## Thursday, November 12, 2009

### Math problems of the week: 2nd grade Everyday Math vs. Singapore Math

I. A sampling of problems from the 2nd grade Everyday Math Student Math Journal, Volume I, "Addition and Subtraction Facts," pp. 20-50.

Use > , <, or =.

6 + 7 ___ 15 - 4
5 + 8 ___ 8 + 5
18 - 9 ___ 5 + 4

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Today is ________________
(month) (day) (year)

The date 1 week from today will be ____________

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Use a number grid.

How many spaces from: 17 to 26? 49 to 28?

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Which is heavier: 1 ounce or 1 pound?

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Write an addition story.

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Play Broken Calculator.
Show 17. Broken key is 2.
Show three ways.

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Draw a rectangle around the digit in the tens place

349
406

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Follow the rule. Fill in the missing numbers.

Rule: + 6

___in___out___
___2____8____
___3 ____9____
___5 _________
___9__________

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Subtract. Use the -9 and -8 shortcuts.

13-9 = ___
14 - 8 = ___
...

II. A sampling of problems from the 2nd grade Singapore Math Primary Mathematics Workbook, Volume I, "Addition and Subtraction," pp. 31-68.

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Compare two sets.
[squared-off picture of 11 flowers, labeled "A," next to squared off picture of 6 flowers, labeled "B"]

11 - 6 = ___
Set A has ___ more flowers than Set B.

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3 + 4 =
30 + 40 =
300 + 400 =

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Subtract
7 - 3 = ___
70 - 30 = ___
700 - 300 = ___

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Subtract.

689
- 32

786
- 73

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7 + 6 =
27 + 6 =
527 + 6 =

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264
+ 36

486
+ 54
...

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A watch costs \$167.
A camera costs \$48 more than the watch.
What is the cost of the camera?
What is the total cost of the camera and the watch?

The total cost of the camera is \$ ____
The total cost of the camera and the watch is \$ _____

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251
-170

358
+416

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David collected 930 stamps.
He had 845 stamps left over after giving some stamps to his friends.
How many stamps did he give to his friends?

III. Extra Credit:

Discuss the phrase "a mile wide and an inch deep."

Everyday Math tells people to "trust the spiral." Do you?

Marcy said...

I worry about kids moving from district to district. What happens when that spiral is interrupted?

TerriW said...

If only they would play "Broken Calculator" all year...

Beth said...

This reminds me of something my daughter was taught in public school 5th grade. The topic was how to tell if one number is divisible by another.

The school gave a whole set of rules for each divisor, which I have now forgotten, and I'm sure my daughter has too. It was something like, "to see if a number is divisible by 6, add the last two digits and see if the sum is divisible by 6." There was a similar, but slightly different, rule for 9, and so on.

I taught my daughter my technique, which is to come up with a number that you know is a divisor, and keep adding the given divisor until you get as close as you can. For instance, to see if 112 is divisible by 9, you could say, "I know that 99 is divisible by 9. 9 + 99 is 108, which differs from 112 by 4, so 112 is not divisible by 9."

The advantage to my method is that it's easy to understand and remember, and it applies the same way to every problem.

The disadvantage to the school's approach is that math becomes a big sack full of unrelated, incomprehensible algorithms which are forgotten five minutes after they're "learned".

My daughter had a similar experience at her private school at the beginning of this year. She spent a lot of time classifying prime numbers as "abundant, deficient, or precise" based on the number of their factors. I'm sure she's forgotten it by now, 6 weeks later.

bky said...

Beth -- divisibility rules are good things. Everyone can tell at a glance if a number is divisible by 2, 5, or 10. With not much more work (sum the digits) you get a test for divisibility by 3 and 9. It is worthwhile explaining why those tests work rather than just throwing them out there, but they are simple and useful and easy to remember (a number is divisible by 3 if and only if its digit sum is divisible by three; in fact the remainders upon dividing by 3 are the same). This is not a jumble of facts, it is a toolkit. Assuming a student is in fact called on to do arithmetic by hand

Beth said...

OK, I'll bite. You say "It is worthwhile explaining why those tests work." So what's the explanation? I don't understand why summing the digits gives you a test for divisibility by 3. I'm sure my daughter doesn't either. That's why it's so forgettable.

Katharine Beals said...

In contrast to divisibility by 2, 5, and 10, the only way I know to show why summing the digits tests divisibility by 3 is through mathematical induction, which I'm guessing is beyond most grade school students (and their teachers).

Anonymous said...

I don't understand the broken calculator question. What are the kids supposed to do?

Anonymous said...

Would you consider including problems from the Russian math books that were translated by the University of Chicago when you do these math problem comparisons? I find this whole topic very interesting.

http://ucsmp.uchicago.edu/Transl.html

Beth said...

I'm okay with the divisibility rules for 2, 5 and 10 -- they're simple enough for a 5th grader to understand. The rule for 3, on the other hand, is really not clear, and if the 5th graders can't understand the proof then they're not ready for the concept, IMHO.

The problem is that kids get used to the idea that math is this impenetrable thicket of rules handed down from on high instead of the logical wonder that it is. They wind up turning off their own logical thinking process because they're afraid of making a mistake (definitely true for my daughter).

For myself, my memory is not very good, and I'm not really comfortable with a mathematical idea unless I can derive it, or at least make sense of it. I can't trust my memory to retrieve a rule I never understood in the first place.

When I was a kid, for years I remembered that there were two formulas for measuring circles, namely 2*pi*r and pi*r^2. I could never remember which formula went with which measurement. I'm embarrassed to tell you how old I was before I realized that pi*r^2 must be the measurement for area, since the radius is squared and area comes in square units.

Allison said...

--classifying prime numbers as "abundant, deficient, or precise" based on the number of their factors.

Prime numbers have two factors: 1, and themselves. e.g. 13 is prime. Its factors are 13 and 1. You can't classify then any other way.

I have a bachelor's in math and have never heard of "abundant, deficient, or precise" numbers.

I can't imagine that "deficient" has any meaning at all here--how can a number be deficient in its number of factors? All numbers have a prime factorization. Composite numbers have more than two numbers in the prime factorization. What could abundant or deficient mean?

Dee in MI said...

I remember learning divisibility tests in seventh grade and have remembered most of them since. I probably used them more when first learning algebra, before I just got a feel for multiples by doing a lot of math. I am, by coincidence, teaching them to my sixth grade son today. They are:

Mult. of 10, 5, and 2, you know

If the last two digits of a number are multiple of 4, then the number is a mulitple of 4 (as in 169,828)

If the last three digits of a number are a multiple of 8, then the number is a multiple of 8 (as in 1,379,136)

If the sum of the digits of a number is a multiple of 3, then the number is a multiple of 3.

If the sum of the digits of a number is a multiple of 9, then the number is a multiple of 9.

If the number is both even and a multiple of 3, then the number is a multiple of 6.

This can make simplifying fractions and factoring polynomials much easier down the road.

This may not connect for all kids, but I definately caught on at the time.

Michael Weiss said...

The rule for divisibility by 3 is not really hard to explain.

Consider a four digit number ABCD. This really means 1000A + 100B + 10C + D.

The key idea is that:
1000A = 999A + A; since 999A is divislble by 3, the remainder when 1000A is divided by 3 is the same as the remainder when A is divided by 3.

Likewise 100B = 99B + B, so the remainder when 100B is divided by 3 is the same as the remainder when B is divided by 3.

Similarly 10C = 9C + C, so (etc.)

Putting it all together you find that the remainder when the four-digit number ABCD is divided by 3 (that is, 1000A+100B+10C+D)is the same as the remainder when A+B+C+D is divided by 3. So to test if ABCD is a multiple of 3, just test the sum of its digits.

Everything I wrote above works if you replace "3" with "9".

Extra credit: Explain why, in Base 6, a number is divisible by 5 if and only if the sum of its digits is divisible by 5. Propose two other divisibility tests in other number bases.

Dee in MI said...

The 4 and and 8 tests aren't hard to understand either.

Recognize first that if you add two multiples of any number, then the sum will also be a multiple of that number.

Since 100 is a multiple of 4, any multiple of a 100 is also a multiple of 4. So the sum of a multiple of 100 and a multiple of 4 will be a multiple of 4.

Since 8 is a multiple of 1000, then the 8 test is similar.

Math man said...

wery good site...

www.matematiketkinliklerim.com

Beth said...

Thanks to everyone who wrote explanations of the divisibility tests! I will think about them and see about teaching them to dd.

To Allison -- I made a mistake. It wasn't just prime numbers that got classified as "abundant" or whatever depending on the number of factors, it was any kind of number, prime or composite. So it makes a little sense.

Anyway, the point I was really trying to make is that it is better to teach kids math in a way they can understand, instead of presenting it as a stream of arbitrary algorithms. I would rather see fewer topics, covered in depth, than more topics covered as rules to memorize.

CassyT said...

TerriW said...

If only they would play "Broken Calculator" all year...

Thank you for the best laugh of the day - Classic!

Anonymous said...

I Googled the abundant and deficient numbers (bonus: perfect and amicable numbers as well):