Nehushtan’s puzzles

Nehushtan

Nehushtan

New member
Joined
Mar 11, 2013
Messages
25
Nehushtan’s puzzles

#1 (Rebus)

BLCCPRE​


#2 (Cryptic)

Stick insect and mammal. (7,3)​


#3

Let \(\displaystyle x\), \(\displaystyle y\), \(\displaystyle z\) be real numbers such that \(\displaystyle x\leqslant y\leqslant z\). Show that \(\displaystyle xz\leqslant xy+yz\).
 
Nehushtan said:
#3

Let \(\displaystyle x\), \(\displaystyle y\), \(\displaystyle z\) be real numbers such that \(\displaystyle x\leqslant y\leqslant z\). Show that \(\displaystyle xz\leqslant xy+yz\).
Click to expand...


Let \(\displaystyle \ k_1, \ k_2 \ \) be real numbers such that \(\displaystyle \ k_1, \ k_2 \ge 0.\)

Let \(\displaystyle \ y = x + k_1.\) *

Let \(\displaystyle \ z = y + k_2.\)

Then it follows that \(\displaystyle \ z = x + k_1 + k_2.\) **

Substitute * and ** into the inequality \(\displaystyle \ \ xz \ \le \ xy \ + \ yz \ \):

\(\displaystyle x(x + k_1 + k_2) \ \le \ ? \ \ x(x + k_1) \ + \ (x + k_1)(x + k_1 + k_2) \)

\(\displaystyle x^2 + k_1x + k_2x \ \le \ ? \ \ x^2 + k_1x + x^2 + k_1x + k_2x + k_1x + (k_1)^2 + k_1k_2\)

\(\displaystyle x^2 + k_1x + k_2x \ \le \ ? \ \ 2x^2 + 3k_1x + k_2x + (k_1)^2 + k_1k_2\)

\(\displaystyle 0 \ \le \ ? \ \ x^2 + 2k_1x + (k_1)^2 + k_1k_2\)

\(\displaystyle 0 \le \ ? \ \ (x + k_1)^2 + k_1k_2 \)

This is true, because neither of the two summands on the right-hand side of the inequality can ever be negative.

Therefore, the original inequality (with its constraints for x, y, and z) is true.
 
Sorry. Instead of giving the number of letters, I’ll just state that the answer to Puzzle #1 is a fictional character. :)

The answer to Puzzle #2 is correct – well done.

The solution to Puzzle #3 looks correct (I’m too lazy to check the details – sorry :p) but the solution I had in mind was

\(\displaystyle x\,\leqslant\,y\,\leqslant\,z\)​

\(\displaystyle \implies\ (y-x)(z-y)\,\geqslant\,0\)

\(\displaystyle \implies\ xy+yz-xz\,\geqslant\,y^2\,\geqslant\,0\)​
 
Last edited:
Maybe I should have presented Puzzle #1 like this: BLC CPRE. :roll:

While you’re thinking about it, here’s my next puzzle.

#4

You have \(\displaystyle N\) balls, all identical in appearance except that one is much heavier than the others. What is the minimum number of times you need to weigh them on a scale balance to find the heavy one?
 
Nehushtan said:


#4

You have \(\displaystyle N\) balls, all identical in appearance except that one is much heavier than the others. What is the minimum number of times you need to weigh them on a scale balance to find the heavy one?
Click to expand...

Ceiling of N/3
 
Nehushtan said:
Maybe I should have presented Puzzle #1 like this: BLC CPRE. :roll:

While you’re thinking about it, here’s my next puzzle.

#4

You have \(\displaystyle N\) balls, all identical in appearance except that one is much heavier than the others. What is the minimum number of times you need to weigh them on a scale balance to find the heavy one?
Click to expand...
N must be an integer > 1, or the problem makes no sense.

\(\displaystyle Let\ W = the\ number\ of\ weighings\ required.\)

\(\displaystyle M\ an\ integer \ge 1\ and\ 2^M \le N < 2^{(M+1)} \implies 1 \le W \le M.\)

Procedure. If N is even, weigh N/2 against N/2. If one pan sinks, that pan contains the heavier ball. If N is odd, weigh (N - 1)/2 against
(N - 1)/2. If one pan sinks, that pan contains the heavier ball, but if the pans balance, the heavier ball is the one not weighed. Repeat as many times as needed. Because the number remaining to be tested is reduced by at least half on each test, the number of tests required cannot be as high as M + 1.

Edit: The problem is not that hard for those who have coded binary searches. The problem would be harder if it was not specified whether the ball that differs from the others is heavier or lighter. Since it is known that, with twelve balls, you can find the odd ball in at most three weighings, the answer may be the same as for this problem, but it makes my head ache to think of designing the algorithm to solve that for general N.
 
Last edited:
Denis said:
Don't lose any sleep over this, Jeff; solution all over the joint!
A very old puzzle, originally 13 coins with 1 heavier.
Googling "13 coins one heavier" gives 14.2 million hits !!
Click to expand...
Well the old puzzle that I was referring to did not specify whether the odd ball is heavier or light. The answer is that it takes 3 weighings with 12 balls, and the procedure is a bit convoluted. I was saying that I did not want to ponder the general solution for any N, where N - 1 are the same and is not known whether the odd one is heavier or lighter, but of course the moment I said that I started to ponder.
 
Nehushtan said:


While you’re thinking about it, here’s my next puzzle.

#4

You have \(\displaystyle N\) balls, all identical in appearance except that one is > > > much heavier < < < than the others.
Click to expand...
Well, if one of the balls is "much heavier," then you can tell by picking them up in your hands and not even use a balance! The problem should not have the phrase "much heavier" in it. I could have one of the balls be "much heavier" to the extent that it weighs more than the other (N - 1) balls combined. I call the phrase "much heavier" sufficiently ambiguous.
 
Last edited:
Sorry, I haven’t been back for a while. I may as well give the solutions to the unsolved puzzles now in case I don’t come back again. :rolleyes:


#4
lookagain said:
Well, if one of the balls is "much heavier," then you can tell by picking them up in your hands and not even use a balance! The problem should not have the phrase "much heavier" in it. I could have one of the balls be "much heavier" to the extent that it weighs more than the other (N - 1) balls combined. I call the phrase "much heavier" sufficiently ambiguous.
Click to expand...
Why do you sound so skeptical? You’ve got it – congratulations! :D The phrase “much heavier” is in fact the key to the puzzle: if it’s much heavier you should be able to tell the difference in weight just by holding the balls in your hand. Besides, I did not say you have to use a scale balance. All I said was: “What is the minimum number of times you need to weigh them on a scale balance to find the heavy one?” Since you can pick out the odd one out by feeling them, you don’t need to use the instrument. Therefore the answer to the puzzle is: 0 times.


#1
Half-Blood Prince (C = half O)
 
Last edited:
Nehushtan said:
#4
Why do you sound so skeptical? ** \(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \)You’ve got it – congratulations! :D The phrase “much heavier” is in fact the key to the puzzle: if it’s much heavier you should be able to tell the difference in weight just by holding the balls in your hand. Besides, I did not say you have to use a scale balance. All I said was: “What is the minimum number of times you need to weigh them on a scale balance to find the heavy one?” Since you can pick out the odd one out by feeling them, you don’t need to use the instrument. Therefore the answer to the puzzle is: 0 times.

Half-Blood Prince (C = half O)
Click to expand...
** That isn't a good puzzle as stated, because, again, the phrase "much heavier" is ambiguous! Good puzzles don't have ambiguity of meanings of words and phrases in them. Now, certain ones may be tricky and/or mislead the solver by having the solver assume certain aspects. However, that is different from making statements that are ambiguous. For future sake, any problem/puzzle submitter should not use subjective phrasing.
 
Last edited:
Hello, Nehushtan!

I have a question . . .

Maybe I should have presented Puzzle #1 like this: BLC CPRE
Click to expand...

**

Is the second word PRINCE (PR in CE)?

**
 
Hey, I’m back! :mrgreen:

My next puzzle is a cryptic clue. :cool:

#5

Some people are vocally unhappy with items tailor-made for them? (8)
 
You must log in or register to reply here.
Top