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2025 AMC 10B Problem 23

Problem 23 of 25HarderNumber Theory

A rectangular grid of squares has 141141 rows and 9191 columns. Each square has room for two numbers. Horace and Vera each fill in the grid by putting the numbers from 11 through 141×91=12,831141 \times 91 = 12{,}831 into the squares. Horace fills the grid horizontally: he puts 11 through 9191 in order from left to right into row 1,1, puts 9292 through 182182 into row 22 in order from left to right, and continues similarly through row 141.141. Vera fills the grid vertically: she puts 11 through 141141 in order from top to bottom into column 1,1, then 142142 through 282282 into column 22 in order from top to bottom, and continues similarly through column 91.91. How many squares get two copies of the same number?

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Solution

At row i,i, column j,j, Horace writes 91(i−1)+j91(i - 1) + j and Vera writes 141(j−1)+i.141(j - 1) + i. Set them equal and simplify to get 9i−14j=−5,9i - 14j = -5, so i=14j−59,i = \tfrac{14j - 5}{9}, an integer exactly when j≡1(mod9).j \equiv 1 \pmod 9. For j=1,10,19,…,91,j = 1, 10, 19, \ldots, 91, that’s 1111 values, and ii runs 1,15,29,…,141,1, 15, 29, \ldots, 141, all within range. So 1111 squares match. Thus, C is the correct answer.
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Tagged: Diophantine Equation · modular arithmetic

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