Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Find the answer United States

Problem:

You have a 10×1010 \times 10 grid of squares. You write a number in each square as follows: you write 1,2,3,,101,2,3, \ldots, 10 from left to right across the top row, then 11,12,,2011,12, \ldots, 20 across the second row, and so on, ending with 100100 in the bottom right square. You then write a second number in each square, writing 1,2,,101,2, \ldots, 10 in the first column (from top to bottom), then 11,12,,2011,12, \ldots, 20 in the second column, and so forth.
When this process is finished, how many squares will have the property that their two numbers sum to 101101?

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

Solution

Solution:

The number in the iith row, jjth column will receive the numbers 10(i1)+j10(i-1)+j and 10(j1)+i10(j-1)+i, so the question is how many pairs (i,j)(i, j) (1i,j101 \leq i, j \leq 10) will have
101=[10(i1)+j]+[10(j1)+i]121=11i+11j=11(i+j). 101 = [10(i-1)+j] + [10(j-1)+i] \quad \Leftrightarrow \quad 121 = 11i + 11j = 11(i+j).
Now it is clear that this is achieved by the ten pairs (1,10),(2,9),(3,8),,(10,1)(1,10), (2,9), (3,8), \ldots, (10,1) and no others.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.