Question number 21. In the equation n times u times the quantity m plus b plus e plus r equal to 33, each letter stands for a different digit from 0 to 9. How many different ways are there to choose the values of the letters? So let's look at the product here. We have 33 is 3 times 11 and that is supposed to be the product of n u with the quantity m plus b plus e plus r. So this forces n u, that product, to be equal to 3 and the sum m plus b plus e plus r has to be equal to 11. And so why do we know that? Well, if it was the other way around, if m plus b plus e plus r was equal to 11, we couldn't actually choose four numbers like that that add to 11. And more importantly, the product of n and u could not be 11. It has to be a number for n and a number for u that is smaller than 11. So n u, the product, is 3 and so the letters n and the letters u here, they belong to the set containing the numbers 1 and 3. So we have two choices here for these letters. And what about the remaining four? So the letters m, b, e, and r have to belong to the set where we have the numbers 9, 8, 7, 6, 5, 4, 3 is taken, we can have a 2, 1 is taken, and then we can have a 0. Okay, and then we have to think about this set and decide is that really the correct choice for the remaining integers or any of them have to be removed from the set. And then looking at these values here, we see that 9 probably should not be in the set because if 9 is in the set and the terms or the values of m, b, e, and r are distinct, then any other number together with 9 would increase the sum past 11. So we we have to erase or get rid of 9. For the same reason, we get rid of 8 because if 8 is in the set, even with 9 already removed, still any three numbers here will put the sum past 11. Okay, we do the same thing with with 7 because 7 and any three of the remaining numbers increase the sum past 11. So what about what about 6? If we add the remaining numbers here, we have 6 plus 5 plus 4 plus 2 plus 0, that is more than 11, so we also get rid of 6. What about 5 plus 4 plus 2? That is exactly 11, and then we have a 0 over here. So the set here is really just containing these four numbers. We should say 5, 4, 2, and 0. So there are four numbers to choose from for each value of the four letters. So here we have 4 times 3 times 2 times 1 or 24 choices for the value of m, b, e, and r all together like that, and so we multiply these two overall. We have 2 times 24 or 48 choices, and so that answer here is letter D.

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