Can You Solve These Math Problems?
Adults like to pretend they know everything. But when it comes to helping a child with homework, not many decide to do it. Of course, those adults who studied well in school will surely solve any math problems. But are there many of them? Let’s see.
Today we’re going to share with you two different, but very interesting math problems that can be solved only by those who didn’t waste time at school.
Can You Solve These Math Problems?
The First Problem
This seemingly simple math problem has caused a whole discussion on the Internet. This is because the answers were different for different readers. Some were convinced that the correct answer was 5. Some were convinced that the result should be one. Others insisted that 7 was the correct answer. Who is right? Or is the answer completely different?
The Second Problem
If you managed to easily deal with the first one, then we offer a much more complicated problem.
Once upon a time, a knight was captured by Sultan Saladin. The ruler announced that he would free the captive and his horse if he received a ransom of 30 thousand gold coins. The knight had neither money nor rich relatives, so he decided to cheat. The man said: “O mighty Saladin, you give me no hope. In my homeland, every smart captive is given the chance to be free. He is offered a riddle. If he solves it, he becomes free, if he doesn’t, the ransom amount is doubled.”
Saladin loved riddles, so he liked the offer: “Okay, so be it. Here’s your riddle. Tomorrow morning you will be given twelve identical-looking coins and scales. One coin will be counterfeit. But nobody knows if it is lighter or heavier than others. You will have three tries to find the fake.”
How to find a false coin among 12 coins in 3 tries? Is it even possible to do this?© Depositphotos
The Answer to the First Problem
To solve the first problem, you need to remember the rule that first we perform multiplication and division, and only then addition and subtraction. If so, our problem will look like this:
6 – 1 * 0 + 2/2 = 6 – 0 + 1 = 7.
The second problem will take a longer time. We also don’t know which coin is heavier: the fake or genuine one.© Depositphotos
The Answer to the Second Problem
- Divide the coins into 3 batches of 4 coins. Weigh the first two on the scales. If they weigh the same, then we are lucky, and the fake coin is in the third batch.
- Then we weigh two coins (any coin from the first and second batch) and two coins from the last one. If the batches weigh the same, the counterfeit coin is one of the remaining two.
- Therefore, then we weigh one of the genuine coins with any of the two remaining ones. If they weigh the same, the fake coin is the one we didn’t touch. If the weight is different, then you already know which one is the fake.
- So, if the weight is different during the second try, then a pair with the fake is also found. © Depositphotos
- If at the first try, one group is heavier, then the fake coin is among these 8 coins. In this case, name all the coins: the first batch (heavy) — 1, 2, 3, 4; the second (light) — 5, 6, 7, 8; the third (genuine ones) — 9, 10, 11, 12.
- During the second try, weigh the coins 1, 9, 10, 11, and 2, 3, 4, 5. If the batches weigh the same, then the fake is 6, 7, or 8. Moreover, after the first try, we have already learned that the second batch is lighter. This means that the fake is lighter than the original. Therefore, we weigh 6 and 7. The one that is lighter is fake. If they weigh the same, then the counterfeit coin is number 8.
- If after the second try we noticed that batch 1, 9, 10, 11 is heavier than batch 2, 3, 4, 5, then the fake coin is either 1 (heavier) or 5 (lighter). It is enough to weigh, for example, coin number 1 with any genuine coin. If number 1 is heavier, then it is fake. If they weigh the same, then the fake one is coin number 5.
- So, if batch 2, 3, 4, 5 turns out to be heavier, the counterfeit coin is heavier, and it is among 2, 3, or 4. In this case, weigh, for example, 2 and 3. The fake one will be heavier. If they weigh the same, coin number 4 is fake.
Did you solve both math problems? The last riddle was not a simple one. Of course, it could be simplified by changing the conditions, but would that be fair? We hope the knight also managed to solve it because he certainly had more motivation. Or maybe there is an easier way, which he found?