An infinite number of mathematicians are standing behind a bar. The first asks the barman for half a pint of beer, the second for a quarter pint, the third an eighth, and so on. How many pints of beer will the barman need to fulfill all mathematicians' wishes?
You are playing a game with your friend Jack. There are digits from 1 to 9. You both will take turn erasing one digit and adding it to your score. The first one to score 15 points will win the game.
I am thinking of a five-digit number such that:
The first and last digits are the same, their submission is an even number and multiplication is an odd number and is equal to the fourth number. Subtract five from it and we obtain the second number. Then divide into exact halves and we get the 3rd number.
I have thought of a number that is made up by using all the ten digits just once. Here are a few clues for you to guess my number:
First digits is divisible by 1.
First two digits are divisible by 2.
First three digits are divisible by 3.
First four digits are divisible by 4.
First five digits are divisible by 5.
First six digits are divisible by 6.
First seven digits are divisible by 7.
First eight digits are divisible by 8.
First nine digits are divisible by 9.
The number is divisible by 10.
There are 100 bulbs in a room. 100 strangers have been accumulated in the adjacent room. The first one goes and lights up every bulb. The second one goes and switches off all the even-numbered bulbs - second, fourth, sixth... and so on. The third one goes and reverses the current position of every third bulb (third, sixth, ninth? and so on.) i.e. if the bulb is lit, he switches it off and if the bulb is off, he switches it on. All the 100 strangers progress similarly.
After the last person has done what he wanted, which bulbs will be lit and which ones will be switched off?
