First, three observations: (1) The number of dollars that the men received for the herd of cattle was a perfect square. (2) The number of sheep they bought was odd, because it was necessary to throw in the dog in order to divide the animals evenly. (3) Since they bought an odd number of sheep at $10.00 per head, the next-to-last digit in the square payment from observation (1) must be odd.

Now, as it happens, in any perfect square, if the next-to-last digit is odd, the last figure must be a 6. The simplest way to see this is to consider that any number can be expressed as 10a + b, where a comprises all the digits except the last and b is the last digit (0 to 9). The square of this number is 100a² + 20ab + b². But 100a² has zeros in the last two places, and 20ab has a zero in the last place and is preceded by an even number. So, in order for the square to have an odd digit in the next-to-last position, b² must have an odd first digit, and the only possibilities are 4² = 16 and 6² = 36.

In either case, the dog cost $6.00, and the rancher owed his partner $2.00 for accepting a $6.00 dog in place of a $10.00 sheep.

The next number is 9. This is Golomb’s sequence, a non-decreasing sequence of integers in which a_n is the number of times that n occurs in the sequence. We’re given a₁ = 1 to start with, and this indicates that the number 1 appears only once in the whole sequence. That means that a₂ is greater than 1. Setting a₂ to 2 immediately tells us that 2 appears twice in the sequence, so a₃ also must be 2. That tells us that 3 occurs twice in the sequence, and so on. In this way each number in the sequence is uniquely determined.