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A fenced, rectangular meadow has dimensions $24$ m by $52$ m. A mathematical farmer has $2019$ m of fence that he wants to use to fence the outside of the field and to partition the field into identical square plots with sides parallel to the edges of the field. Determine the largest possible number of square plots into which she can divide the field and how much fencing would be left over.

For the part of field already fenced, I can partition it into 4 by 4 squares as gcd(24, 52)=4. And then I can partition the 4 by 4 squares into smaller squares. But I don't know how much fence I should use to fence the outside of the field.

Andy
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  • The perimeter of the field is $152$ m. Doesn't that mean you need $152$ m. of fence for the outside? I don't understand what you mean by "the part of the field already fenced." The first sentence says "a fenced rectangular field." This seems like either the field is already fenced, and no more fence is needed for the outside, or that the field is to be fenced, and you need $152$ m. of fence. If only part of the field is fenced, we need to know how much. – saulspatz Jun 06 '18 at 18:29
  • @saulspatz In my understanding, the first sentence said the area with dimensions 24 m by 52 m is already fenced at the beginning. Now I can use the fence of 2019 m to extend the fenced area of field. Actually I thought the same but then I don't understand why they mentioned "fenced" rectangular meadow and then said "to fence the outside of the field". – Andy Jun 06 '18 at 19:06

2 Answers2

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Hint:

The perimeter of the meadow $=52+52+24+24=152\ m$

So the remaining fence $=2019-152=1867\ m$

Now consider horizontal fence inside $=h$

consider vertical fence $=v$

Now ler number of rows and columns of squares be $h+1$ and $v+1$

Note that $24h+52v\le 1867\ m$

So the horizontal side length $=\frac{24}{v+1}$ and vertical side length $=\frac{52}{h+1}$

Since all the required region are squares $\frac{24}{v+1}=\frac{52}{h+1}$

I will leave the rest to you....

tien lee
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The right and the bottom edge of the meadow together need $76$ m of fencing, so that there are $2019-76=1943$ m of fencing left to fence the left and top edges of the little squares.

If there are $n$ squares of side length $s$ along the long side of the meadow and $m$ squares along the short side then we necessarily have ${n\over m}={13\over6}$, hence $$n=13k ,\quad m=6k,\quad s={4\over k}$$ for a certain $k\in{\mathbb N}_{\geq1}$. In this way we obtain $78k^2$ little squares, which need $$78k^2\cdot 2s=624 k\leq1943$$ meters of individual fencing. The largest admissible $k$ is $3$, so that we obtain $702$ little squares, and $71$m of fencing are left over.

Christian Blatter
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