#answers to riddles

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# Math Riddles

##### Riddle 8

**Three Brothers on a Farm**

Three brothers live in a farm. They agreed to buy new seeds: Adam and Ben would go and Charlie stayed to protect fields. Ben bought 75 sacks of wheat in the market whereas Adam bought 45 sacks. At home, they split the sacks equally. Charlie had paid 1400 dollars for the wheat. How much dollars did Ben and Adam get of the sum, considering equal split of the sacks?

Every farmer’s part is 1/3(45+75) = 40 sacks.

Charlie paid $1400 for 40 sacks, then 1 sack costs $1400/40 = $35/sack.

Adam got $35*(45-40)=35*5 = $175.

Ben got $35*(75-40)=35*35 = $1225. **Answer:** Ben $1225, Adam $175

##### Riddle 9

**The Insurance Salesman**

An insurance salesman walk up to house and knocks on the door. A woman answers, and he asks her how many children she has and how old they are. She says I will give you a hint. If you multiply the 3 children’s ages, you get 36. He says this is not enough information. So she gives a him 2 nd hint. If you add up the children’s ages, the sum is the number on the house next door. He goes next door and looks at the house number and says this is still not enough information. So she says she’ll give him one last hint which is that her oldest of the 3 plays piano.

Why would he need to go back to get the last hint after seeing the number on the house next door?

Because the sum of their ages ( the number on the house) is ambiguous and could refer to more than 1 trio of factors.

Answer to Riddle

If you list out the trio of factors that multiply to 36 and their sums, you get.

- 1 1 36 = 38
- 1 2 18 = 21
- 1 3 12 = 16
- 1 4 9 = 14
- 6 6 1 = 13
- 2 2 9 = 13
- 2 3 6 = 11
- 3 3 4 = 10

Since the number on the house next door is not enough information there must be more than 1 factor trio that sums up to it, leaving two possibilities: < 6, 6, 1>. <2, 2, 9>. When she says her ‘oldest’ you know it can *not* be <6,6,1> since she would have two ‘older’ sons not an ‘oldest’.

##### Riddle 10

This is a famous one. The classic Monty hall riddle!

You are confronted by 3 doors. Behind one of them is a car, behind the two others, you will only see a goat. Now, if you correctly pick the car, you win the car. Otherwise, if you get one of the 2 goats, you don’t get the car.

So, pick any door. It doesn’t matter which one, but we will suppose that you picked door #2, as an example.

Now, after you have picked a door and *before* finding out what is actually behind it, you are shown a goat behind one of the other doors.(Remember there has to be a goat in 1 of the doors that you *have not* picked. )

Let’s say you choose door #2, as shown above. For example’s sake, let’s say there’s a goat in door 1. The question and the riddle is. should you switch the door that you picked? In other words, in this example, should you now choose door 3? Or, should you stick with your first choice (door #2)?

There actually is a mathematically correct answer to this riddle: You should indeed *change your choice*. If you don’t believe me, just try out our free online Monty hall simulation .

##### Riddle 11

If 9999 = 4, 8888 = 8, 1816 = 6, 1212 = 0, then 1919 =

Look at how many *closed areas* there are.

- 9999 has 4 closed areas (the top of the ‘9’)
- 8888 has 8 closed areas, the top and bottom parts of the 8 and there are no other digits
- 1816 has 3 closed areas, (top and bottom of 8 and bottom of 6, and it has 2 other digits ( 3*2=6)
- 1212 has 0 closed areas,(0*4=0)

##### Riddle 12

An athlete is able to jump FOREVER. However, everytime that she jumps she gets a bit more tired, and *every jump* goes $$ \frac 1 2 $$ as far as her prior jump. Now, for her very first jump, she goes $$ \frac 1 2 $$ of a foot.

On her second jump, she goes $$ \frac 1 4 $$ of a foot, and so on and so forth. The beginning of her journey is shown in the gif below.

How many jumps does it take for her to travel 1 foot?

She will **never** get to the 1 foot mark because you keep adding smaller and smaller amounts!

### Other Good Riddles

Quick summary of riddle: Zeno of Elea (490-425 BC) is known for creating many paradoxes which were debated by mathematicians for centuries. His riddle involving Achilles, the character from Homer’s *Iliad* and a tortoise went something like:

The tortoise challenged Achilles to a race and Achilles, full of typical hubris, accepted and even gave the Tortoise a 10 foot head start. Before the race started, the tortoise told Achilles that the reason Achilles would lose is that even though Achilles would be catching up, the tortoise would always be moving ahead. Therefore, Achilles would always be covering a fraction of the distance between the two. Let’s say he covered half of the distance in 1 second (5 feet) and then in the next he covered half of the **new** distance, the remaining 5 feet plus the Tortoise’s new distance. In the end, the tortoise convinced Achilles that he could not win the race because although he would be getting closer and closer, he would still always be covering smaller and smaller fractions of the total distance between the two. Therefore, Achilles forfeited the race. To read this riddle in a modern narrative form click here .