number that you can write using all the vowels exactly once?
fIvE thOUsAnd (5000).
80.62 %
56 votes