B1. A teacher makes a large fraction from the numbers 1 to 12. She writes part of the numbers in the numerator and puts $x$-signs between those numbers. The remaining numbers she puts in the denominator, again with $x$-signs between those numbers. Both in the numerator and in the denominator, there is at least one number. She does this in such a way that the fraction is equal to the smallest possible integer. What is this number?
B1. 231

We can factorize both the numerator and the denominator of the fraction into prime numbers. Factors that occur in both the numerator and the denominator cancel each other out. What remains is the simplified form of the fraction. The prime factors that can occur are 2, 3, 5, 7, and 11. For each of these prime numbers, we keep track of how often they occur as prime factors in the numbers 1 through 12. If a prime number occurs an odd number of times, the factors cannot all cancel each other out and the prime factor must occur in the smallest integer we are looking for.

The prime numbers 7 and 11 occur only once, namely in the numbers 7 and 11 themselves. The prime number 5 occurs twice: once as 5 and once as a factor of the number 10. The prime number 3 occurs five times: once as a factor in the numbers 3, 6, and 12, and twice as a factor in the number 9. The prime number 2 occurs ten times: once as a factor in the numbers 2, 6, and 10, twice as a factor in the numbers 4 and 12, and three times as a factor in the number 8.

Thus, we see that the integer we are looking for must at least have the factors 3, 7, and 11 and must therefore be at least \(3 \times 7 \times 11 = 231\). This turns out to work, namely as follows:

$$
\frac{4 \times 7 \times 9 \times 10 \times 11 \times 12}{1 \times 2 \times 3 \times 5 \times 6 \times 8}
$$