# 7th class question my daughter asked and need answer if possible

My daughter has asked me to solve this question but I am unable to than I thought to post it here may be someone help.

Q : The watchman works 4 days a week and has a rest on the fifth day. He had been resting on sunday and began working on Monday. After how many days will his rest fall on sundays?

A) 31
B) 12
c) 34
D) 7


Someone explain to me how?

Don't know what should be the proper tag for this question.

Thanks

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## migrated from mathematica.stackexchange.comMar 17 '13 at 18:05

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@GitGud, the question is, after how many days... – Andreas Caranti Mar 17 '13 at 18:10
@AndreasCaranti The plural sundays misled me. – Git Gud Mar 17 '13 at 18:10
@GitGud, by the way, I also misread the question in the beginning. – Andreas Caranti Mar 17 '13 at 18:13
You can help by providing your daughter with an old calendar, and have her mark off the rest days explicitly, until she first bumps into Sunday. That is a perfectly legitimate solution. If the work needs to be handed in, the calendar with the marked off rest days gives the proof. Then one can explore the markings to see whether there is structure there that makes it possible to reach the answer more quickly. – André Nicolas Mar 17 '13 at 18:27
"The watchman works 4 days a week and has a rest on the fifth day." So on the sixth day, does he work or not? If so, isn't incorrect to say he works 4 days a week? – LarsH Mar 18 '13 at 5:30

There are slicker ways, but the question can fairly easily be solved by brute force. A good lesson for her to learn is that it rarely hurts to experiment a bit with the data and look for patterns. Let’s take a look at what happens over the first two weeks or so:

$$\begin{array}{c} \text{Su}&\text{Mo}&\text{Tu}&\text{We}&\text{Th}&\text{Fr}&\text{Sa}&\text{Su}&\text{Mo}&\text{Tu}&\text{We}&\text{Th}&\text{Fr}&\text{Sa}&\text{Su}&\text{Mo}\\ \text{R}&\text{W}&\text{W}&\text{W}&\text{W}&\text{R}&\text{W}&\text{W}&\text{W}&\text{W}&\text{R}&\text{W}&\text{W}&\text{W}&\text{W}&\text{R} \end{array}$$

Here $\text{R}$ is a rest day, and $\text{W}$ is a work day. Look at the sequence of rest days: Sunday, Friday, Wednesday, Monday. Clearly they are moving backwards two days in the week for each cycle of five days. To get to Sunday again, they have to continue to move back through Saturday, Thursday, and Tuesday, after which the next cycle has its rest day on Sunday again. That’s seven cycles altogether, a total of $7\cdot5=35$ days. His working Monday is the first of those $35$ days, so it will be another $34$ days until he has his next Sunday rest day.

A slightly more sophisticated approach is to realize that to get a rest day on Sunday, he must go through a whole number of $5$-day work cycles and a whole number of weeks. In other words, the number of days from one Sunday rest day to the next must be both a multiple of $5$ and a multiple of $7$. The least common multiple of $5$ and $7$ is $35$, so Sunday rest days must fall $35$ days apart, and the next one will therefore fall $34$ days after the Monday start.

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@brian-m-scott, All the answers were good but my Daughter picked your answer easily which is well explained. Thanks – Pirzada Mar 17 '13 at 19:09
@Rashid: You’re welcome. And good luck to her with the math class. – Brian M. Scott Mar 17 '13 at 19:19

The idea is that the cycle of $5$ days of work/rest, and the weekly cycle of $7$ days will complete at the same time in a number of days equal to the least common multiple of $5$ and $7$. In this case this is simply the product $5 \times 7 = 35$. (In general, the least common multiple of two numbers equals their product, divided by their greatest common divisor.) So after $34$ days the rest day will fall again on a Sunday.

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You can help by providing your daughter with an old calendar, and have her mark off the rest days explicitly, until she first bumps into Sunday. That is a perfectly legitimate solution. If the work needs to be handed in, the calendar with the marked off rest days is a complete mathematical proof. Then one might want to explore the markings to see whether there is structure there that makes it possible to reach the answer more quickly. The point of the above approach is that your student will be in concrete control every step of the way, she will know precisely what's going on.

Perhaps better, from the point of view of learning, is for the student to produce her own calendar. It may be a good idea to label the days $1$, $2$, $3$, $\dots$, $30$, $31$, $32$, and so on, so that numerical patterns in the marked off days can be more readily detected. But this is certainly not necessary.

The following is a much more abstract version, which should only be done after the concrete manipulation with the calendar. Think of the first Sunday as Day $1$. Then the next Sunday is Day $8$, the one after that is Day $15$, then $22$, $29$, $36$, $43$, and so on.

Now make a list of the rest days. The first one, we are told, was on Day $1$. The next one is Day $6$, then Day $11$, then $16$, then $21$, and so on. Continue until we bump into a Sunday. It will be soon: if we continue the rest days, we get $26$, $31$, $36$, got it!

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For completeness, I'll post a more abstract solution.

Let's call sunday the 0th day, which is a resting day. One cycle of working and resting takes 5 days. Let $k$ be the number of cycles, then the question is to find the minimum positive $k$ for which $5k=7n$, or $5k\equiv0 \pmod 7$. Since 7 is prime, $\mathbf{Z}/7\mathbf{Z}$ is a field and has no zero divisors, so $k\equiv0 \pmod 7$, so $k=7$ is the minimal solution, which takes $5\cdot7-1=34$ days.

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The rest day is every fifth day: 5, 10, 15, 20, 25, 30, 35,...

A Sunday is every seventh day: 7, 14, 21, 28, 35...

First day that is both rest day and Sunday is 35 which is after 34 days.

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The question is broken. A week has 7 days. If "The watchman works 4 days a week and has a rest on the fifth day", we have no way of knowing whether the fifth day is the first of three rest days, or if instead, his rest days are scattered through the week.

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This was my first thought too. The word 'week' has no place in the first sentence of that question. – Aric TenEyck Mar 18 '13 at 2:01

well it goes backwards by two days each time. just count how many times it takes for you to reach the first sunday and multiply by 5, this is becuase he works 5 days a week. but you will have to subtract 1 since we dont count the sunday. :-)

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Hint $\$ From the prior day (Sunday), the following Sundays are multiples of $7$ days away, and the following rest days are multiples of $5$ days away, so they both coincide on days that are common multiples of $7$ and $5$ days away, the least being $\rm\: lcm(7,5) = 35\,$ days. Since we are counting from Monday, not Sunday, we need to subtract $1,\,$ yielding $34$ days.

Remark $\$ Said arithmetically, if we label days by integers on the number line, with the prior day (Sunday) being the origin $\,0,\,$ then Sundays are numbered $\rm\,7n\,$ and rest days are numbered $\rm\,5k.\:$ Therefore a Sunday and rest day coincide only for days numbered $\rm\,7n = 5k,\,$ i.e. for days whose number is a common multiple of $\,7\,$ and $\,5.$

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Every fifth day is a rest day. And the rest day is falling one day before the present day . Like if on Sunday it is rest day then the next rest day is Friday . So just multiply 5 by the no.. of days in a week and subtract 1 because last day is Sunday.

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