Math and Logic Puzzles: Redux

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Post Post #96 (isolation #0) » Tue Oct 29, 2019 12:29 am

Post by vizIIsto »

I've got a fun little maths problem you might want to get your head around... and chances are you probably already know about the puzzle. Welp!

A long time ago lived a very rich trader in the desert who had three sons. The rich trader had a whopping seventeen camels. The trader decided that after his death, his oldest son should inherit half of his camels, the middle one would get a third and the youngest son received a ninth of the seventeen camels.

Your task is to calculate how you're going to divide the camels among the three sons, and tell me how you got there. It'd be very helpful if you didn't have to cut up some of the camels.. :good:
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Post Post #98 (isolation #1) » Tue Oct 29, 2019 12:53 am

Post by vizIIsto »

In post 97, Mitillos wrote:1/2 + 1/3 + 1/9 = 17/18
Yes.
But it's not the answer I want to hear. How will you divide the camels among the sons?
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Post Post #100 (isolation #2) » Tue Oct 29, 2019 1:35 am

Post by vizIIsto »

In post 99, Mitillos wrote:Lend the dead man a camel.
That still doesn't answer the question how you will divide the seventeen camels among the three sons. Give a complete answer with reasoning.
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Post Post #102 (isolation #3) » Tue Oct 29, 2019 6:12 am

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In post 101, Mitillos wrote:Since you insist: You can't. By the time the will takes effect, the trader is dead. It is impossible to change the number of camels, assuming you have provided all pertinent information, thus at least one camel will have to be chopped up.
If we allow silly solutions like lending a dead man a camel, he now has 18 camels, which can be divided as you have requested. The sum I gave shows that one camel will be left over, which can then be returned to the lender.
And that's exactly how to solve the problem.
1/2 of 18 is 9; 1/3 of 18 is 6; 1/9 of 18 is 2. 9 + 6 + 2 = 17. You don't need to 'lend a camel' to solve it, but adding an extra camel is how you will be able to divide the 17 camels fair and square.
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Post Post #108 (isolation #4) » Wed Jan 01, 2020 6:11 am

Post by vizIIsto »

I forgot that this thread existed. I made a maths geeks thread in GD (by the way, check it out!) and posted a problem there which I'm sure fits in perfectly well here. Here's the problem:

This question appeared on an exam. It went as follows:

"The product of the three integers x, y and z is 192.
z = 4, and p is equal to the average of x and y.
What is the minimum value of p?"


The four answers were: a) 6 b) 7 c) 8 d) 9,5

But... No one gave the right answer.
What is the right answer?


To make it a little bit easier in case you struggle to find the right answer, here's three hints. But don't read them unless you really need some help, and one by one!

Spoiler: Hint 1
Nobody had the right answer. Doesn't that seem a little suspicious to you?

Spoiler: Hint 2
You're probably this far that you've solved xyz=192 and z = 4 means x*y=48. Now, try to consider all possible combinations of two integers which multiply to 48.

Spoiler: Hint 3
The reason that no one gave the right answer is because there's something wrong with the question. Or the answers. Or both!
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Post Post #111 (isolation #5) » Wed Jan 01, 2020 7:17 am

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You know, with the Professor Layton puzzle feeling I got by giving three hints, let's go all out by giving you guys a Layton-esque answer to your submissions.

Spoiler: Plotinus w/o hints & Ircher
WRONG.

If this was indeed the right answer, wouldn't it be striking that you would have been the only student to give the right answer?
Read the question carefully, and come to the conclusion that this answer, as obvious as may be, is incorrect.

Spoiler: Plotinus with hints
CORRECT!

No one said that the integers couldn't be negative, right?
This was later figured out by the test makers, who admitted their mistake and gave everybody the full points for the question.
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