A Multiple Problem

I am a two-digit number. When I am divided by 8, there is a remainder of 3. When I am divided by 9, there is a remainder of 4. I am less than 80. What number am I?

Jane, Indonesia

One way is to guess and check. Make an intelligent guess and check if both conditions are met. I can guess 27. 27 divided by 8 gives 3 remainder 3. 27 divided by 9 gives no remainder. So is the number 27?

I think it takes time to guess this way. Let's use logical reasoning. The number is 3 more than a multiple 8 and 4 more than a multiple of 9.

Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.

Three more than a multiple of 8: 11, 19, 27, 35, 43, 59, 67, 75 (no need to try 83 and beyond)

Multiple of 9: 9, 18, 27, 36, 45, 54, 63, 72

Four more than multiple of 9: 13, 22, 31, 40, 49, 58, 67, 76

You got the solution, right?

How about a different way? If you know some algebra, the number is 8m+3 or 9n+4 where m and n are whole numbers. 8m+3 = 9n+4 or 8m = 9n + 1 which gives a possible solution of m = 8 and n = 7. Thus, the number which is 8m+3 can be easily found.

What if the last condition that it is less than 80 is not given?

Can you make up a similar interesting problem for the others to solve?

Pre-School Numeracy & Assessment

We are writing you from Santiago, Chile. We are using Earlybirds Kindergarten Mathematics. The doubt is that in Book A, to evaluate classification, is it necessary to evaluate all the previous steps such as "different things", "things that are used together", "things that do not belong" etc.? Or can we evaluate only the final concept, that is sorting by the three attributes.

We have to tell you that this is our first year using your method in Mathematics, and, at the beginning , we had some problems, because of the language (our students don´t speak English at home, they learn it only at school). But now, we have used the book for 4 months, and we think the students, and we, the teachers, have learned a lot from it.

Anamaría, Chile

In Kindergarten, we do not want to overwhelm children with assessment and evaluation. In this unit students have learnt how to match things according to some attribute (being able to say two animals are the same despite differences in size, color, orientation; being able to say two things are the same even though they are drawn differently; matching by colors; matching by patterns). Some where in between they apply this to pick the odd item out among, say, four items. Later, they learn to classify according to more abstract attributes such as function (what they are used for). Again, the apply these skills to pick the odd one out. Finally, the use the skills to classify items into two groups according to given criteria and their own criteria. In the evaluation, we assess and evaluate if the children are able to apply these skills in picking the odd one out and in classification.

There is no need to evaluate every sub-skill. However, if any child cannot complete the main task (odd one out and classification) then teachers may want top check if they can do the sub-skills.

Test Items from Singapore National Tests

I understand that somewhere in your website I can access past primary-level mathematics tests, but I cannot find them. Do they exist, and can I get access to them? I am learning about Singapore Math.

Lee, a math coach in Utah

These are not available online. Released items from the tests are compiled into a book every year. For example, the most recent book consists of released items from 2004 to 2008 tests. About three-fourths of all items are released each year. This book is not available outside Singapore. I believe this is an agreement between the copyright holder of the test items and the publishers. In Singapore, they are easily available in any Popular Bookshops. If you are outside Singapore, get a friend to help you buy a copy.

The national test at the end of primary schooling is the Primary School Leaving Examination. Students are tested on English, Mathematics, Mother Tongue Language and Science. For mathematics. most students do the Mathematics test. A small proportion do the Foundation Mathematics test which focuses more on basics and less on problem solving.

Students complete 15 multiple-choice items (20 points), 20 short-answer items (30 points) and 13 long-answer items (50 points) which includes some challenging tasks. The first paper (50 min) includes the multiple-choice items and 15 short-answer ones. The second paper (1 h 40 min) includes the other items. Students can use a calculator in the second paper. This format will be used for the first time in 2009. Previously, these items were in one 2 h 15 min paper and students do not use calculators (and the numbers are not tedious to compute).

Word Problem

I'm the mathematics coordinator of 3rd and 4th grades of my school in Chile. We are using the My Pals are Here Series. But I have a doubt, concerned to the fact that even though we are focused on solving word problems using the models methology and it makes sense to the girls when we are working together, they are still having problems when they are working alone, specially during tests. They start right away with operations but most of the time it is wrong, because they didn't visualize the entire problem. Do you have any suggestion? Should we continue with next chapter or work more on problem solving with additions and subtractions?

Question posted by Paula, a mathematics co-ordinator in Chile

As you have mentioned, when the students start straight away with the operations they are often wrong. They need to comprehend the problems well. Drawing a model will help them understand how the information are related. In simple one-step problem, it may not necessary to do so. But in a problem with a lot of information, this becomes essential for average students. Otherwise although they can read the word, they do not comprehend the information.

Also in multi-step problems, the students may not have the ability to monitor their thinking. This is metacognition. When we teach word problems, we should model and coach rather than explain. That way, we help them in developing the ability to think through the many steps in a problem.

Speed Problem...Again

A car needs 7 hours to travel from Town X to Town Y. A motorcycle needs 8 hours to travel from Town Y to Town X. The car leaves Town X for Town Y and the motorcycles leaves from Town Y to Town X at the same time. How long will it take for the car and the motorcycle to meet?

Angie

Speed Problems are frequently brought up. There are earlier entries discussing Speed Problems. See below.

So, how long will it take for the car and the motorcyle to meet. The standard joke is that we hope they don't!

That aside, we need to assume that the speed of the two vehicles are constant. If that is so then in an hour, the car travels 1/7 the distance in an hour and the motorcycle travels 1/8 the distance in an hour. The problem is solved when the distance travelled by the car and motorcyle add up to 1 whole. In an hour, total distance covered by both is (1/7 + 1/8) of XY. This works out to 15/56 of XY. In 2 hours, it is (2/7 + 2/8) of XY or 30/56 of XY. In 3 hours, 45/56. In 4 hours, 60/56. They would have passed each other in 4 hours. Can I leave it to you to complete the last step of the solution. It is by no means trivial but there are enough leads already.

Request for Presentation Slides

I am a 5th grade teacher in Fayetteville, NC. I had the amazing opportunity to attend your session at the NCTM Conference in Washington, DC a few weeks ago and was truly inspired! If possible, would you be able to email me a copy of your handouts and powerpoints used in your session? I would greatly appreciate it! Thank you for your time and amazing inspiration!

Laura, an American teacher

The presentation slides at the NCTM Annual Meeting & Exposition are available at http://math.nie.edu.sg/t3/downloads-conference.htm Look for the conference that you are interested in and click on the pdf. The slides should download. The slides for my other presentations are also available here.

A Problem from a Hong Kong School

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

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.