first of all, find the number of days in 4 years, which would be 3x365 + 366 (one leap year)
the days that a law can be passed has to be divisible by 10, 12 and 7
the lowest common multiple would be 420
How many multiples of 420 are there in your total number of days?
The parliament of the land of Archronia consists of two houses. The Parliament was elected in 1995 for a period of four years beginning on Monday, January 1st, 1996, when the two houses had their first sessions. According to the rules, the meetings of the first house must occur every ten days for the duration of the term, and the meetings of the second house must occur every twelve days. For example, the second meetings of the first and second houses were held on the 11th and 13th of January respectively. A new law can be passed only when both houses meet on a Monday. How many opportunities will the parliament members have to pass new laws during this four year term?
4 answers
The number of days after January 1, 1996 must be divisible by 10, 12 and 7. It must also be less than 365x4 = 1460.
The Lowest Common Multiple of those numbers is 2^2*3*5*7 = 420 (i.e, they can pass a law every 60 weeks.)The number of opportunitites is the integer part of 1460/420, which is three.
The Lowest Common Multiple of those numbers is 2^2*3*5*7 = 420 (i.e, they can pass a law every 60 weeks.)The number of opportunitites is the integer part of 1460/420, which is three.
3, thanx
thank-you very much
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