A Number Theory Problem from the 43rd Tournament of Towns

Problem. Find the largest number n such that for any prime number p greater than 2 and less than n, the difference np is also a prime number.

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4 Comments

  1. Puzzled:

    n-3, n-5, n-7 are all prime. But at least one of three consecutive terms of an arithmetic progression is divisible by three, so it must be exactly 3. And the largest candidate can be found from n-7=3. Since 10 satisfies the conditions it is the answer.

  2. Puzzled:

    More accurately, we have three consecutive terms of an arithmetic progression with common difference 2. So exactly one of those is divisible by 3.

  3. Tanya Khovanova's Math Blog » Blog Archive » Three, Five, and Seven have Different Remainders When Divided by Three:

    […] There are many cute math problems that use the trivial fact announced in the title. For example, I recently posted the following problem from the 43rd Tournament of Towns. […]

  4. Tanya Khovanova's Math Blog » Blog Archive » Three, Five, and Seven have Different Remainders When Divided by Three - New Marathi Live:

    […] There are many cute math problems that use the trivial fact announced in the title. For example, I recently posted the following problem from the 43rd Tournament of Towns. […]

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