Friendly Factor Pairs
The number 6 has two pairs of factors: 2x3 and 1x6. These two pairs of factors are called friendly because the sum of one pair is equal to the difference of the other pair, ie. 2+3 = 6-1 = 5.
The number 6 has two pairs of factors: 2x3 and 1x6. These two pairs of factors are called friendly because the sum of one pair is equal to the difference of the other pair, ie. 2+3 = 6-1 = 5.
Consider a natural number N and the sum of its digits S. Let’s call a divisor of S a Digit Sum Divisor of N.
From the list of numbers 1 to n, cross out every other number starting with the smallest and repeat the process with the remaining numbers until only one number is left. What number would that be for a given n?
Here’s one way to put a pair of 1s, a pair of 2s, and a pair of 3s in a row such that the pair of 1s have one number between them, the pair of 2s have two numbers between them, and the pair of 3s have three numbers between them:
Starting with 3, let’s double it and add one (or 2n+1), then modulate the answer by 14 (divide by 14 and take the remainder). If we repeat these two steps, we get the first cycle.
Starting with a natural number at the top, we can split it at each level such that the two lower neighbours always add up to the higher neighbour. Let’s repeat this process until a row can’t be split using whole numbers in the following row anymore.
Consider the set of natural numbers 1 to 8. For this set of numbers, 6 is called the balance because the numbers before 6 add up to the same sum as the numbers after 6. In other words, 1+2+3+4+5 = 7+8.
The original problem asks to cut a square by drawing straight lines between any number of midpoints of sides and corners of the square such that a resulting region is 1/5 of the area of the whole square.
Continuing from last week’s problem about counting squares, let’s look at the related problem of counting the number of rectangles in a grid, with extensions
Consider the widely known problem of counting the number of squares in a square grid. Today we will try to extend the problem.