22. Given the integers from 1 to 25 , Ajibola wants to remove a few and then split those that remain into two groups so that the products of the integers in each group are equal. Ajibola removes the smallest possible number of integers in order to achieve this. What is the sum of the numbers which Ajibola removes?
A 75
B 79
C 81
D 83
E 89
Solution
B

Thinking of the prime factorisation of the products of the numbers of both groups, they have to be the same, since the products are equal. Therefore, prime numbers that have only one multiple among the numbers from 1 to 25 have to be removed. These prime numbers are 13, 17, 19 and 23. 7 has three multiples among the numbers from 1 to $25: 7,14$ and 21 . It is left to the reader to show that all the primes other than $7,13,17,19$ and 23 , that occur in the prime factorisation occur to an even power. So the smallest number of primes which need to be removed is five and these are $7,13,17,19$ and 23 . Their sum is 79 . The remaining 20 numbers can be split in two groups that have equal products.
For example:
$$
3 \times 5 \times 8 \times 14 \times 15 \times 18 \times 20 \times 22 \times 24=2 \times 4 \times 6 \times 9 \times 10 \times 11 \times 12 \times 16 \times 21 \times 25=2^{11} \times 3^{5} \times 5^{3} \times 7 \times 11
$$