A six-digit number of the form abcabc (for example, 249,249) always has several certain prime number fac-tors. Which prime numbers are always factors of a number of this form? Explain your reasoning.
A six-digit number of the form abcabc can be expressed as 1001 * abc.
We know that 1001 = 7 * 11 * 13.
Therefore, any number of the form abcabc will always be divisible by 7, 11, and 13.
This can be proven by considering the divisibility rules for these prime numbers:
1. Divisibility by 7:
A number is divisible by 7 if the difference between twice the last digit and the rest of the number is divisible by 7. In the case of abcabc, the difference is abc - abc = 0, which is divisible by 7.
2. Divisibility by 11:
A number is divisible by 11 if the difference between the sum of alternate digits (starting from the leftmost digit) and the rest of the number is divisible by 11. In the case of abcabc, the sum of alternate digits is a + c + a = 2a + c, and the rest of the number is b + a + b = 2b + a. The difference is (2a + c) - (2b + a) = a + c - 2b, which is divisible by 11.
3. Divisibility by 13:
A number is divisible by 13 if the difference between four times the last digit and the rest of the number is divisible by 13. In the case of abcabc, the difference is (4 * abc) - abc = 3abc, which is divisible by 13.
Therefore, the prime numbers 7, 11, and 13 are always factors of a number of the form abcabc.