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June 26, 2022, 12:28:08 pm

Author Topic: Tough question from a past sac paper.  (Read 1966 times)  Share 

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Tough question from a past sac paper.
« on: February 19, 2022, 05:36:06 pm »
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This question was in an old sac paper from my school. I tried using simultaneous equations but my solutions didn't work as the dates would occur over two months. I asked the teacher for the solution and he said "he didn't know".

Any ideas?

fun_jirachi

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Re: Tough question from a past sac paper.
« Reply #1 on: February 19, 2022, 06:51:17 pm »
+1
A particular year Y between 1 and X will correspond with \(\frac{365Y}{X}\) ie. on a converted calendar from 1 to X 1485 AD would be \(\frac{365 \times 1485}{X}\).

You also want (\frac{365 \times 2003}{X}\) to be in the same month -- it may be helpful to look at the number of days in each month on a regular calendar, and the index of the first day of any particular month (ie. 1st March is the 60th day of the year on a 365-day calendar) and see if it's at all feasible to have 2003 end up before the last day of that same month.

You can do a bit of algebra to link these two ideas up. Afterwards, see what you can do to optimise the year you just got (this only forms a loose upper bound). -- don't want to give too much of the problem away since this is the sort of problem that will extend you that extra bit.
« Last Edit: February 19, 2022, 06:56:40 pm by fun_jirachi »
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Re: Tough question from a past sac paper.
« Reply #2 on: February 19, 2022, 07:29:10 pm »
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A particular year Y between 1 and X will correspond with \(\frac{365Y}{X}\) ie. on a converted calendar from 1 to X 1485 AD would be \(\frac{365 \times 1485}{X}\).

You also want (\frac{365 \times 2003}{X}\) to be in the same month -- it may be helpful to look at the number of days in each month on a regular calendar, and the index of the first day of any particular month (ie. 1st March is the 60th day of the year on a 365-day calendar) and see if it's at all feasible to have 2003 end up before the last day of that same month.

You can do a bit of algebra to link these two ideas up. Afterwards, see what you can do to optimise the year you just got (this only forms a loose upper bound). -- don't want to give too much of the problem away since this is the sort of problem that will extend you that extra bit.

Okay, I got an answer by solving:

(365x2003/Y) = 31 for Y, and got Y=23583.7097, or AD 23,584. Is this correct?

Either way, could you provide the formal solution? Since I don't how it couldn't be possible that it occurs in another month, not just January.


fun_jirachi

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Re: Tough question from a past sac paper.
« Reply #3 on: February 19, 2022, 11:02:07 pm »
+2
I realised almost as soon as I posted you didn't actually have to establish an upper bound (written in my working out but not strictly necessary, could just skip to the exact step where you found the answer).

You have a possible answer, but you can do a lot better than that -- so here's my working out from before. 2003 doesn't have to be in January, which is what you've got there.

In the interest of a short explanation (and you can have a look at this on your own -- I think deriving this for yourself is good practice because you've overshot by a long way in your solution), for these two years to be in the same converted month, 2003 must be no later than March. Hence, we want 2003 to be as late as possible on the 31st of March which happens to be the 90th day of the year ie. since the days get rounded down, we effectively want to be as close to 91 as possible. Our hypothetical year \(X\) is thus equal to \(\lim_{x \to 91} \frac{2003 \times 365}{x} \approx 8034\).

To verify that 1485 is in the same month, we can sub in \(\frac{1485 \times 365}{8034} = 67.47\), which so happens to be March 8.
« Last Edit: February 19, 2022, 11:16:04 pm by fun_jirachi »
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Re: Tough question from a past sac paper.
« Reply #4 on: February 20, 2022, 11:58:58 am »
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I realised almost as soon as I posted you didn't actually have to establish an upper bound (written in my working out but not strictly necessary, could just skip to the exact step where you found the answer).

You have a possible answer, but you can do a lot better than that -- so here's my working out from before. 2003 doesn't have to be in January, which is what you've got there.

In the interest of a short explanation (and you can have a look at this on your own -- I think deriving this for yourself is good practice because you've overshot by a long way in your solution), for these two years to be in the same converted month, 2003 must be no later than March. Hence, we want 2003 to be as late as possible on the 31st of March which happens to be the 90th day of the year ie. since the days get rounded down, we effectively want to be as close to 91 as possible. Our hypothetical year \(X\) is thus equal to \(\lim_{x \to 91} \frac{2003 \times 365}{x} \approx 8034\).

To verify that 1485 is in the same month, we can sub in \(\frac{1485 \times 365}{8034} = 67.47\), which so happens to be March 8.
I understand how you arrived at the answer, but I can't figure out how to prove that 2003 must be no later than March other than by brute forcing every single date. I've been trying to solve this problem for a few days now. Can you provide that proof?

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Re: Tough question from a past sac paper.
« Reply #5 on: February 20, 2022, 01:29:09 pm »
+2
The latest day in April that 2003 can be is \(\lim_{x\to 121} \frac{365 \times 2003}{x} \approx 6042\) (1st of May is the 121st day of the year). This would result in 1485 being on the \(\frac{1485 \times 365}{6042} = 89.70\)th day, which is still in March. Any earlier date for 2003 would still result in 1485 being in March or earlier. This is automatically invalidated because 2003 and 1485 have to be in the same month.

We then look at March 31st, which is the calculation I did in the previous post. This so happens to be valid and is the correct answer unless I'm mistaken.

The fact that 1485 and 2003 have to be in the same month drastically reduces the number of dates you have to look at. You do not under any circumstances have to brute force 365 days.
« Last Edit: February 20, 2022, 01:31:11 pm by fun_jirachi »
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Re: Tough question from a past sac paper.
« Reply #6 on: February 20, 2022, 05:11:49 pm »
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The latest day in April that 2003 can be is \(\lim_{x\to 121} \frac{365 \times 2003}{x} \approx 6042\) (1st of May is the 121st day of the year). This would result in 1485 being on the \(\frac{1485 \times 365}{6042} = 89.70\)th day, which is still in March. Any earlier date for 2003 would still result in 1485 being in March or earlier. This is automatically invalidated because 2003 and 1485 have to be in the same month.

We then look at March 31st, which is the calculation I did in the previous post. This so happens to be valid and is the correct answer unless I'm mistaken.

The fact that 1485 and 2003 have to be in the same month drastically reduces the number of dates you have to look at. You do not under any circumstances have to brute force 365 days.
Ah that makes sense. I think I worked it out and attached my solution.

No idea how students were supposed to do that in a sac. Is this question hard or actually easy? Would never have been able to figure it out without your help, thank you.

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Re: Tough question from a past sac paper.
« Reply #7 on: February 20, 2022, 08:09:42 pm »
+1
Ah that makes sense. I think I worked it out and attached my solution.

No idea how students were supposed to do that in a sac. Is this question hard or actually easy? Would never have been able to figure it out without your help, thank you.

I think it's tough to grasp, but at the end of the day, the concepts are comparatively easy. At any rate, it seems somewhat dodgy that a question that lacks clarity in any respect makes its way into a formal examination.

I would hope you don't get anything similar in other exams, but should you come across something similar, it's not a bad idea to attempt to fully comprehend the question before even starting on the maths.
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