A mom shared that her P1 daughter in Hong Kong was posed this mathematics problem:

AB + B = BA

A = ________B = ________

How do you teach a P1 child algebra?

Lily

AB + B = BA

A = ________B = ________

How do you teach a P1 child algebra?

Lily

*In formal algebra, ab means the product of a and b. I do not think that is what the problem is about. In this problem, I believe A and B represent digits and AB is a two-digit number which when added to B gives a two-digit number which has the tens and ones digit of the original number (AB) reversed. In this case the letters are not used in the same convention as formal algebra. I would recommend that this problem is presented orally to grade one children. The teacher might say, "Each letter (or shape) stands for a digit. The same letter stands for the same digit. Different letters stand for different digits. I have a number, the tens digit is A and the ones digit is B (Teacher writes down AB). When it is added to a number B (Teacher writes down AB + B), the total is a number with B as its tens digit and A as its ones digit. (Teacher writes down AB + B = BA) Find the digits A and B."*

*I would illusrate with an example say 12. What is A in this illustration? What is B? A is 1 and B is 2. So, in this problem AB + B (which is 12 + 2) is supposed to be BA (21). But it is not, right? So AB is not 12. I suppose guess and check is the best strategy for grade one children to use.*

*Advanced students or older students may reason this way: Is A an odd or even digit? Yes, it must be an even digit. Why? Did you notice that B + B = A. Sure it could be 1A as well. But not 2A or 3A or 4A and so on, right? Since the final sum is BA, AB + B must involve renaming. Why? Otherwise the tens digit in the sum is the A, isn't it? So B must be 6, 7, 8 or 9. And A is one less than B. Think about this one! Hence, A is 5 when B is 6, A is 7 when B is 8 - both not possible. Why?*

*Hence, A = 6 and B = 7 or A = 8 and B = 9. Checking 67 + 7 = 74 and 89 + 9 = 98. I think the solution is A = 8 and B = 9.*

*Secondary students may solve it algebraically: 10A + B + B = 10B + A or 9A = 8B or the ratio of A : B = 8 : 9. For digits, A has to be 8 and B has to be 9.*

*I don't think P1 children are expected to do the algebraic solution or even the reasoning based on number properties. They are most likely able to solve it by guess-and-check.*

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