Maths Olympiad Prep

Track / Stage 6 / 360 of 400 #1360 of 1964

Problem 1360

National Olympiad, first round
Number theory Difficulty 6.7 Find the answer

Farmer John has a 47×5347 \times 53 rectangular square grid. He labels the first row 1,2,⋯ ,471, 2, \cdots, 47, the second row 48,49,⋯ ,9448, 49, \cdots, 94, and so on. He plants corn on any square of the form 47x+53y47x + 53y, for non-negative integers x,yx, y. Given that the unplanted squares form a contiguous region RR, find the perimeter of RR.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

1. Understanding the Problem:
Farmer John has a 47×5347 \times 53 rectangular grid. He plants corn on any square of the form 47x+53y47x + 53y for non-negative integers xx and yy. We need to find the perimeter of the contiguous region RR formed by the unplanted squares.

2. Expressibility:
A number is expressible if it can be written in the form 47a+53b47a + 53b for non-negative integers aa and bb. We need to determine which numbers are not expressible.

3. Chicken McNugget Theorem:
The Chicken McNugget theorem states that for two coprime integers mm and nn, the largest number that cannot be expressed as am+bnam + bn (where aa and bb are non-negative integers) is mn−m−nmn - m - n. Here, m=47m = 47 and n=53n = 53, which are coprime.

Therefore, the largest number that cannot be expressed is:
47⋅53−47−53=2491−100=2391 47 \cdot 53 - 47 - 53 = 2491 - 100 = 2391

4. Non-expressible Numbers:
All numbers from 11 to 23912391 that cannot be expressed as 47a+53b47a + 53b are unplanted. However, since we are dealing with a grid, we need to consider the numbers modulo 4747 and 5353.

5. Column-wise Analysis:
We analyze the grid column-wise. Each column cc (where 1≤c≤461 \leq c \leq 46) contains numbers of the form c+47kc + 47k for k=0,1,2,…k = 0, 1, 2, \ldots. We need to determine when c+47kc + 47k is expressible.

6. Modulo Condition:
Suppose c+47kc + 47k is expressible. Then, there exist non-negative integers aa and bb such that:
c+47k=47a+53b c + 47k = 47a + 53b
Rearranging, we get:
c=47(a−k)+53b c = 47(a - k) + 53b
This implies c≡53b(mod47)c \equiv 53b \pmod{47}. Since 53≡6(mod47)53 \equiv 6 \pmod{47}, we have:
c≡6b(mod47) c \equiv 6b \pmod{47}
Therefore, b≡c6(mod47)b \equiv \frac{c}{6} \pmod{47}.

7. Unplanted Squares in Each Column:
For each column cc, the number of unplanted squares is 9c(mod53)9c \pmod{53}. We need to count the unplanted squares in each column and determine the perimeter of the contiguous region RR.

8. Perimeter Calculation:
The unplanted squares form 8 "staircases" connected edge-to-edge. Each staircase has a maximum height and width. The perimeter of each staircase is 2(h+w)2(h + w).

Summing up the perimeters of all staircases:
2(5+45)+2(6+46)+2(6+47)+2(6+48)+2(6+49)+2(6+50)+2(6+51)+2(5+43)=850 2(5+45) + 2(6+46) + 2(6+47) + 2(6+48) + 2(6+49) + 2(6+50) + 2(6+51) + 2(5+43) = 850

9. Overcounting Correction:
We overcount twice the edges where the staircases are connected. The total length of these edges is 1+2+⋯+7=281 + 2 + \dots + 7 = 28. Therefore, we subtract 2⋅282 \cdot 28 from the total perimeter.

Final perimeter:
850−2⋅28=850−56=794 850 - 2 \cdot 28 = 850 - 56 = 794

The final answer is 794\boxed{794}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.