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.

[…] 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. […]

## 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.

3 August 2022, 6:03 am## 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.

4 August 2022, 6:24 am## 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. […]

18 August 2022, 11:12 am