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4

As JMP pointed out, this is a modified It appears that the professor has simply forgotten to cross the minus symbol for this to work correctly.


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8

It is: For example:


12

It appears to me this is Because And


-1

If I understand the question correctly, at the end of the 15th day has consumed 7.5 pills because he is allowed half a pill a day. Now, for the remaining 15 days he needs another 7.5 pills. On the 16th day, assuming in package are 5 whole pills and 5 half pills, on that day he has an equal chance of picking a whole or half pill.


0

Put the three bags on the scale. Take one coin from one bag and put it on your right side, put a mark on that bag, take one more coin from a different bag and put it on your left side. By so doing you know which bag contains the fake coins.


2

Starting at $(0,0)$, we do the followings: If a whole pill is taken, we move one unit to the right. If a half pill is taken, we move one unit upwards. Therefore, if we are at point $(x,y)$ we have: $(15-x)$ whole pills left $(x-y)$ half pills left $(15-y)$ total pills left Now we define $f(x,y)$ as the probability that we reach point $(x,y)$. Obviously, ...


0

Colombo’s method works like this: Label the bag with number 1, 2, and 3. Take $i$ coin(s) from bag #$i$ and put them all on a scale. You will get $6 lb + k oz$. Here bag #$k$ is the one containing fake coins. This method only works if there is only one bag containing fake coins. My method will be: Label the bags with number 0, 1, 2. Take $2^{i}$ coin(s) ...


2

Despite the minuses, this was a really nice puzzle. Here is the first set Here is the second set


0

Another one of those puzzles you can brute force, and therefore would have a coding answer. The answer is: Try online Brute force code: function permut(string) { if (string.length < 2) return string; // This is our break condition var permutations = []; // This array will hold our permutations for (var i = 0; i < string.length; i++) { ...


0

If we set x=9 then we have x*(x-4)*(x-8)=45 (x-1)(x-5)(x-7)=64 (x-2)(x-3)(x-6)=126 so I obtain 951*842*763=610966146 My opinion is that if we obtain minimum results from the multiplications of monomials, then we will obtain the maximum product of the three 3-digit numbers.


3

The answer is as @avi already mentioned. But here is my justification: The conclusion is that Now we know that the 9 digits are splitted in 3 groups It now remains to form the three factors by picking one digit in each group. Using a similar reasoning As a result, the three factors formed for the three groups are such that


2

We can calculate: $ (a*100 + b*10 + c)*(d*100 + e*10 + f)*(g*100 + h*10 + i)= \\ = adg1000000 + (adh+aeg+bdg)100000 + (adi+aeh+bdh+afg+beg+cdg)10000 + (aei+bdi+afh+beh+cdh+bfg+ceg)1000 + (afi+bei+cdi+bfh+ceh+cfg)100 + (bfi+cei+cfh)10 + cfi $ If we now maximise coefficients in order: $max(adg)$ a = 9 d = 8 g = 7 $max(adh+aeg+bdg) = max(72h+63e+56b)$ h ...


0

By simply trying out a couple of combinations, I found: Some of the other combinations I tried are: My methodology: Edit: FYI, I did the calculations by hand, rather than with a program, so I didn't try all options.


15

The answer is: Here's the intuition: Finally, here's


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