“We have five extra-curricular groups at school<” a Young Pioneer began. “They are political, literary, photographic, chess, and choral groups. The political group meets every other day, the literary every third day, the photographic every fourth day, the chess every fifth day, and the choral every sixth day. These five groups first met on January 1 and thencefortth meetings were held according to schedule. The question is how many times all five meet on the one and same day in the first quarter (January 1 excluded)?”
“Was it a Leap Year?”
“No.”
“in other words, there were 90 days in the first quarter.”
“Right.”
“Let me add another question,” the professor broke in. “It’s this: how many days were there when none of the groups met in the first quarter?”
“So, there’s a catch to it? There will be no other evening when all the five group meet and no evening when some do not meet. That’s clear!”
“Why?”
“Don’t know. But I have a feeling there is a catch.”
“Comrades!” said the man who had suggested the game. “We won’t reveal the results now. Let’s have more time to think about them. Professor will announce the answers at supper.”
Cheers,
Nalin Pithwa
Mathematics Hothouse 
4 Comments
The LCM of 2,3,4,5,6 is 60 so therefore they will meet again on the 61st day (taking Jan 1 into account.)
Hope it’s not supper time yet.
yes, fine.
Days when nobody met are prime numbers between 6 and 90, and 49 (prime numbers are divisible by 1 and themselves only, and 49 is the square of 7, and the only perfect square that fits the bill) plus 1- the 8th day, 12th day, 14th day and so on.
very good…thanks for participating online 🙂