N is a natural number greater than 1. A and B are single-digit natural numbers, with A ≥ B, such that for any value of N, (A + B)N has the same units digit as A + B and (A × B)N has the same units digit as (A × B). How many pairs of values of A and B exist satisfying these conditions?

To solve this problem, we need to understand the properties of the units digit of a number.

Step 1: Understand the problem The problem is asking for pairs of single-digit natural numbers A and B (with A ≥ B) that satisfy two conditions:

1. The units digit of (A + B)N is the same as the units digit of A + B for any natural number N greater than 1.
2. The units digit of (A × B)N is the same as the units digit of A × B for any natural number N greater than 1.

Step 2: Analyze the conditions For the first condition, we know that the units digit of a sum is determined by the units digits of the numbers being added. Therefore, the units digit of (A + B) will always be the same, regardless of the value of N.

For the second condition, we know that the units digit of a product is determined by the units digits of the numbers being multiplied. However, when a number is raised to a power, the units digit can change. For example, the units digit of 2^2 (4) is different from the units digit of 2^3 (8). Therefore, we need to find pairs of A and B such that the units digit of (A × B) does not change when raised to a power.

Step 3: Find the solution We know that the units digit of a number raised to a power cycles through a pattern. For example, the units digit of 2 raised to a power cycles through the pattern 2, 4, 8, 6. However, if the units digit of a number is 0, 1, 5, or 6, the units digit does not change when the number is raised to a power.

Therefore, the pairs of A and B that satisfy both conditions are those where A and B are each either 0, 1, 5, or 6. There are 10 such pairs: (0, 0), (1, 0), (1, 1), (5, 0), (5, 1), (5, 5), (6, 0), (6, 1), (6, 5), and (6, 6).