New answers tagged number-theory
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.
1
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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