All the numbers are written in a array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to . What is the number in the center?
Quick testing shows thatis a valid solution. , and the numbers follow the given condition. The center number is found to be .
First let the numbers bewith the numbers around the outsides and in the middle. We see that the sum of the four corner numbers is . If we switch and , then the corner numbers will add up to and the consecutive numbers will still be touching each other. The answer is .
Consecutive numbers share an edge. That means that it is possible to walk from to by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity:
But there are only four even numbers in the set, so the five darker squares must contain the odd numbers, which sum to Therefore if the sum of the numbers in the corners is , the number in the center must be , which is answer .
The sum of an infinite geometric series is a positive number , and the second term in the series is . What is the smallest possible value of
The sum of an infinite geometric series is of the form:where is the first term and is the ratio whose absolute value is less than 1.
We know that the second term is the first term multiplied by the ratio. In other words:
Thus, the sum is the following:
Since we want the minimum value of this expression, we want the maximum value for the denominator, . The maximum x-value of a quadratic with negative is .
Plugging into the quadratic yields:
Therefore, the minimum sum of our infinite geometric sequence is . (Solution by akaashp11)
After observation we realize that in order to minimize our sum with being the reciprocal of r. The common ratio has to be in the form of with being an integer as anything more than divided by would give a larger sum than a ratio in the form of .
The first term has to be , so then in order to minimize the sum, we have minimize .
The smallest possible value for such that it is an integer that's greater than is . So our first term is and our common ratio is . Thus the sum is or . Solution 2 by No_One
We can see that if is the first term, and is the common ratio between each of the terms, then we can getAlso, we know that the second term can be expressed as notice if we multiply by , we would getThis quadratic has real solutions if the discriminant is greater than or equal to zero, orThis yields that or . However, since we know that has to be positive, we can safely conclude that the minimum possible value of is .