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?
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 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.
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To determine the number of opportunities the parliament members have to pass new laws during the four-year term, we need to calculate the number of times both houses meet on a Monday.
First, let's determine how many times the first house meets in a four-year term. Since the first house meets every ten days, we divide the total number of days in four years by ten:
4 years * 365 days/year = 1460 days
1460 days / 10 days/meeting = 146 meetings
Therefore, the first house will meet 146 times during the four-year term.
Next, let's determine how many times the second house meets in a four-year term. Since the second house meets every twelve days, we divide the total number of days in four years by twelve:
4 years * 365 days/year = 1460 days
1460 days / 12 days/meeting = 121.67 meetings
However, we need to consider that the number of meetings should be rounded down to account for integers. Therefore, the second house will meet 121 times during the four-year term.
Now, we need to find the number of times both houses meet on a Monday. We divide the total number of meetings by the least common multiple (LCM) of 10 and 12, which is 60:
Total number of meetings = 146 (first house) + 121 (second house) = 267 meetings
267 meetings / LCM(10, 12) = 267 meetings / 60 = 4.45
However, since we cannot have fractional meetings, we need to round down the number to the nearest whole number. Therefore, both houses will meet on a Monday a total of 4 times during the four-year term.
Therefore, the parliament members will have 4 opportunities to pass new laws during this four-year term when both houses meet on a Monday.