Question number 21, there are 10 different positive integers and exactly 5 of them are divisible by 5 and exactly 7 of them are divisible by 7. Let m be the greatest of these 10 numbers. What is the minimal possible value of m? Let's begin by writing down some numbers divisible by 5, so 5, 10, 15, 20, 25, 30, and so on. These are divisible by 5. Then some numbers divisible by 7, 7, 14, 21, 28, 35, 42, and so on. And then some numbers divisible by both, so these overlap at 35, at 70, 105, and so on. And then let's construct our example. So I will underline here the numbers that I want. I want to double up and have a list of 10, so let's begin with the 2 numbers here divisible by 5 and 7, and then I need 3 more numbers divisible by 5 and 5 more numbers divisible by 7, so I will choose 7, 14, 21, 28, 35 is taken up already, so let's do 42. So we see here that in our example, the number 70 is the value of m, which is not listed, and there is a smaller value we should discuss. So the question now becomes, can we have m is equal to 63? So can we improve our list like that? So let's see what happens if we would try. We would see that replacing this with 63, which is divisible by 7, but not divisible by 5, would give us 7 numbers divisible by 7, but then I would have to add in one more number to the list that is divisible by 5, and that would have to be one of these over here, and then my list would not contain 10 numbers. It would contain 11 numbers. So no, that is not possible. The list would be too large, and you can see that replacing any of the numbers that's divisible by 7 and 5 leads to the same problem, and if we go past 70, then 105 is the next number on the list, but that clearly is not better than 70. So the minimum value here of m is none of these, and the answer is E.

positive integers

divisible

minimum value

largest number

combinations