day 57: sharpening the tools

BTIW:

1. I took time to read the Bible readings for today, yesterday, and Monday.

2. I didn’t use the heaviest weight ever for my push presses, and I strung together 20 double unders.

TINTWO:

1. Communication

from AIME 1988 Problem 06

It is possible to place positive integers into the vacant twenty-one squares of the 5 times 5 square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).

Solution (general)

First, let a = the number to be placed in the first column, fourth row. Let b = the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of a and b:

\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & & \\\hline 3a & & & & \\\hline 2a & & ...

Next, let a + b + c = the number to be placed in the second column, fourth row. We can determine the entire second column and fourth row in terms of a, b, and c:

\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & & & \\\hline 3a & 3a + b + 3c & & & ...

We have now determined at least two values in each row and column. We can finish the table without introducing any more variables:

\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & 4a + 2b + 8c & 4a + 3b + 12c & 4a + 4b + 16c \\\hlin...

We now have a system of equations.

3a + b + 3c = 74
2a + 4b + 8c = 186a + 2b + 2c = 103

Solving, we find that

(a,b,c) = (13,50, - 5).

The number in the square marked by the asterisk is

4a + 3b + 12c = \boxed{142}

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s