Math riddles

cleanlogicmathsimple

Create a number using only the digits 4,4,3,3,2,2,1 and 1. So I can only be eight digits. You have to make sure the ones are separated by one digit, the twos are separated by two digits the threes are separated with three digits and the fours are separated by four digits.
41312432.
86.32 %
41 votes
logicmathclean

Consider the following explanation for why 1=2: 1. Start out Let y = x 2. Multiply through by x xy = x2 3. Subtract y2 from each side xy - y2 = x2 - y2 4. Factor each side y(x-y) = (x+y)(x-y) 5. Divide both sides by (x-y) y = x+y 6. Divide both sides by y y/y = x/y + y/y 7. And so... 1 = x/y + 1 8. Since x=y, x/y = 1 1 = 1 + 1 8. And so... 1 = 2 How is this possible?
Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not followinng basic mathematical rules, we are able to get strange results like these.
86.32 %
41 votes
logiccleanclevermath

At a dinner party, many of the guests exchange greetings by shaking hands with each other while they wait for the host to finish cooking. After all this handshaking, the host, who didn't take part in or see any of the handshaking, gets everybody's attention and says: "I know for a fact that at least two people at this party shook the same number of other people's hands." How could the host know this? Note that nobody shakes his or her own hand.
Assume there are N people at the party. Note that the least number of people that someone could shake hands with is 0, and the most someone could shake hands with is N-1 (which would mean that they shook hands with every other person). Now, if everyone at the party really were to have shaken hands with a different number of people, then that means somone must have shaken hands with 0 people, someone must have shaken hands with 1 person, and so on, all the way up to someone who must have shaken hands with N-1 people. This is the only possible scenario, since there are N people at the party and N different numbers of possible people to shake hands with (all the numbers between 0 and N-1 inclusive). But this situation isn't possible, because there can't be both a person who shook hands with 0 people (call him Person 0) and a person who shook hands with N-1 people (call him Person N-1). This is because Person 0 shook hands with nobody (and thus didn't shake hands with Person N-1), but Person N-1 shook hands with everybody (and thus did shake hands with Person 0). This is clearly a contradiction, and thus two of the people at the party must have shaken hands with the same number of people. Pretend there were only 2 guests at the party. Then try 3, and 4, and so on. This should help you think about the problem. Search: Pigeonhole principle
86.32 %
41 votes
logicmath

A clock chimes 5 times in 4 seconds. How many times will it chime in 10 seconds?
11 times. It chimes at zero and then once every second for 10 seconds.
86.32 %
41 votes
logicmathclean

You are visiting NYC when a man approaches you. "Not counting bald people, I bet a hundred bucks that there are two people living in New York City with the same number of hairs on their heads," he tells you. "I'll take that bet!" you say. You talk to the man for a minute, after which you realize you have lost the bet. What did the man say to prove his case?
This is a classic example of the pigeonhole principle. The argument goes as follows: assume that every non-bald person in New York City has a different number of hairs on their head. Since there are about 9 million people living in NYC, let's say 8 million of them aren't bald. So 8 million people need to have different numbers of hairs on their head. But on average, people only have about 100,000 hairs. So even if there was someone with 1 hair, someone with 2 hairs, someone with 3 hairs, and so on, all the way up to someone with 100,000 hairs, there are still 7,900,000 other people who all need different numbers of hairs on their heads, and furthermore, who all need MORE than 100,000 hairs on their head. You can see that additionally, at least one person would need to have at least 8,000,000 hairs on their head, because there's no way to have 8,000,000 people all have different numbers of hairs between 1 and 7,999,999. But someone having 8,000,000 is an essential impossibility (as is even having 1,000,000 hairs), So there's no way this situation could be the case, where everyone has a different number of hairs. Which means that at least two people have the same number of hairs.
86.20 %
51 votes
logicmathsimpleclean

How to measure exactly 4 gallon of water from 3 gallon and 5 gallon jars, given, you have unlimited water supply from a running tap.
Step 1. Fill 3 gallon jar with water. ( 5p – 0, 3p – 3) Step 2. Pour all its water into 5 gallon jar. (5p – 3, 3p – 0) Step 3. Fill 3 gallon jar again. ( 5p – 3, 3p – 3) Step 4. Pour its water into 5 gallon jar untill it is full. Now you will have exactly 1 gallon water remaining in 3 gallon jar. (5p – 5, 3p – 1) Step 5. Empty 5 gallon jar, pour 1 gallon water from 3 gallon jar into it. Now 5 gallon jar has exactly 1 gallon of water. (5p – 1, 3p – 0) Step 6. Fill 3 gallon jar again and pour all its water into 5 gallon jar, thus 5 gallon jar will have exactly 4 gallon of water. (5p – 4, 3p – 0)
85.98 %
60 votes
logicmath

If, Fernando + Alonso + McLaren = 6 Fernando x Alonso = 2 Alonso x McLaren = 6 Then, McLaren x Fernando = ?
3 or 0.75 Explanation: Rewriting the last 2 equations in terms of Alonso, Fernando = 2/Alonso McLaren = 6/Alonso Replacing above values in equation "Fernando + Alonso + McLaren = 6" 2/Alonso + Alonso + 6/Alonso =6 (2 + Alonso^2 + 6)/Alonso = 6 8 + Alonso^2 = 6Alonso Alonso^2 - 6Alonso + 8 = 0 (Alonso - 4) (Alonso - 2) = 0 Therefore; Alonso = 4 or 2 Let's take value of Alonso as 2 Fernando = 2/2 = 1 McLaren = 6/2 = 3 Therefore; McLaren x Fernando = 3 x 1 = 3 Let's take value of Alonso as 4 Fernando = 2/4 = 0.5 McLaren = 6/4 = 1.5 Therefore; McLaren x Fernando = 1.5 x 0.5 = 0.75
85.94 %
50 votes
logiccleanclevermathstory

A man told his son that he would give him $1000 if he could accomplish the following task. The father gave his son ten envelopes and a thousand dollars, all in one dollar bills. He told his son, "Place the money in the envelopes in such a manner that no matter what number of dollars I ask for, you can give me one or more of the envelopes, containing the exact amount I asked for without having to open any of the envelopes. If you can do this, you will keep the $1000." When the father asked for a sum of money, the son was able to give him envelopes containing the exact amount of money asked for. How did the son distribute the money among the ten envelopes?
The contents or the ten envelopes (in dollar bills) hould be as follows: $1, 2, 4, 8, 16, 32, 64, 128, 256, 489. The first nine numbers are in geometrical progression, and their sum, deducted from 1,000, gives the contents of the tenth envelope.
85.94 %
50 votes
logicmath

Think of a number. Double it. Add ten. Half it. Take away the number you started with. What is your number?
Your number is 5.
85.67 %
39 votes
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