Ten primes, each less than 3000, form an arithmetic progression. Find them.

The constant difference d needs to be divisible by 2, 3, 5, and 7.

Assume it wasn't divisible by 2. Then, a+d and a+2d will have a different remainder, and since there are only 2 remainders possible, one of them must be zero, so the number must be divisible by 2. (we can see a+d and a+2d>2)

Similarly, assume it's not divisible by 3. If a=3, then a+3d is divisible by 3, and not prime. Otherwise, a+d, a+2d, and a+3d have different remainders, so one must be 0, and that number is divisible by 3.

Continuing on in this process, we can show the constant difference must be divisible by 2, 3, 5 and 7. Thus, it must be a multiple of 210. (For primes under 3000, this constant difference must be 210 or 420)

Posted by Gamer on 2008-09-30 17:40:12