How many numbers between 1 and 100 can you pick at most, so that none of them divide another?
You can choose fifty numbers at most: 51, 52, 53, … , 100.
In order to see that you cannot choose more than fifty, express each number in the form 2ⁿ×m, where m is an odd number. Since no two numbers can have the same m in their expressions, and there are only fifty odd numbers between 1 and 100, the statement of the problem follows.
I am big on Saturday and Sunday.
I am small on Tuesday, Wednesday, and Thursday.
I am not on Monday and Friday.
What am I?
The answer is the letter S.
White to play and mate Black in 20 moves.
The moves are as follows:
In the three murder cases below, you can read the testimonies of all suspects. For each case, find who the killer is, knowing that no 2 people are in the same row or column, and that the killer was alone in a room with the victim.
The solutions are shown below.
One night, a man received a call from the Police. The Police told the man that his wife was murdered, and that he should get to the crime scene as soon as possible. The man immediately hung up the phone and drove his car for 20 minutes. As soon as he got to the crime scene, the Police arrested him, and he got convicted for murder.
How did the Police know that the man committed the crime?
The police did not tell the man where the crime scene was.
Brain Drop is a new puzzle podcast by Brian Hobbs, released on a (mostly) weekly basis. In each episode, Brian presents 3 new puzzles and shares the solutions of the puzzles from the previous week. He uses professional voicework, music, and sound effects, to set up the mood and make his show more entertaining. Click the banner below to check out Brain Drop and see if you can answer Brian’s latest set of puzzles!
Find all 10-digit numbers with the following property:
Let the number be ABCDEFGHIJ. The number of all digits is 10:
A + B + C + D + E + F + G + H + I + J = 10
Therefore, the sum of all digits is 10. Then:
0×A + 1×B + 2×C + 3×D + 4×E + 5×F + 6×G + 7×H + 8×I + 9×J = 10
We see that F, G, H, I, J < 2. If H = 1, I = 1, or J = 1, then the number contains at least 7 identical digits, clearly 0s. We find A > 6 and E = F = G = 0. It is easy to see that this does not lead to solutions, and then H = I = J = 0.
If G = 1, we get E = F = 0. There is a 6 in the number, so it must be A. We get 6BCD001000, and easily find the solution 6210001000.
If G = 0, F can be 0 or 1. If F = 1, then there must be a 5 in the number, so it must be A. We get 5BCDE10000. We don’t find any solutions.
If F = 0, then the number has at least five 0s, and therefore A > 4. However, since F = G = H = I = J = 0, the number does not have any digits larger than 4, and we get a contradiction.
Thus, the only solution is 6210001000.
There are 5 points in a square 1×1. Show that 2 of the points are within distance 0.75.
Split the unit square into 4 small squares with side lengths 0.5. At least one of these squares will contain 2 of the points. Since the diagonals of the small squares have lengths less than 0.75, these 2 points must be within such distance.
Please confirm you want to block this member.
You will no longer be able to:
Please allow a few minutes for this process to complete.
