Competition problem — Number theory

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Number theory Difficulty 7.0 National Olympiad Prove it Brazil

Christine has a deck of cards numbered from $1$ to $25$ . She asked her friend Dorothy to choose six cards from the deck. Christine wrote down the chosen numbers and put the cards back in the deck. She then asked Dorothy to choose six cards again and, again, she wrote down the chosen numbers.

a. The first six numbers that Dorothy had chosen have the property that the difference between any two of the chosen numbers is a multiple of $4$ and only one of them is not a prime. Find the six numbers.

b. In the second time around, Dorothy chose the numbers in a way that for each pair of the numbers except one, one of the numbers divide the other. Find the largest of the six numbers.

Solution

a. The prime numbers between $1$ and $25$ are $2$ , $3$ , $5$ , $7$ , $11$ , $13$ , $17$ , $19$ , $23$ . The ones of the form $4k + 1$ , $k \in \mathbb{Z}$ , are $5$ , $13$ and $17$ ; the ones of the form $4k + 3$ are $3$ , $7$ , $11$ , $19$ and $23$ . Since there are supposed to be five prime numbers, the numbers are $3$ , $7$ , $11$ , $15$ , $19$ and $23$ .

b. The largest number must be divisible by at least other four of the numbers, so it has at least five divisors (these four and itself). The only numbers between $1$ and $25$ with at least five divisors are $16$ and $24$ . So the largest number is $16$ or $24$ . If it is $16$ , then $1$ , $2$ , $4$ and $8$ are four of the other numbers. The other lies between two consecutive powers of $2$ ; but it cannot divide the largest of them and it is not a multiple of the smallest, which is a contradiction.

On the other hand, there are a few examples with $24$ as last number as, say, $1$ , $2$ , $4$ , $8$ , $16$ , $24$ and $1$ , $2$ , $4$ , $6$ , $12$ , $24$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.